Heat and Thermodynamics

The diameter of molecule is, $d=5 \times 10^{-10} \mathrm{~m}$

Temperature, $\mathrm{T} 41^{\circ} \mathrm{C}=(41+273) \mathrm{K}=314 \mathrm{~K}$

Pressure, $\mathrm{P}=1.38 \times 10^5 \mathrm{~Pa}$

Boltzmann constant, $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$

The standard formula for the mean free path $\lambda$ is:

$ \lambda=\frac{\mathrm{k}_{\mathrm{B}} \mathrm{~T}}{\sqrt{2} \pi \mathrm{~d}^2 \mathrm{P}} $

Substituting the values,

$ \lambda=\frac{\left(1.38 \times 10^{-23}\right) \times 314}{\sqrt{2} \times \pi \times\left(5 \times 10^{-10}\right)^2 \times\left(1.38 \times 10^5\right)} $

$\Rightarrow $ $\lambda=\frac{10^{-23} \times 314}{\sqrt{2} \times \pi \times 25 \times 10^{-20} \times 10^5}$

$\Rightarrow $ $\lambda=\frac{314}{\sqrt{2} \times \pi \times 25 \times 10^{-8}}$

Using $\pi \approx 3.14$ :

$ \lambda=\frac{314}{\sqrt{2} \times 3.14 \times 25 \times 10^{-8}} $

$\Rightarrow $ $\lambda=\frac{4}{\sqrt{2}} \times 10^{-8}=2 \sqrt{2} \times 10^{-8} \mathrm{~m}$

Hence, the correct option is D.

To determine the correct graph, we must analyse the phase changes and temperature changes that occur when heat is supplied to water starting from $-20^{\circ} \mathrm{C}$.

Step 1 : Heating Ice ( $-20^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ )

Initially, the substance is ice. As heat is supplied, the temperature of the ice increases linearly from $-20^{\circ} \mathrm{C}$ to 0 ${ }^{\circ} \mathrm{C}$. This is represented by a sloping line on the graph.

$ \mathrm{Q}_1=\mathrm{m}_{\text {ice }} \mathrm{c}_{\text {ice }}(\mathrm{T}+20) $

Step 2 : Melting (Phase Change at $0^{\circ} \mathrm{C}$ )

At $0{ }^{\circ} \mathrm{C}$, the ice begins to melt into water. During a phase change, the temperature remains constant even as heat is added (latent heat of fusion). This is represented by a horizontal line at $0^{\circ} \mathrm{C}$.

$ Q_2=m_{i c e} L_f $

Step 3 : Heating Liquid Water ( $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ )

Once all the ice has melted, further heat increases the temperature of the liquid water linearly from $0^{\circ} \mathrm{C}$ up to its boiling point at $100^{\circ} \mathrm{C}$. This is another sloping line.

$ \mathrm{Q}_3=\mathrm{m}_{\text {water }} \mathrm{c}_{\text {water }} \mathrm{T} $

Step 4 : Boiling (Phase Change at $100^{\circ} \mathrm{C}$ )

At $100^{\circ} \mathrm{C}$, the water begins to turn into steam. The temperature stays constant during this process (latent heat of vaporization). This is represented by a second horizontal line at $100^{\circ} \mathrm{C}$.

$ \mathrm{Q}_4=\mathrm{m}_{\text {water }} \mathrm{L}_{\mathrm{v}} $

Step 5 : Heating Steam ( $100^{\circ} \mathrm{C}$ to $120^{\circ} \mathrm{C}$ )

After all the water has turned into steam, the temperature of the steam increases from $100^{\circ} \mathrm{C}$ toward $120^{\circ} \mathrm{C}$. This is represented by a final sloping line.

$ \mathrm{Q}_5=\mathrm{m}_{\text {vapour }} \mathrm{C}_{\text {vapour }}(\mathrm{T}-100) $

So, the correct graph must show :

1. 1. A start point below $0^{\circ} \mathrm{C}$ (at $-20^{\circ} \mathrm{C}$ ).

2. 2. A horizontal plateau at $0^{\circ} \mathrm{C}$ (melting).

3. 3. A horizontal plateau at $100^{\circ} \mathrm{C}$ (boiling).

Hence, the correct option is (C).

Mass of the ice is $\mathrm{m}_{\mathrm{i}}=10 \mathrm{~kg}$,

Initial temperature of ice is $\mathrm{T}_{\mathrm{i}}=-10^{\circ} \mathrm{C}$,

The specific heat of ice is $\mathrm{c}_{\mathrm{i}}=2100 \frac{\mathrm{~J}}{\mathrm{~kg} \cdot{ }^{\circ} \mathrm{C}}$,

The latent heat of fusion of ice is $\mathrm{L}_{\mathrm{f}}=3.36 \times 10^5 \mathrm{~J} / \mathrm{kg}$

The mass of water is $\mathrm{m}_{\mathrm{w}}=100 \mathrm{~kg}$,

The initial temperature of water is $\mathrm{T}_{\mathrm{w}}=25^{\circ} \mathrm{C}$,

The specific heat capacity of water is $\mathrm{c}_{\mathrm{w}}=4200 \frac{\mathrm{~J}}{\mathrm{~kg} \cdot{ }^{\circ} \mathrm{C}}$

Let the final equilibrium temperature is $\mathrm{T}_{\mathrm{f}}$.

The heat gained by ice is for three process,

$Q_{\text {gained }}=Q_1$ (Heating ice to $\left.0^{\circ} \mathrm{C}\right)+Q_2$ (Melting ice at $\left.0^{\circ} \mathrm{C}\right)+Q_3$ (Heating the melted ice (water) to $T_f$ )

$\Rightarrow $ $\mathrm{Q}_1=\mathrm{m}_{\mathrm{i}} \mathrm{c}_{\mathrm{i}} \Delta \mathrm{T}=10 \times 2100 \times(0-(-10))=210000 \mathrm{~J}$

$\Rightarrow $ $\mathrm{Q}_2=\mathrm{m}_{\mathrm{i}} \mathrm{L}_{\mathrm{f}}=10 \times 3.36 \times 10^5=3360000 \mathrm{~J}$

$\Rightarrow $ $\mathrm{Q}_3=\mathrm{m}_{\mathrm{i}} \mathrm{c}_{\mathrm{w}}\left(\mathrm{T}_{\mathrm{f}}-0\right)=10 \times 4200 \times \mathrm{T}_{\mathrm{f}}=42000 \mathrm{~T}_{\mathrm{f}}$

$ Q_{\text {gained }}=210000+3360000+42000 \mathrm{~T}_{\mathrm{f}}=3570000+42000 \mathrm{~T}_{\mathrm{f}} $

The water cools from $25^{\circ} \mathrm{C}$ to $\mathrm{T}_{\mathrm{f}}$. So the heat lost by water is,

$ \mathrm{Q}_{\text {loss }}=\mathrm{m}_{\mathrm{w}} \mathrm{c}_{\mathrm{w}}\left(25-\mathrm{T}_{\mathrm{f}}\right)=100 \times 4200 \times\left(25-\mathrm{T}_{\mathrm{f}}\right)=420000\left(25-\mathrm{T}_{\mathrm{f}}\right) $

$\Rightarrow $ $ \mathrm{Q}_{\text {loss }}=10500000-420000 \mathrm{~T}_{\mathrm{f}} $

Using conservation of energy, $\mathrm{Q}_{\text {gain }}=\mathrm{Q}_{\text {loss }}$

$ 3570000+42000 \mathrm{~T}_{\mathrm{f}}=10500000-420000 \mathrm{~T}_{\mathrm{f}} $

$\Rightarrow $ $462000 \mathrm{~T}_{\mathrm{f}}=6930000$

$\Rightarrow $ $ \mathrm{T}_{\mathrm{f}}=\frac{6930000}{462000}=15^{\circ} \mathrm{C} $

So the decrease in the temperature of the water:

$ \Delta \mathrm{T}_{\text {water }}=\mathrm{T}_{\text {initial }}-\mathrm{T}_{\mathrm{f}}=25^{\circ} \mathrm{C}-15^{\circ} \mathrm{C}=10^{\circ} \mathrm{C} $

Therefore, the decrement to the temperature of water is $10^{\circ} \mathrm{C}$.

Hence, the correct option is (B).

The work done in a cyclic process represented on a $\mathrm{p}-\mathrm{V}$ diagram is equal to the area enclosed by the closed loop.

The path of the cycle is $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{A}$ in an anti-clockwise direction.

In a $\mathrm{p}-\mathrm{V}$ diagram, a clockwise cycle represents positive work done by the system $(\mathrm{W}<0)$.

The equation of curve is $(V-2)^2=4$ ap $($ path $A \rightarrow B \rightarrow C)$.

• At $\mathrm{V}=2, \mathrm{p}=0$ (Point B is on the V -axis).

• At $\mathrm{V}=1$ and $\mathrm{V}=3, \mathrm{p}=\frac{(1)^2}{4 \mathrm{a}}=\frac{1}{4 \mathrm{a}}$. This is the pressure at points A and C .

The total work done W is the area under the straight line minus the area under the curve:

$ \mathrm{W}=\int_1^3 \mathrm{pdV}-\operatorname{Area}(\mathrm{CA}) $

Since the line CA is at $\mathrm{p}=\frac{1}{4 \mathrm{a}}$, its area is :

$ \operatorname{Area}(\mathrm{CA})=\mathrm{p} \times \Delta \mathrm{V}=\frac{1}{4} \mathrm{a} \times(3-1)=\frac{2}{4 \mathrm{a}}=\frac{1}{2 \mathrm{a}} $

Work done along the curve is,

$ \int\limits_1^3 \mathrm{pdV}=\int_1^3\left[\frac{(\mathrm{~V}-2)^2}{4 \mathrm{a}}\right] \mathrm{dV}=\frac{1}{4 \mathrm{a}}\left[\frac{(\mathrm{~V}-2)^3}{3}\right]_1^3 $

$\Rightarrow $ $ \int\limits_1^3 \mathrm{pdV}=\frac{1}{4 \mathrm{a}}\left[\frac{(3-2)^3}{3}-\frac{(1-2)^3}{3}\right]=\frac{1}{4 \mathrm{a}}\left[\frac{1}{3}-\left(-\frac{1}{3}\right)\right]=\frac{1}{4 \mathrm{a}} \times \frac{2}{3}=\frac{1}{6 \mathrm{a}} $

So, the net work done in the cycle is

$ W_{\text {net }}=\frac{1}{6 a}-\frac{1}{2 a}=\frac{1-3}{6 a}=-\frac{1}{3 a} $

Therefore, the total work done in closed path is $-\frac{1}{3 a}$. Hence, the correct option is (D).

The path (AB) from state $P_1$ to state $P_2$ is a vertical line which represents isochoric process where volume remains constant.

Initial Pressure: $\mathrm{P}_1=21.7 \mathrm{~Pa}$, Initial volume: $\mathrm{V}_1=1 \mathrm{~m}^3$.

Final Pressure: $P_2=30 \mathrm{~Pa}$, Final volume: $V_2=1 \mathrm{~m}^3$

For any isochoric process, since there is no change in volume ( $\mathrm{dV}=0$ ), the work done by the gas is zero :

$ \mathrm{W}_{\mathrm{AB}}=\int \mathrm{P} \mathrm{dV}=0 $

Applying the $1^{\text {st }}$ law of thermodynamics

The heat involved ( $\mathrm{Q}=\alpha$ ) in a constant volume process is equal to the change in internal energy ( $\Delta \mathrm{U}$ ) :

$ Q=\Delta U+W $

$\Rightarrow $ $ \mathrm{Q}=\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T}=\alpha \ldots \text { (i) } $

From the ideal gas law ( $\mathrm{PV}=\mathrm{nRT}$ ), we know that $\mathrm{nT}=\frac{\mathrm{PV}}{\mathrm{R}}$. Therefore, the change

$ \mathrm{n} \Delta \mathrm{~T}=\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{R}} \ldots (ii)$

From both the equations,

$ \alpha=\mathrm{C}_{\mathrm{v}}\left(\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{R}}\right) $

Substituting the given values :

$ V_1=V_2=1 \mathrm{~m}^3 ; \mathrm{P}_2-\mathrm{P}_1=30-21.7=8.3 \mathrm{~Pa} ; \mathrm{C}_{\mathrm{v}}=21 \mathrm{~J} / \mathrm{Kmol} ; \mathrm{R}=8.3 \mathrm{~J} / \mathrm{molK} $

$\alpha=21 \times\left(\frac{(30 \times 1)-(21.7 \times 1)}{8.3}\right)$

$\Rightarrow $ $\alpha=21 \times\left(\frac{8.3}{8.3}\right)$

Therefore, the heat involved in the process from $P_1$ to $P_2$ is 21 Joules.

Hence, the correct option is (A).

Heat supplied to water:

$Q=mc\Delta T$

Here, $m=4\,\text{kg}$, $c=4.2\times 10^3\,\text{J kg}^{-1}\text{K}^{-1}$, and $\Delta T = 20-4 = 16\,\text{K}$.

So,

$Q=4\times 4.2\times 10^3 \times 16 = 268800\,\text{J}$

Since heating is done at atmospheric pressure, water expands and does work:

$W = P\Delta V$

Initial volume:

$V_1=\frac{m}{\rho_1}=\frac{4}{1000}=0.004\,\text{m}^3$

Final volume:

$V_2=\frac{m}{\rho_2}=\frac{4}{998}=0.004008016\,\text{m}^3$

Change in volume:

$\Delta V=V_2-V_1=0.004008016-0.004=8.016\times 10^{-6}\,\text{m}^3$

Work done:

$W = 10^5 \times 8.016\times 10^{-6} = 0.8016\,\text{J}$

By first law of thermodynamics,

$\Delta U = Q - W = 268800 - 0.8016 = 268799.1984\,\text{J} \approx 268799.2\,\text{J}$

Correct option: A (268799.2 J)

Case -1 : Isothermal Expansion $(\Delta \mathrm{T}=0)$

For an ideal gas undergoing isothermal expansion from $V_1$ to $V_2$ :

$ \mathrm{W}_{\mathrm{iso}}=\mathrm{nRT} \ln \left(\frac{\mathrm{~V}_2}{\mathrm{~V}_1}\right) $

Given :

• $\mathrm{n}=1$ mole

• $\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$

• $\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{2 \mathrm{~V}}{\mathrm{~V}}=2$

• $\ln 2=0.693$

$ W=1 \times R \times 300 \times 0.693=207.9 R $

Case - 2 : Adiabatic Expansion ( $\Delta \mathrm{Q}$ )

For an adiabatic process, the work done is given by:

$ \mathrm{W}_{\mathrm{adia}}=\frac{\mathrm{nR}\left(\mathrm{~T}_1-\mathrm{T}_2\right)}{\gamma-1} $

The gas is diatomic, so the adiabatic constant $\gamma=\frac{7}{5}=1.4$.

$ \mathrm{T}_1=300 \mathrm{~K} $

Work done is the same magnitude ( $\mathrm{W}=207.9 \mathrm{R}$ ).

$ 207.9 \mathrm{R}=\frac{1 \times \mathrm{R}\left(300-\mathrm{T}_2\right)}{1.4-1} $

$\Rightarrow $ $207.9=\frac{300-\mathrm{T}_2}{0.4}$

$\Rightarrow $ $83.16=300-\mathrm{T}_2$

$\Rightarrow $ $\mathrm{T}_2=216.84 \mathrm{~K}$

$\Rightarrow $ $\mathrm{T}\left({ }^{\circ} \mathrm{C}\right)=216.84-273.15$

$ \mathrm{T}\left({ }^{\circ} \mathrm{C}\right) \approx-56.31^{\circ} \mathrm{C} $

The final temperature is approximately $-56^{\circ} \mathrm{C}$.

Hence, the correct option is (A).

According to the first law of thermodynamics,

$ \Delta Q=\Delta U+\Delta W $

Where,

The heat supplied to a system $=\Delta \mathrm{Q}=126 \mathrm{~J}$

The change in internal energy $=\Delta U=3 n R T$

The work done by the system $=\Delta \mathrm{W}$

Number of moles $n=1$ mole of Helium.

Change in temperature $=\Delta \mathrm{T}=4^{\circ} \mathrm{C}=4 \mathrm{~K}$

Since the piston is light, movable, and frictionless, the gas pressure remains equal to the atmospheric pressure $= 10^5 \mathrm{~Pa}$.

Area $=\mathrm{A}=17 \mathrm{~cm}^2=17 \times 10^{-4} \mathrm{~m}^2$

Using the formula provided :

$ \Delta \mathrm{U}=3 \cdot \mathrm{n} \cdot \mathrm{R} \cdot \Delta \mathrm{~T} $

$\Rightarrow $ $\Delta \mathrm{U}=3 \cdot 1 \cdot 8.314 \cdot 4 \mathrm{~J}$

$ \Delta \mathrm{U}=99.768 \mathrm{~J} $

From the first law :

$ \Delta \mathrm{W}=\Delta \mathrm{Q}-\Delta \mathrm{U} $

$\Rightarrow $ $\Delta \mathrm{W}=126-99.768=26.232 \mathrm{~J}$

Work done by a gas at constant pressure is :

$ \Delta \mathrm{W}=\mathrm{P} \Delta \mathrm{~V}=\mathrm{P} \times(\mathrm{Ax}) $

Where x is the distance the piston moves.

$ 26.232=10^5 \times\left(17 \times 10^{-4}\right) \times \mathrm{x} $

$\Rightarrow $ $26.232=170 \times x$

$\Rightarrow $ $x=\frac{26.232}{170} \approx 0.1543 \mathrm{~m}$

$\Rightarrow $ $\mathrm{x}=0.1543 \times 100 \mathrm{~cm} \approx 15.5 \mathrm{~cm}$

Initial state (at the bottom)

Volume $=\mathrm{V}_1=2.9 \mathrm{~cm}^3$

Temperature $=\mathrm{T}_1=17^{\circ} \mathrm{C}=17+273=290 \mathrm{~K}$

Pressure $=\mathrm{P}_1$

Total pressure at the bottom is atmospheric pressure $\left(\mathrm{P}_0\right)$ plus the hydrostatic pressure $(\mathrm{h} \rho \mathrm{g})$.

$ P_1=10^5+10^3 \times 10 \times 5=1.5 \times 10^5 \mathrm{~Pa} $

Final state (at the surface)

Temperature $=\mathrm{T}_2=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$

Pressure $=P_2=P_0$ (Only atmospheric pressure acts at the surface.)

$ \mathrm{P}_2=10^5 \mathrm{~Pa} $

According to the Ideal Gas Law :

$ \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2} $

$\Rightarrow $ $\frac{\left(1.5 \times 10^5\right) \times 2.9}{290}=\frac{\left(10^5\right) \times V_2}{300}$

$\Rightarrow $ $\frac{1.5 \times 2.9}{290}=\frac{V_2}{300}$

$\Rightarrow $ $V_2=\frac{1.5 \times 2.9 \times 300}{290}$

$\Rightarrow $ $ \mathrm{V}_2=4.5 \mathrm{~cm}^3 $

So, as the bubble rises, the external pressure decreases significantly while the temperature increases slightly. Both factors contribute to the expansion of the air bubble, resulting in a final volume of $4.5 \mathrm{~cm}^3$.

Therefore, the correct option is (D).

For an ideal gas, the collision frequency of a molecule is given by the kinetic theory of gases :

$ \mathrm{Z}=\sqrt{2} \pi \mathrm{~d}^2 \mathrm{v}_{\text {avg }} \mathrm{n} $

Where :

• d is the molecular diameter (or size).

• $\mathrm{V}_{\text {avg }}$ is the average speed of the gas molecules.

• n is the number density (number of molecules per unit volume).

The average speed $\left(\mathrm{v}_{\text {avg }}\right)$ of gas molecules depends on the absolute temperature ( T ) and the mass of a single molecule (m) according to the formula :

$ \mathrm{v}_{\mathrm{avg}}=\sqrt{\frac{8 \mathrm{kT}}{\pi \mathrm{~m}}} $

Where k is the Boltzmann constant.

Substituting this into the collision frequency formula gives :

$ \mathrm{Z}=\sqrt{2} \pi \mathrm{~d}^2\left(\sqrt{\frac{8 \mathrm{kT}}{\pi \mathrm{~m}}}\right) \mathrm{n} $

It is given that the temperatures (T), pressures, and number densities (n) are exactly the same for both gases A and B.

$ \mathrm{Z} \propto \frac{\mathrm{~d}^2}{\sqrt{\mathrm{~m}}} $

$ \Rightarrow \frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\left(\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{~d}_{\mathrm{B}}}\right)^2 \times \sqrt{\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{~m}_{\mathrm{A}}}} $

Molecular size of A is half that of $\mathrm{B} \Rightarrow \mathrm{d}_{\mathrm{A}}=\frac{1}{2} \mathrm{~d}_{\mathrm{B}} \Rightarrow \frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=\frac{1}{2}$

Mass of molecule A is four times that of $\mathrm{B} \Rightarrow \mathrm{m}_{\mathrm{A}}=4 \mathrm{~m}_{\mathrm{B}} \Rightarrow \frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}}=\frac{1}{4}$.

Collision frequency in gas B is $\mathrm{Z}_{\mathrm{B}}=32 \times 10^{18} / \mathrm{s}$.

$ \frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\left(\frac{1}{2}\right)^2 \times \sqrt{\frac{1}{4}} $

$\Rightarrow $ $\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{1}{4} \times \frac{1}{2}$

$\Rightarrow $ $\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{1}{8} \Rightarrow \mathrm{Z}_{\mathrm{A}}=\frac{\mathrm{Z}_{\mathrm{B}}}{8}=\frac{32 \times 10^{18}}{8}=4 \times 10^{18} / \mathrm{s}$

Using law of heat conduction the rate of heat flow $\left(H=\frac{d Q}{d t}\right)$ through a material is proportional to the crosssectional area (A) and the temperature difference ( $\Delta \mathrm{T}$ ), and inversely proportional to the length (L) :

$ H=\frac{k A \Delta T}{L} $

Where k is the thermal conductivity of the material.

We can rearrange this equation to look like Ohm's Law for electricity $\left(\mathrm{I}=\frac{\Delta \mathrm{V}}{\mathrm{R}}\right)$ :

$ H=\frac{\Delta T}{L / k A} $

Here, the term $L / k A$ acts as the thermal resistance $\left(R_{t h}\right)$ of the rod. Heat current $(H)$ is analogous to electrical current (I), and temperature difference ( $\Delta \mathrm{T}$ ) is analogous to potential difference ( $\Delta \mathrm{V}$ ).

$ \mathrm{R}_{\mathrm{th}}=\frac{\mathrm{L}}{\mathrm{kA}} $

It is given that rods $x$ and $y$ have equal dimensions, meaning they have the same length $(L)$ and same crosssectional area (A).

We are also given that the thermal conductivity of $\operatorname{rod} \mathrm{x}\left(\mathrm{k}_{\mathrm{x}}\right)$ is three times that of $\operatorname{rod} \mathrm{y}\left(\mathrm{k}_{\mathrm{y}}\right)$.

Let $\mathrm{k}_{\mathrm{y}}=\mathrm{k}$, then $\mathrm{k}_{\mathrm{x}}=3 \mathrm{k}$.

So, the thermal resistances of rod x and y will be,

$ \mathrm{R}_{\mathrm{y}}=\frac{\mathrm{L}}{\mathrm{k}_{\mathrm{y}} \mathrm{~A}}=\frac{\mathrm{L}}{\mathrm{kA}}=\mathrm{R} $

$\Rightarrow $ $ \mathrm{R}_{\mathrm{x}}=\frac{\mathrm{L}}{\mathrm{k}_{\mathrm{x}} \mathrm{~A}}=\frac{\mathrm{L}}{3 \mathrm{kA}}=\frac{1}{3}\left(\frac{\mathrm{~L}}{\mathrm{kA}}\right)=\frac{\mathrm{R}}{3} $

We can now treat the given structure as an electrical circuit with specific resistances.

The heat current splits into two parallel paths BC and BD at B .

$ \mathrm{R}_{\mathrm{BCE}}=\mathrm{R}_{\mathrm{BC}}+\mathrm{R}_{\mathrm{CE}}=\mathrm{R}+\mathrm{R}=2 \mathrm{R} $

$\Rightarrow $ $ \mathrm{R}_{\mathrm{BDE}}=\mathrm{R}_{\mathrm{BD}}+\mathrm{R}_{\mathrm{DE}}=\frac{\mathrm{R}}{3}+\frac{\mathrm{R}}{3}=\frac{2 \mathrm{R}}{3} $

For parallel resistors $R_1$ and $R_2$, the equivalent resistance is $\frac{R_1 R_2}{R_1+R_2}$ :

$ \mathrm{R}_{\mathrm{BE}}=\frac{(2 \mathrm{R}) \times\left(\frac{2 \mathrm{R}}{3}\right)}{(2 \mathrm{R})+\left(\frac{2 \mathrm{R}}{3}\right)} $

$\Rightarrow $ $ R_{B E}=\frac{4 R^2 / 3}{\frac{6 R+2 R}{3}}=\frac{4 R^2}{8 R}=\frac{R}{2} $

From $A$ to $F$ with three equivalent resistors: $R_{A B}, R_{B E}$, and $R_{E F}$.

The total equivalent resistance ( $R_{\text {total }}$ ) is :

$ \mathrm{R}_{\text {total }}=\mathrm{R}_{\mathrm{AB}}+\mathrm{R}_{\mathrm{BE}}+\mathrm{R}_{\mathrm{EF}} $

$\Rightarrow $ $\mathrm{R}_{\text {total }}=\frac{\mathrm{R}}{3}+\frac{\mathrm{R}}{2}+\mathrm{R}$

$\Rightarrow $ $ \mathrm{R}_{\text {total }}=\frac{11 \mathrm{R}}{6} $

We know the overall temperature difference between A and F :

$ \Delta \mathrm{T}_{\text {total }}=\mathrm{T}_{\mathrm{A}}-\mathrm{T}_{\mathrm{F}}=100-40=60^{\circ} \mathrm{C} $

The total steady-state heat current (H) flowing through the entire system is :

$ H=\frac{\Delta T_{\text {total }}}{R_{\text {total }}}=\frac{60}{11 R / 6}=\frac{360}{11 R} $

Consider the segment from A to B :

$ H=\frac{T_A-T_B}{R_{A B}} $

$\begin{aligned} \Rightarrow \frac{360}{11 \mathrm{R}} & =\frac{100-\mathrm{T}_{\mathrm{B}}}{\mathrm{R} / 3} \\ \Rightarrow \frac{360}{11 \mathrm{R}} & =\frac{3\left(100-\mathrm{T}_{\mathrm{B}}\right)}{\mathrm{R}} \\ \Rightarrow \frac{120}{11} & =100-\mathrm{T}_{\mathrm{B}}\end{aligned}$

$\Rightarrow $ $\mathrm{T}_{\mathrm{B}}=100-\frac{120}{11}=\frac{1100-120}{11}=\frac{980}{11}$

$\Rightarrow $ $ \mathrm{T}_{\mathrm{B}} \approx 89.09^{\circ} \mathrm{C} $

Consider the segment from E to F :

$ \mathrm{H}=\frac{\mathrm{T}_{\mathrm{E}}-\mathrm{T}_{\mathrm{F}}}{\mathrm{R}_{\mathrm{EF}}} $

$\begin{aligned} & \Rightarrow \frac{360}{11 R}=\frac{T_E-40}{R} \\ & \Rightarrow \frac{360}{11}=T_E-40\end{aligned}$

$\Rightarrow $ $\mathrm{T}_{\mathrm{E}}=40+\frac{360}{11}=\frac{440+360}{11}=\frac{800}{11}$

$\Rightarrow $ $ \mathrm{T}_{\mathrm{E}} \approx 72.72^{\circ} \mathrm{C} $

Therefore, the temperatures at junction points B and E are close to $89^{\circ} \mathrm{C}$ and $73^{\circ} \mathrm{C}$ respectively.

Hence, the correct option is (C).

It is given that there is no exchange of heat in this process, which means $\Delta \mathrm{Q}=0$, i.e., this is an adiabatic process.

For an ideal gas undergoing a reversible adiabatic process, the relationship between Temperature (T) and Volume (V) is given by the equation:

$ \mathrm{T} \cdot \mathrm{~V}^{\gamma-1}=\text { constant } $

Where $\gamma$ is the adiabatic index or the ratio of specific heats $\left(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\right)$. This $\gamma$ value is different depending atomicity of the gases (monoatomic, diatomic, polyatomic).

$ \mathrm{T}_1 \cdot \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \cdot \mathrm{~V}_2^{\gamma-1} $

Final volume $\left(V_2\right)$ is 8 times the initial volume $\left(V_1\right) \Rightarrow V_2=8 V_1$

Final temperature $\left(T_2\right)$ is $\left(\frac{1}{4}\right)^{\text {th }}$ the initial temperature $\left(T_1\right) \Rightarrow T_2=\frac{T_1}{4}$

$ \mathrm{T}_1 \cdot \mathrm{~V}_1^{\gamma-1}=\left(\frac{\mathrm{T}_1}{4}\right) \cdot\left(8 \mathrm{~V}_1\right)^{\gamma-1} $

$\Rightarrow $ $\mathrm{V}_1^{\gamma-1}=\frac{1}{4} \cdot 8^{\gamma-1} \cdot \mathrm{~V}_1^{\gamma-1}$

$\Rightarrow 1=\frac{1}{4} \cdot 8^{\gamma-1} $

$\Rightarrow 4=8^{\gamma-1} $

$\Rightarrow 2^2=2^{3(\gamma-1)} $

$\Rightarrow 2=3(\gamma-1) $ 2=3 \gamma-3 $

$\Rightarrow 5=3 \gamma $

$\Rightarrow \gamma=\frac{5}{3}$

The value of $\gamma$ is directly related to the degrees of freedom (f) of the gas molecules by the formula :

$ \gamma=1+\frac{2}{f} $

$\Rightarrow \frac{5}{3}=1+\frac{2}{f}$

$\Rightarrow \frac{5}{3}-1=\frac{2}{f}$

$\Rightarrow \frac{2}{3}=\frac{2}{f}$

$\Rightarrow f=3$

A gas with 3 degrees of freedom (all translational, no rotational or vibrational modes activated) is a monoatomic gas.

$\mathrm{NH}_3$ (Ammonia) is polyatomic molecule (non-linear).

$\mathrm{O}_2$ (Oxygen) is a diatomic molecule.

$\mathrm{CO}_2$ (Carbon Dioxide) is polyatomic molecule (linear).

He (Helium) is a noble gas, exists as single atoms. It is a monoatomic gas. Therefore, the given gas is a monoatomic gas (He).

Hence, the correct option is (D).

According to the Kinetic Theory of Gases, the r.m.s speed of an ideal gas is given by :

$ \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} $

Where T is the absolute temperature and M is the molecular mass (in kg ).

It is given that the r.m.s. speed of oxygen $\left(\mathrm{O}_2\right)$ at $\mathrm{T}_{\mathrm{O}_2}$ is equal to the r.m.s. speed of hydrogen $\left(\mathrm{H}_2\right)$ at $\mathrm{T}_{\mathrm{H}_2}$.

$ \mathrm{V}_{\mathrm{rms}\left(\mathrm{O}_2\right)}=\mathrm{V}_{\mathrm{rms}\left(\mathrm{H}_2\right)} $

$\Rightarrow $ $\sqrt{\frac{3 \mathrm{RT}_{\mathrm{O}_2}}{\mathrm{M}_{\mathrm{O}_2}}}=\sqrt{\frac{3 \mathrm{RT}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{H}_2}}}$

$\Rightarrow $ $ \frac{\mathrm{T}_{\mathrm{O}_2}}{\mathrm{M}_{\mathrm{O}_2}}=\frac{\mathrm{T}_{\mathrm{H}_2}}{\mathrm{M}_{\mathrm{H}_2}} $

Temperature of oxygen is $\mathrm{T}_{\mathrm{O}_2}=47+273=320 \mathrm{~K}$.

Molar masses are $\mathrm{M}_{\mathrm{O}_2}=32 \times 10^{-3} \mathrm{~kg}$ and $\mathrm{M}_{\mathrm{H}_2}=2 \times 10^{-3} \mathrm{~kg}$.

$ \frac{320}{32}=\frac{\mathrm{T}_{\mathrm{H}_2}}{2} $

$\Rightarrow $ $10=\frac{\mathrm{T}_{\mathrm{H}_2}}{2} \Rightarrow \mathrm{~T}_{\mathrm{H}_2}=20 \mathrm{~K}$

$\Rightarrow $ $\mathrm{T}_{\mathrm{H}_2}=20-273$

$\Rightarrow $ $ \mathrm{T}_{\mathrm{H}_2}=-253^{\circ} \mathrm{C} $

Therefore, the rms speed of oxygen molecules at $47^{\circ} \mathrm{C}$ is equal to that of the hydrogen molecules kept at $-253^{\circ} \mathrm{C}$.

Hence, the correct option is (B).

The heat energy ( Q ) required to raise the temperature of a mass ( m ) is given by $\mathrm{Q}=\mathrm{mc} \Delta \mathrm{T}$

Where c is the specific heat capacity and $\Delta \mathrm{T}$ is the change in temperature.

Temperature change is $\Delta \mathrm{T}=87^{\circ} \mathrm{C}-27^{\circ} \mathrm{C}=60^{\circ} \mathrm{C}$

Specific heat is $\mathrm{c}=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$.

The heat required per unit time is $\mathrm{P}_{\mathrm{req}}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}} \cdot \mathrm{c} \cdot \Delta \mathrm{T}$

Water flows at 5.0 litres $/ \min$.

Since the density of water is $1 \mathrm{~kg} / \mathrm{L}$, the mass flow rate is $5.0=\frac{5.0}{60} \mathrm{~kg} / \mathrm{sec}$

$ P_{\mathrm{req}}=\left(\frac{5.0}{60}\right) \times 4200 \times 60 \frac{\mathrm{~J}}{\mathrm{~s}}=21000 \mathrm{~J} / \mathrm{s} $

Let the rate of consumption of gas be $\mathrm{R} \mathrm{g} / \mathrm{s}$.

The heat produced per second is,

$ \mathrm{P}_{\text {prod }}=\mathrm{R} \times \mathrm{H} $

Given heat of combustion of gas $\mathrm{H}=5.0 \times 10^4 \mathrm{~J} / \mathrm{g}$

$ \mathrm{P}_{\text {prod }}=\mathrm{R} \times 50000 \mathrm{~J} / \mathrm{s} $

Assuming 100% efficiency (because no heat loss is given), $\mathrm{P}_{\text {prod }}=\mathrm{P}_{\text {req }}$

$ R \times 50000=21000 $

$\Rightarrow \mathrm{R}=\frac{21000}{50000}=0.42 \mathrm{~g} / \mathrm{s}$

Therefore, the gas must be consumed at a rate of $0.42 \mathrm{~g} / \mathrm{s}$ to maintain the temperature rise.

Hence, the correct option is (D).

To find the total heat supplied ( Q ), we use the $1^{\text {st }}$ law of thermodynamics :

$ Q=\Delta U+W $

Where, $\Delta \mathrm{U}$ is the change in internal energy and W is the work done by the gas.

The internal energy of an ideal gas depends on its degrees of freedom (f). For a diatomic gas, there are 3 translational and 2 rotational degrees of freedom ( $f=5$ ). However, it is given that the gas has rotational modes only, which means we only consider $f=2$.

The formula for internal energy change is :

$ \Delta U=\frac{n \mathrm{fR} \Delta T}{2} $

The amount of gas is 1 mole, $\mathrm{n}=1$ mole, the degree of freedom is $\mathrm{f}=2$, the universal gas constant is $\mathrm{R}= \frac{8.3 \mathrm{~J}}{\mathrm{~mol} \cdot \mathrm{~K}}$ and the change in temperature of gas is $\Delta \mathrm{T}=1.2^{\circ} \mathrm{C}=1.2 \mathrm{~K}$

$ \Delta \mathrm{U}=\frac{1 \times 2 \times 8.3 \times 1.2}{2} $

$\Rightarrow $ $ \Delta U=8.3 \times 1.2=9.96 \mathrm{~J} $

Work done by the gas in a piston system is given by:

$ \mathrm{W}=\mathrm{P} \Delta \mathrm{~V} $

Since the mass of the piston is neglected and it moves slowly, the internal pressure P equals the atmospheric pressure ( $\mathrm{P}_{\mathrm{atm}}$ ).

The change in volume $\Delta \mathrm{V}$ is $\Delta \mathrm{V}=\mathrm{A} \times \Delta \mathrm{x}$, where the cross-sectional area is A and displacement of the piston is $\Delta \mathrm{x}$.

The atmospheric pressure is $\mathrm{P}=100 \mathrm{kPa}=10^5 \mathrm{~Pa}$

The area of cross-section is $\mathrm{A}=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2$

The displacement of piston $\mathrm{d}=25 \mathrm{~mm}=25 \times 10^{-3} \mathrm{~m}$

$ \Delta V=A \times \Delta x=\left(4 \times 10^{-4}\right) \times\left(25 \times 10^{-3}\right)=100 \times 10^{-7}=10^{-5} \mathrm{~m}^3 $

So, the work done by the gas is;

$ \mathrm{W}=10^5 \times 10^{-5}=1 \mathrm{~J} $

Using $1^{\text {st }}$ law of thermodynamics,

$ \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} $

$\Rightarrow $ $ Q=9.96 \mathrm{~J}+1 \mathrm{~J}=10.96 \mathrm{~J} $

Therefore, the amount of heat supplied by the gas is 10.96 J.

In an adiabatic process, no heat is exchanged with the surroundings ( $\mathrm{Q}=0$ ).

For an ideal gas undergoing an adiabatic change, the relationship between pressure and volume is :

$ \mathrm{PV}^\gamma=\text { constant } $

Where $\gamma$ is the ratio of specific heats $\left(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\right)$.

For a monoatomic gas, $\gamma=53$.

The initial pressure of gas is $\mathrm{P}_1=\mathrm{P}$.

The initial volume of gas is $\mathrm{V}_1=\mathrm{V}$

The final volume of gas is $V_2=27 \mathrm{~V}$

Using the adiabatic relation $\mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma$ :

$ \mathrm{P} \cdot \mathrm{~V}^{5 / 3}=\mathrm{P}_{\text {final }} \cdot(27 \mathrm{~V})^{5 / 3} $

$\Rightarrow $ $\mathrm{P}_{\text {final }}=\mathrm{P}\left(\frac{\mathrm{V}}{27 \mathrm{~V}}\right)^{5 / 3}=\mathrm{P}\left(\frac{1}{27}\right)^{5 / 3}$

$ \mathrm{P}_{\text {final }}=\mathrm{P}\left(\frac{1}{3^3}\right)^{5 / 3}=\mathrm{P}\left(\frac{1}{3}\right)^5=\frac{\mathrm{P}}{243} $

From the $1^{\text {st }}$ law of thermodynamics $(Q=\Delta U+W)$, since $Q=0$, we have:

$ \Delta U=-W $

The work done ( W ) in an adiabatic process is:

$ \mathrm{W}=\frac{\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2}{\gamma-1} $

Therefore:

$ \Delta U=\frac{P_2 V_2-P_1 V_1}{\gamma-1} $

Substitute the given values, $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{PV}, \mathrm{P}_2 \mathrm{~V}_2=\left(\frac{\mathrm{P}}{243}\right) \times(27 \mathrm{~V})=\frac{\mathrm{PV}}{9}$

$\Delta U=\frac{\frac{P V}{9}-P V}{\frac{5}{3}-1}$

$\Rightarrow $ $\Delta U=\frac{-\frac{8}{9} P V}{2 / 3}$

$\Rightarrow $ $\Delta \mathrm{U}=-\frac{8}{9} \mathrm{PV} \times \frac{3}{2}$

$\Rightarrow $ $ \Delta \mathrm{U}=-\frac{4}{3} \mathrm{PV} $

Therefore, the change in internal energy is $-\frac{4}{3} \mathrm{PV}$.

Hence, the correct option is (C).

In left compartment heat is added. The gas expands, increasing its pressure and volume.

In right compartment heat is not added because the wall is adiabatic, but the piston moves toward it because the pressure on the left increases. Since the walls and the piston are adiabatic, the gas in the right compartment undergoes an adiabatic compression.

The piston stops moving when the pressure in both compartments is equal. Therefore, the final pressure in the right compartment $\left(\mathrm{P}_{\mathrm{R}}\right)$ will also be $\frac{27 \mathrm{P}}{8}$.

For an adiabatic process, the relationship between pressure $(\mathrm{P})$ and volume $(\mathrm{V})$ is:

$ \mathrm{P} \mathrm{~V}^\gamma=\text { constant } $

Where $\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=1.5$.

$ P_1 V_1^\gamma=P_2 V_2^\gamma $

$\Rightarrow $ $ \frac{V_2}{V_1}=\left(\frac{P_1}{P_2}\right)^{1 / \gamma} $

The initial pressure in right compartment is $\mathrm{P}_1=\mathrm{P}$ and its initial volume is $\mathrm{V}_1=9$ litres

The final pressure in right compartment is $\mathrm{P}_2=\frac{27 \mathrm{P}}{8}$

Substituting these values:

$ \frac{V_{\text {final }}}{9}=\left(\frac{P}{27 P / 8}\right)^{1 / 1.5} $

$\Rightarrow $ $\frac{\mathrm{V}_{\text {final }}}{9}=\left(\frac{8}{27}\right)^{2 / 3}$

$\Rightarrow $ $\frac{V_{\text {final }}}{9}=\frac{4}{9}$

$\Rightarrow $ $ V_{\text {final }}=4 \text { litres } $

Therefore, the final volume in the right-hand side compartment is 4 litres.

Hence, the correct option is (B).

The internal energy $(U)$ of $n$ moles of an ideal gas at temperature $T$ is given by $U=n C_v T$, where $C_v$ is the molar specific heat at constant volume.

For the mixture, total initial internal energy $\mathrm{U}_{\text {initial }}=\mathrm{U}_1+\mathrm{U}_2$

Total final internal energy $\mathrm{U}_{\text {final }}=\left(\mathrm{n}_1+\mathrm{n}_2\right) \mathrm{C}_{\mathrm{v}, \text { mix }} \mathrm{T}_{\text {mix }}$

Since both gases are monoatomic, their molar specific heats are the same,

$ C_{v 1}=C_{v 2}=C_{v, \operatorname{mix}}=\frac{3}{2} R $

Therefore, using conservation of energy (total initial internal energy is equal to final internal energy).

$ \mathrm{n}_1 \mathrm{C}_{\mathrm{v}} \mathrm{~T}_1+\mathrm{n}_2 \mathrm{C}_{\mathrm{v}} \mathrm{~T}_2=\left(\mathrm{n}_1+\mathrm{n}_2\right) \mathrm{C}_{\mathrm{v}} \mathrm{~T}_{\operatorname{mix}} $

$\Rightarrow $ $ \mathrm{T}_{\mathrm{mix}}=\frac{\mathrm{n}_1 \mathrm{~T}_1+\mathrm{n}_2 \mathrm{~T}_2}{\left(\mathrm{n}_1+\mathrm{n}_2\right)} $

Moles of first gas is $\mathrm{n}_1=2$

Temperature of first gas is $\mathrm{T}_1=\mathrm{T}$

Moles of second gas $\mathrm{n}_2=6$

Temperature of second gas $\mathrm{T}_2=2 \mathrm{~T}$

On substituting the values into expression for the temperature of mixture,

$ \mathrm{T}_{\mathrm{mix}}=\frac{(2 \times \mathrm{T})+(6 \times 2 \mathrm{~T})}{2+6} $

$\Rightarrow $ $\mathrm{T}_{\operatorname{mix}}=\frac{2 \mathrm{~T}+12 \mathrm{~T}}{8}$

$\Rightarrow $ $\mathrm{T}_{\mathrm{mix}}=\frac{14 \mathrm{~T}}{8}$

$\Rightarrow $ $ \mathrm{T}_{\mathrm{mix}}=\frac{7}{4} \mathrm{~T} $

Therefore, the correct option is (D).

Using Ideal Gas Law:

PV = nRT

Where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is absolute temperature. Since the system is closed, the total number of moles ( $\mathrm{n}_{\text {total }}$ ) remains constant before and after heating.

Initial State:

Both vessels (each of volume V ) are at $\mathrm{P}_0=90 \mathrm{kPa}$ and $\mathrm{T}_0=400 \mathrm{~K}$.

$ \mathrm{n}_{\text {total }}=\frac{\mathrm{P}_0 \mathrm{~V}}{\mathrm{RT}_0}+\frac{\mathrm{P}_0 \mathrm{~V}}{\mathrm{RT}_0}=\frac{2 \mathrm{P}_0 \mathrm{~V}}{\mathrm{RT}_0} \ldots (i)$

Final State: Pressure becomes $\mathrm{P}_{\mathrm{f}}$ in both vessels. Vessel 1 stays at $\mathrm{T}_1=400 \mathrm{~K}$, and vessel 2 is at $\mathrm{T}_2=500 \mathrm{~K}$.

$ \mathrm{n}_{\text {total }}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}}{\mathrm{RT}_1}+\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}}{\mathrm{RT}_2}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}+\frac{1}{\mathrm{~T}_2}\right) \ldots (ii)$

From both the equations,

$ \frac{2 \mathrm{P}_0 \mathrm{~V}}{\mathrm{RT}_0}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}+\frac{1}{\mathrm{~T}_2}\right) \Rightarrow \frac{2 \mathrm{P}_0}{\mathrm{~T}_0}=\mathrm{P}_{\mathrm{f}}\left(\frac{1}{\mathrm{~T}_1}+\frac{1}{\mathrm{~T}_2}\right) $

Substituting the values ( $\mathrm{P}_0=90, \mathrm{~T}_0=400, \mathrm{~T}_1=400, \mathrm{~T}_2=500$ ):

$ 2 \times \frac{90}{400}=\mathrm{P}_{\mathrm{f}}\left(\frac{1}{400}+\frac{1}{500}\right) $

$\Rightarrow $ $\frac{180}{400}=\mathrm{P}_{\mathrm{f}}\left(\frac{5+4}{2000}\right)$

$\Rightarrow $ $0.45=\mathrm{P}_{\mathrm{f}}\left(\frac{9}{2000}\right)$

$\Rightarrow $ $ P_f=\frac{0.45 \times 2000}{9}=\frac{900}{9}=100 \mathrm{kPa} $

Therefore, the final pressure of the vessel is 100 kPa . Thus, the correct option is (A).

For an ideal gas, we use the equation

$ PV = nRT $

Given in the question:

$ PT^3 = \text{constant} $

Let

$ PT^3 = k \Rightarrow P = \frac{k}{T^3} $

Now substitute this value of $P$ in the ideal gas equation:

$ \frac{k}{T^3} \cdot V = nRT $

So,

$ V = \frac{nRT^4}{k} $

Hence,

$ V \propto T^4 $

Now coefficient of volume expansion is defined as

$ \beta = \frac{1}{V}\frac{dV}{dT} $

Since $V \propto T^4$, let

$ V = cT^4 $

Then,

$ \frac{dV}{dT} = 4cT^3 $

Therefore,

$ \beta = \frac{1}{cT^4}\cdot 4cT^3 = \frac{4}{T} $

So the correct answer is

$ \boxed{\frac{4}{T}} $

Option C.

For an ideal gas, the change in internal energy is

$ \Delta U = n C_v (T_f - T_i) $

Now, we use the relation

$ \gamma=\frac{C_p}{C_v} $

and also

$ C_p-C_v=R $

From these two relations,

$ \gamma C_v - C_v = R $

$ C_v(\gamma-1)=R $

$ C_v=\frac{R}{\gamma-1} $

Substituting in the expression of $\Delta U$,

$ \Delta U=n\left(\frac{R}{\gamma-1}\right)(T_f-T_i) $

So Statement I is true.

Now check Statement II.

For an ideal gas with $f$ degrees of freedom,

$ C_v=\frac{f}{2}R $

and

$ C_p=C_v+R=\frac{f}{2}R+R=\left(\frac{f}{2}+1\right)R $

Hence,

$ \gamma=\frac{C_p}{C_v} =\frac{\left(\frac{f}{2}+1\right)R}{\frac{f}{2}R} =1+\frac{2}{f} $

So Statement II is also true.

But Statement II is not the direct explanation of Statement I.

Statement I follows mainly from:

$ \Delta U=nC_v\Delta T \quad \text{and} \quad C_p-C_v=R $

The relation

$ \gamma=1+\frac{2}{f} $

is true, but it is not required to explain Statement I.

Therefore, the correct option is:

$ \boxed{\text{Option B}} $

Let us check each statement one by one.

Statement A

Zeroth law of thermodynamics gives concept of temperature

This is correct.

The zeroth law says that if two systems are separately in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This gives the basis for the concept of temperature.

So, A is true.

Statement B

First law of thermodynamics gives concept of internal energy

This is also correct.

The first law is

$ \Delta Q = \Delta U + \Delta W $

or equivalently,

$ \Delta U = \Delta Q - \Delta W $

This law introduces the thermodynamic state function internal energy.

So, B is true.

Statement C

In isothermal expansion of ideal gas, $\Delta Q \neq \Delta W$

This is incorrect.

For an ideal gas in an isothermal process, temperature remains constant, so change in internal energy is zero:

$ \Delta U = 0 $

From the first law,

$ \Delta Q = \Delta U + \Delta W $

Hence,

$ \Delta Q = \Delta W $

So the statement says $\Delta Q \neq \Delta W$, which is wrong.

Thus, C is false.

Statement D

Product of intensive and extensive variables is extensive

This is correct.

Recall:

• Intensive variable: does not depend on amount of substance, like pressure, temperature.

• Extensive variable: depends on amount of substance, like volume, mass, internal energy.

If an intensive quantity is multiplied by an extensive quantity, the result is generally extensive.

Example:

$ P \times V $

Here, pressure $P$ is intensive and volume $V$ is extensive. Their product behaves as an extensive quantity.

So, D is true.

Statement E

The ratio of any extensive variable to mass will be an extensive variable

This is incorrect.

Mass is itself an extensive variable.

So,

$ \frac{\text{extensive variable}}{\text{mass}} $

gives a specific property, which is an intensive variable.

Example:

$ \text{specific volume} = \frac{V}{m} $

This is an intensive property.

So, E is false.

Final conclusion

True statements are:

$ A,\ B,\ D $

Therefore, the correct option is:

$ \boxed{\text{Option C: A, B and D Only}} $

For an ideal gas, the average translational kinetic energy of one molecule is

$ \overline{K} = \frac{3}{2}kT $

where $k$ is Boltzmann constant and $T$ is absolute temperature.

So, if the average kinetic energy of $\mathrm{H}_2$ and $\mathrm{O}_2$ molecules is the same, then from

$ \frac{3}{2}kT_{\mathrm{H}_2} = \frac{3}{2}kT_{\mathrm{O}_2} $

we get

$ T_{\mathrm{H}_2} = T_{\mathrm{O}_2} $

Hence, Assertion A is true.

Now, the r.m.s. speed is given by

$ v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}} $

This depends on the mass $m$ of one molecule or molar mass $M$.

Since $\mathrm{H}_2$ and $\mathrm{O}_2$ have different molar masses, at the same temperature their r.m.s. speeds will not be the same.

In fact,

$ v_{\mathrm{rms}} \propto \frac{1}{\sqrt{M}} $

So lighter $\mathrm{H}_2$ molecules move faster than heavier $\mathrm{O}_2$ molecules at the same temperature.

Therefore, Reason R is false.

So, the correct option is:

$ \boxed{\text{Option C: A is true but R is false}} $

For linear expansion, if the temperature changes from $T_a$ to $T_b$, then the increase in length is

$ \Delta L = L_a \alpha (T_b-T_a) $

where $L_a$ is the length at temperature $T_a$.

Now let the length at temperature $T_1$ be $L_1$.

Then for the rise from $T_1$ to $T_2$,

$ \Delta L_1 = L_1 \alpha (T_2-T_1) $

Given

$ T_2-T_1=\Delta T $

so,

$ \Delta L_1 = L_1 \alpha \Delta T $

Thus the length at temperature $T_2$ becomes

$ L_2 = L_1 + \Delta L_1 = L_1(1+\alpha \Delta T) $

Now for the rise from $T_2$ to $T_3$, the increase in length is

$ \Delta L_2 = L_2 \alpha (T_3-T_2) $

We are given

$ T_3+T_1=2T_2 $

So,

$ T_3-T_2 = T_2-T_1 = \Delta T $

Hence,

$ \Delta L_2 = L_2 \alpha \Delta T $

Substitute $L_2 = L_1(1+\alpha \Delta T)$:

$ \Delta L_2 = L_1(1+\alpha \Delta T)\alpha \Delta T $

But

$ L_1 \alpha \Delta T = \Delta L_1 $

Therefore,

$ \Delta L_2 = \Delta L_1(1+\alpha \Delta T) $

So the correct answer is

$ \boxed{\Delta L_1[1+\alpha \Delta T]} $

Hence, Option D is correct.

For the given process,

$ P = P_0\left(1+\left(\frac{V_0}{V}\right)^2\right)^{-1} = \frac{P_0}{1+\left(\frac{V_0}{V}\right)^2} $

Both samples contain $2$ moles of ideal gas, so at any stage,

$ PV = nRT = 2RT $

Hence,

$ T = \frac{PV}{2R} $

We are told that the two samples attain the same pressure during the process.

So first let us find the volume of each sample when the pressure is equal.

Since pressure depends only on $V$ through the given relation,

$ P = \frac{P_0}{1+\left(\frac{V_0}{V}\right)^2} $

for a given pressure, the corresponding volume is uniquely fixed. Therefore, if both samples attain the same pressure, their final volumes must be the same.

Let this common final volume be $V$.

Then their temperatures at that pressure will be

$ T = \frac{PV}{2R} $

So if both have same $P$ and same $V$, their temperatures would also be same.

But the question is asking for $T_B - T_A$, which indicates we should compare the temperatures of the two samples at their given initial volumes under the same process relation.

So let us calculate temperature for each sample using the given $P$-$V$ relation.

Using

$ T = \frac{PV}{2R} $

and

$ P = \frac{P_0}{1+\left(\frac{V_0}{V}\right)^2} $

we get

$ T = \frac{P_0 V}{2R\left(1+\left(\frac{V_0}{V}\right)^2\right)} $

Now for sample $A$:

Initial volume,

$ V_A = V_0 $

So pressure is

$ P_A = \frac{P_0}{1+\left(\frac{V_0}{V_0}\right)^2} = \frac{P_0}{1+1} = \frac{P_0}{2} $

Thus,

$ T_A = \frac{P_A V_A}{2R} = \frac{\left(\frac{P_0}{2}\right)V_0}{2R} = \frac{P_0V_0}{4R} $

Now for sample $B$:

Initial volume,

$ V_B = 3V_0 $

So pressure is

$ P_B = \frac{P_0}{1+\left(\frac{V_0}{3V_0}\right)^2} = \frac{P_0}{1+\frac{1}{9}} = \frac{P_0}{\frac{10}{9}} = \frac{9P_0}{10} $

Thus,

$ T_B = \frac{P_B V_B}{2R} = \frac{\left(\frac{9P_0}{10}\right)(3V_0)}{2R} = \frac{27P_0V_0}{20R} $

Now,

$ T_B - T_A = \frac{27P_0V_0}{20R} - \frac{P_0V_0}{4R} $

Taking LCM:

$ T_B - T_A = \frac{27P_0V_0 - 5P_0V_0}{20R} = \frac{22P_0V_0}{20R} = \frac{11P_0V_0}{10R} $

Therefore,

$ \boxed{T_B - T_A = \frac{11P_0V_0}{10R}} $

So the correct option is:

$ \boxed{\text{Option B}} $

To find the ratio of final pressure ($ P_f $) to initial pressure ($ P_i $), we use the adiabatic condition:

$ P_i V_i^\gamma = P_f V_f^\gamma $

We aim to express the pressure ratio:

$ \frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^\gamma $

Given that the final volume $ V_f $ is $ \frac{1}{8} $ of the initial volume $ V_i $, this simplifies to:

$ \left(\frac{V_i}{V_f}\right) = 8 $

Thus, the pressure ratio becomes:

$ \frac{P_f}{P_i} = 8^\frac{5}{3} $

Calculating $ 8^\frac{5}{3} $ gives:

$ \frac{P_f}{P_i} = 32 $

To find the rise in temperature of the water as it falls from a height, we start by equating the potential energy loss to the heat gained, assuming no heat is lost from the water in the pool.

The potential energy lost by the water can be expressed as:

$ mgh $

where:

$ m $ is the mass of water,

$ g $ is the acceleration due to gravity (10 m/s²),

$ h $ is the height (200 m).

The heat gained by the water when temperature changes is given by:

$ ms \Delta T $

where:

$ s $ is the specific heat capacity of water (4200 J/(kg K)),

$ \Delta T $ is the change in temperature.

Setting the potential energy loss equal to the heat gained:

$ mgh = ms \Delta T $

We can simplify this equation by canceling out the mass $ m $:

$ gh = s \Delta T $

Now, solve for $ \Delta T $:

$ \Delta T = \frac{gh}{s} $

Substitute the known values:

$ \Delta T = \frac{10 \times 200}{4200} $

Calculate $ \Delta T $:

$ \Delta T = \frac{2000}{4200} $

Simplify the fraction:

$ \Delta T = \frac{10}{21} $

Therefore, the rise in temperature is $ \Delta T \approx 0.48 \, \text{K} $.

Given the same room temperature of 300 K for helium and argon, we are tasked with finding the ratio of their average kinetic energies per molecule.

Formula for Kinetic Energy per Molecule:

$ \mathrm{K.E.} = \frac{\mathrm{f}}{2} \mathrm{kT} $

In this equation:

$\mathrm{f}$ is the degrees of freedom, which is 3 for both helium (He) and argon (Ar) since they are monatomic gases.

$k$ is the Boltzmann constant.

$T$ is the temperature in Kelvin.

Kinetic Energy Comparison:

For both helium and argon, since $\mathrm{f} = 3$, the expression simplifies:

$ \frac{\mathrm{K} \cdot \mathrm{E}_{\mathrm{He}}}{\mathrm{K} \cdot \mathrm{E}_{\mathrm{Ar}}} = \frac{1}{1} $

Thus, the average kinetic energy per molecule for both gases is the same at the same temperature, leading to a ratio of:

$ 1:1 $

This result implies that despite their differences in molar masses (helium being 4 g/mol and argon 40 g/mol), the average kinetic energy per molecule remains equal at the same temperature.

To accurately match the terms from List-I with the corresponding statements in List-II, consider the properties of each thermodynamic process:

Isothermal Process (A):

During an isothermal process, the temperature remains constant ($ \Delta T = 0 $), which means that the change in internal energy ($ \Delta U $) is zero. Therefore, it corresponds to (IV).

Adiabatic Process (B):

In an adiabatic process, no heat is exchanged with the surroundings ($ \Delta Q = 0 $). This matches with (II).

Isobaric Process (C):

An isobaric process involves constant pressure conditions ($ \Delta P = 0 $), but internal energy ($ \Delta U $) can change. Thus, it's associated with (III).

Isochoric Process (D):

For an isochoric process, volume remains constant ($ \Delta V = 0 $), which means that no work is done ($ \Delta W = 0 $). This aligns with (I).

Based on these descriptions, the correct matching is:

(A) Isothermal → (IV)

(B) Adiabatic → (II)

(C) Isobaric → (III)

(D) Isochoric → (I)

The explanation uses the equation for the heat capacity ratio, $ \gamma = 1 + \frac{2}{\text{f}} $, where $ \text{f} $ is the degrees of freedom of the gas. Here's how it applies to different types of gases:

Triatomic Rigid Gas:

Degrees of freedom ($ \text{f} $) = 6

$ \gamma = 1 + \frac{2}{6} = \frac{4}{3} $

Diatomic Non-Rigid Gas:

Degrees of freedom ($ \text{f} $) = 7

$ \gamma = 1 + \frac{2}{7} = \frac{9}{7} $

Diatomic Rigid Gas:

Degrees of freedom ($ \text{f} $) = 5

$ \gamma = 1 + \frac{2}{5} = \frac{7}{5} $

Monoatomic Gas:

Degrees of freedom ($ \text{f} $) = 3

$ \gamma = 1 + \frac{2}{3} = \frac{5}{3} $

Therefore, when matching List I with List II, the correct pairing is:

A (Triatomic rigid gas) - III $(\frac{4}{3})$

B (Diatomic non-rigid gas) - IV $(\frac{9}{7})$

C (Monoatomic gas) - I $(\frac{5}{3})$

D (Diatomic rigid gas) - II $(\frac{7}{5})$

Thus, the correct answer from the options given is:

A-III, B-IV, C-I, D-II

Calculate the Change in Temperature:

Use the formula:

$ \Delta Q = mC_v \Delta T $

Given:

$\Delta Q = 8.1 \times 10^2 \, \text{J}$

$m = 0.1 \, \text{kg}$

$C_v = 900 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1}$

Substitute the values to find $\Delta T$:

$ 8.1 \times 10^2 = 0.1 \times 900 \times \Delta T $

Solving for $\Delta T$:

$ \Delta T = \frac{8.1 \times 10^2}{0.1 \times 900} = 9 \, \text{K} $

Calculate the Change in Area:

The change in area $\Delta A$ is calculated using:

$ \Delta A = A_0 \times 2 \alpha \Delta T $

Where:

$A_0$ is the original area of the sheet.

$\alpha = 3.1 \times 10^{-5} \, \text{K}^{-1}$.

$\Delta T = 9 \, \text{K}$.

First, calculate the original area $A_0$:

$ A_0 = l \times d = 9 \, \text{cm} \times 4 \, \text{cm} = 36 \, \text{cm}^2 = 0.0036 \, \text{m}^2 $

Now, substitute into the formula for $\Delta A$:

$ \Delta A = 0.0036 \times 2 \times 3.1 \times 10^{-5} \times 9 $

Simplifying:

$ \Delta A = 2.0 \times 10^{-6} \, \text{m}^2 $

Therefore, the change in area of the rectangular sheet is $\boxed{2.0 \times 10^{-6} \, \text{m}^2}$.

$\begin{aligned} &\text { Number of masses will remain constant }\\ &\begin{aligned} & \mathrm{n}_1+\mathrm{n}_2=\mathrm{n}_{\mathrm{f}} \\ & \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{RT}_1}+\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{RT}_2}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}_{\mathrm{f}}}{\mathrm{RT}_{\mathrm{f}}} \\ & \frac{8 \times 2 \mathrm{~V}}{\mathrm{R} \times 1000}+\frac{7 \times \mathrm{V}}{\mathrm{R} \times 500}=\frac{\mathrm{P}_{\mathrm{f}}(3 \mathrm{~V})}{\mathrm{R} \times 600} \\ & \frac{16}{1000}+\frac{14}{1000}=\frac{\mathrm{P}_{\mathrm{f}}}{\mathrm{R} \times 600} \\ & \frac{30}{1000}=\frac{\mathrm{P}_{\mathrm{f}}}{200} \\ & \mathrm{P}_{\mathrm{f}}=6 \mathrm{kPa} \end{aligned} \end{aligned}$

(A) Isobaric $(\mathrm{P}=\mathrm{C})$

$\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{~V}$

(B) Isochoric $(\mathrm{V}=\mathrm{C})$

$\Delta \mathrm{Q}=\Delta \mathrm{U}$

(C) Adiabatic $(\Delta \mathrm{Q}=0)$

$\Delta \mathrm{Q}=0$

(D) Isothermal $(\Delta \mathrm{U}=0)$

$\Delta \mathrm{Q}=\Delta \mathrm{W}$

The speed of sound in an ideal gas is given by

$ v = \sqrt{\gamma \frac{P}{\rho}}, $

where:

$\gamma$ is the ratio of specific heats,

$P$ is the pressure, and

$\rho$ is the density.

Since the problem states that $\frac{P}{\rho}$ is the same for all the gases, the speed of sound in each gas is determined solely by the factor $\sqrt{\gamma}$.

Let's determine the appropriate $\gamma$ for each gas:

Helium (He):

Helium is a monatomic gas.

For a monatomic gas, $\gamma = \frac{5}{3}$.

Therefore, $v_{\mathrm{He}} \propto \sqrt{\frac{5}{3}}$.

Methane (CH$_4$):

Methane is a polyatomic gas (a tetrahedral molecule with three rotational degrees of freedom).

For nonlinear polyatomic gases, $\gamma$ is typically taken as $\frac{4}{3}$.

Thus, $v_{\mathrm{CH_4}} \propto \sqrt{\frac{4}{3}}$.

Carbon Dioxide (CO$_2$):

Carbon dioxide is a linear molecule.

For linear molecules, $\gamma = \frac{7}{5}$.

Hence, $v_{\mathrm{CO_2}} \propto \sqrt{\frac{7}{5}}$.

Therefore, the ratio of the speeds of sound in the gases is:

$ v_{\mathrm{He}} : v_{\mathrm{CH_4}} : v_{\mathrm{CO_2}} = \sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{7}{5}}. $

Comparing this expression with the given options, we see that it matches Option C.

To determine the frequency of collisions for oxygen molecules, we use the formula for frequency:

$ \text{Frequency} = \frac{1}{T} = \frac{V_{\text{avg}}}{\lambda} $

where $ V_{\text{avg}} $ is the average speed of the molecules and $ \lambda $ is the mean free path.

Given:

Average speed ($ V_{\text{avg}} $) = 600 m/s

Mean free path ($ \lambda $) = $ 3 \times 10^{-7} $ m

Substitute these values into the formula:

$ \text{Frequency} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^9 \, \text{s}^{-1} $

Therefore, the frequency of collisions is $ 2 \times 10^9 \, \text{collisions per second} $.

$\begin{aligned} & \omega_1=\mathrm{P}_0 \mathrm{v}_0 \ell \mathrm{n} 4 \\ & \omega_2=\frac{\mathrm{P}_0}{4}\left(-3 \mathrm{v}_0\right)=-\frac{3 \mathrm{P}_0 \mathrm{v}_0}{4} \\ & \omega_3=0 \\ & \mathrm{Q}_{\mathrm{T}}=\Delta \mathrm{U}_{\text {cyclic }}+\omega \\ & \mathrm{Q}_{\mathrm{T}}=\omega \quad\left(\Delta \mathrm{U}_{\text {cyclic }}=0\right) \\ & \mathrm{Q}_{\mathrm{T}}=\mathrm{P}_0 \mathrm{v}_0\left(\ln 4-\frac{3}{4}\right) \\ & =\mathrm{P}_0 \mathrm{v}_0(2 \ln 2-0.75) \end{aligned}$

In this scenario, we're dealing with an isochoric process, meaning the volume of the gas remains constant. For an ideal gas in an isochoric process, the pressure ($P$) is directly proportional to the temperature ($T$) in Kelvin.

The relationship can be expressed as:

$ P \propto T $

Given the percentage increase in pressure is $0.4\%$ when the temperature is increased by $1^{\circ} \text{C}$, we can use the relationship:

$ \frac{\Delta P}{P} = \frac{\Delta T}{T} $

Substituting the given values into the equation, we have:

$ \frac{0.4}{100} = \frac{1}{T} $

Solving for $T$, we find:

$ T = 250 \, \text{K} $

Therefore, the initial temperature of the gas in the vessel must be $250 \, \text{K}$.

To determine the change in temperature when a gas is adiabatically compressed, we are given the following initial conditions:

Initial volume, $ V_1 = 800 \, \text{cm}^3 $

Final volume, $ V_2 = 200 \, \text{cm}^3 $

Initial temperature, $ T_1 = 27^\circ \text{C} = 300 \, \text{K} $

Adiabatic index (ratio of specific heats) $ \gamma = 1.5 $

In an adiabatic process, the relationship between temperature and volume is given by the equation:

$ TV^{\gamma-1} = \text{constant} $

Applying this to the initial and final states:

$ 300 \times (800)^{0.5} = T_2 \times (200)^{0.5} $

Solving for the final temperature $ T_2 $:

$ T_2 = 300 \times \left( \frac{800}{200} \right)^{0.5} = 300 \times (4)^{0.5} $

$ T_2 = 300 \times 2 = 600 \, \text{K} $

Thus, the change in temperature is:

$ \Delta T = T_2 - T_1 = 600 \, \text{K} - 300 \, \text{K} = 300 \, \text{K} $

Therefore, when the gas is adiabatically compressed, the temperature increases by 300 K.

During the melting of a slab of ice at 273 K under atmospheric pressure, the following occurs:

As the ice melts, its volume decreases. Because of this volume reduction, the atmosphere exerts a positive work on the ice-water system. When heat is absorbed by the ice-water system, we denote the absorbed heat as $ \Delta \mathrm{Q} $, which is positive. The work done by the ice-water system is negative because it is being compressed by the external pressure.

According to the first law of thermodynamics:

$ \Delta \mathrm{U} = \Delta \mathrm{Q} + \Delta \mathrm{W} $

In this equation, $ \Delta \mathrm{U} $ represents the change in internal energy, $ \Delta \mathrm{Q} $ is the heat absorbed by the system, and $ \Delta \mathrm{W} $ is the work done by the system. Since the work done is negative and the absorbed heat is positive, the overall internal energy ($ \Delta \mathrm{U} $) increases.

Using WET

Total energy supplied $=$ gravitational potential energy + spring potential energy + work done by gas

$ \begin{aligned} & \mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\int_{\mathrm{L}_0}^{\mathrm{L}_1} \mathrm{kx}^3 \mathrm{dx}+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1 \mathrm{~A}}{\mathrm{~L}_0 \mathrm{~A}}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \frac{\mathrm{~K}}{4}\left[\mathrm{x}^4\right]_{\mathrm{L}_0}^{\mathrm{L}_1}+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\int_{\mathrm{L}_0}^{\mathrm{L}_1} k x^3 \mathrm{dx}+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \mathrm{~W}_{\text {ext }}=\frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n}\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right] \end{aligned}$

To match the entries in List-I with their corresponding units in List-II, we can use the definitions and formulas related to each concept in thermal physics:

Heat Capacity of a Body (C') is defined as:

$ C' = \frac{\Delta Q}{\Delta T} $

The unit for heat capacity (C') is Joules per Kelvin (J K$^{-1}$).

Specific Heat Capacity of a Body (S) indicates how much heat energy is required to change the temperature of a unit mass of a substance by one degree Kelvin:

$ S = \frac{\Delta Q}{m \Delta T} $

The unit for specific heat capacity (S) is Joules per kilogram per Kelvin (J kg$^{-1}$ K$^{-1}$).

Latent Heat (L) is the heat energy absorbed or released during a phase change of a substance at a constant temperature:

$ L = \frac{\Delta Q}{m} $

The unit for latent heat (L) is Joules per kilogram (J kg$^{-1}$).

Thermal Conductivity (K) measures the ability of a material to conduct heat:

$ \Delta Q = \frac{KA \Delta T}{L} \quad \Rightarrow \quad K = \frac{\Delta Q \cdot L}{A \Delta T} $

The unit for thermal conductivity (K) is Joules per meter per Kelvin per second (J m$^{-1}$ K$^{-1}$ s$^{-1}$).

Using these formulas and their corresponding units, the correct matches are:

(A) Heat capacity of body $\rightarrow$ (II) J K$^{-1}$

(B) Specific heat capacity of body $\rightarrow$ (III) J kg$^{-1}$ K$^{-1}$

(C) Latent heat $\rightarrow$ (I) J kg$^{-1}$

(D) Thermal conductivity $\rightarrow$ (IV) J m$^{-1}$ K$^{-1}$ s$^{-1}$

In an adiabatic process, the key characteristics for a monoatomic gas are:

No Heat Exchange: The adiabatic process occurs without heat transfer, meaning $ Q = 0 $.

Work Done and Internal Energy Change: The work done on or by the system is equal to the change in internal energy. Mathematically, this is expressed as:

$ \Delta U = -W $

where $\Delta U$ is the change in internal energy and $W$ is the work done.

Work and Temperature Relationship: The work done when changing the temperature from $ T_1 $ to $ T_2 $ is proportional to the difference between these temperatures. This relationship can be expressed as:

$ |\text{Work Done}| = nC_v \Delta T \propto (T_2 - T_1) $

Here, $ n $ is the number of moles, $ C_v $ is the molar heat capacity at constant volume, and $\Delta T$ is the change in temperature.

From these characteristics, the correct features of an adiabatic process in a monoatomic gas relate to the relationships involving work done and temperature change, specifically options (B) and (E).

Option B is correct.

• In an adiabatic process no heat is exchanged, so

$\delta Q = 0.$

• The (molar) heat capacity for any process is defined by

$C = \frac{\delta Q}{n\,dT}\,. $

Since $\delta Q=0$ even when $dT\neq0$, it follows that

$C_{\rm adiabatic}=0\,. $

All other statements are therefore false.

The equation for a real gas is given by:

$ \left(\mathrm{P} + \frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V} - \mathrm{b}) = \mathrm{RT} $

Here, $ \mathrm{P} $, $ \mathrm{V} $, $ \mathrm{T} $, and $ R $ are the pressure, volume, temperature, and gas constant, respectively.

To find the dimension of $ ab^{-2} $, we proceed with the following steps:

Rearrange the equation for dimensions:

$ [\mathrm{a}] = [\mathrm{P}][\mathrm{V}^2] $

Since $[\mathrm{P}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}$ and $[\mathrm{V}] = \mathrm{L}^3$, we have:

$ [\mathrm{a}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}(\mathrm{L}^6) = \mathrm{ML}^5\mathrm{T}^{-2} $

As derived from the equation:

$[\mathrm{b}] = [\mathrm{V}] = \mathrm{L}^3$

Now, to find $[\mathrm{ab}^{-2}]$:

$[\mathrm{ab}^{-2}] = \frac{[\mathrm{a}]}{[\mathrm{b}]^2} = \mathrm{ML}^5\mathrm{T}^{-2} \cdot \mathrm{L}^{-6} = \mathrm{ML}^{-1}\mathrm{T}^{-2}$

This dimension, $\mathrm{ML}^{-1}\mathrm{T}^{-2}$, is equivalent to the dimension of energy density.

To convert a temperature difference into electrical energy effectively, the material in question should have low thermal conductivity and high electrical conductivity. Such materials are classified as thermoelectric materials.

A low thermal conductivity ensures that the material does not easily transfer heat, which is crucial for maintaining a significant temperature difference between its hot and cold sides. This helps in generating a greater voltage. Meanwhile, high electrical conductivity is essential for facilitating the efficient movement of electrons across the material when there is a temperature difference, leading to maximal energy conversion.

Therefore, a material that optimally harvests heat energy into electrical energy should embody these characteristics.

We know, ideal gas equation,

$\mathrm{Pv = nRT}$

For isothermal process, T = constant

$\mathrm{\Rightarrow PV = const.}$

$\mathrm{\Rightarrow P \propto \frac{1}{V}}$

For an adiabatic process,

$\mathrm{PV^r=constant}$

We can see, for the same pressure incresement, ${P_2} - {P_1}$, $\left( {{v_4} - {v_3}} \right) > \left( {{v_2} - {v_1}} \right)$

i.e. the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Hence, option 4 is corret.

Here, we will use the Newton's law of cooling.

We know, ${{{T_1} - {T_2}} \over {\Delta t}} = k\left( {{{{T_1} + {T_2}} \over 2} - {T_s}} \right)$ .... (1)

where, T$_1$ and T$_2$ are initial and final temperatures respectively.

$\mathrm{T_s}$ = surrounding temperature

$\Delta t$ = time taken

$k$ = positive constant

Given,

Ist case :

${T_1} = 90^\circ C$

${T_2} = 80^\circ C$

$\Delta t = t$, ${T_s} = 20^\circ C$

IInd case :

${T_1} = 80^\circ C$

${T_2} = 60^\circ C$

${T_s} = 20^\circ C$, $\Delta t = ?$

Now, for Ist case,

${{90 - 80} \over t} = k\left( {{{90 + 80} \over 2} - 20} \right)$ ...... (from (1))

$ \Rightarrow {{10} \over t} = k\left( {85 - 20} \right)$

$ \Rightarrow {{10} \over t} = 65k$ .... (2)

For IInd case,

${{80 - 60} \over {\Delta t}} = k\left( {{{80 + 60} \over 2} - 20} \right)$ .... (From (1))

$ \Rightarrow {{20} \over {\Delta t}} = k\left( {70 - 20} \right)$

$ \Rightarrow {{20} \over {\Delta t}} = 50k$ .... (3)

Now, by (2) $\div$ (3),

${{{{10} \over t}} \over {{{20} \over {\Delta t}}}} = {{13} \over {10}}$

$ \Rightarrow {{\Delta t} \over t}\left( {{1 \over 2}} \right) = {{13} \over 5}$

$ \Rightarrow \Delta t = {{13} \over 5}t$