Heat and Thermodynamics

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$

As A to B is can adiabatic path,

So, $\Delta {Q_{A \to B}} = o$ .... (1)

So, now we need to find the heat absorbed for the isothermal path (B to C)

We know, change in internal energy

$\Delta u = m{c_v}\Delta T$

$ \Rightarrow \Delta u = o$ (for isothermal as $\Delta T = o$)

Now, using Ist law of thermodynamics

$\Delta Q = \Delta u + w$

$ \Rightarrow \Delta Q = o + w \Rightarrow \Delta Q = w$

So, $\Delta {Q_{B \to C}} = {W_{B \to C}}$

$ = nRT\ln \left( {{{{v_C}} \over {{v_B}}}} \right)$

$ \Rightarrow \Delta {Q_{B \to C}} = nR(450)\ln \left( {{4 \over 3}} \right)$

$ \Rightarrow \Delta {Q_{B \to C}} = 450nR\ln \left( {{4 \over 3}} \right)$ .... (2)

So net heat absorbed,

$Q = \Delta {Q_{A \to B}} + \Delta {Q_{B \to C}}$

$ \Rightarrow Q = 450nR\ln \left( {{4 \over 3}} \right)$ ... (from 1 and 2)

Net heat absorbed per mole,

$ \Rightarrow {Q \over n} = 450R(\ln 4 - \ln 3)$.

Option 2 is correct.

$\Delta \mathrm{W}=-\Delta \mathrm{U}=-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

We are given that the ratio of the vapour densities of the two gases is

$\frac{4}{25}.$

Since vapour density is proportional to the molecular mass, we can write

$\frac{M_1}{M_2} = \frac{4}{25},$

where $M_1$ and $M_2$ are the molecular masses of the gases.

The root mean square (r.m.s.) velocity of a gas is given by

$v_{\text{rms}} = \sqrt{\frac{3RT}{M}},$

where:

$R$ is the universal gas constant,

$T$ is the temperature,

$M$ is the molecular mass of the gas.

Since both gases are at the same temperature, the ratio of their r.m.s. velocities is

$\frac{v_{\text{rms,1}}}{v_{\text{rms,2}}} = \sqrt{\frac{M_2}{M_1}}.$

Substitute the ratio of molecular masses:

$\frac{v_{\text{rms,1}}}{v_{\text{rms,2}}} = \sqrt{\frac{25}{4}} = \frac{5}{2}.$

Thus, the ratio of their r.m.s. velocities is

$\frac{5}{2}.$

The correct answer is Option A: $\frac{5}{2}.$

$\begin{aligned} & (\mathrm{KE})_{\text {Transsational }}=\left[\frac{3}{2} \mathrm{KT}\right] \times \text { no. of molecule } \\ & \text { No. of molecule }=\left[\frac{50}{44} \times 6.023 \times 10^{23}\right] \\ & (\mathrm{KE})_{\text {Transsational }}=4108.644 \mathrm{~J} \end{aligned}$

We know, for an ideal gas,

${V_{rms}} = \sqrt {{{3RT} \over M}} $

$ \Rightarrow {V_{ms}} = V_{rms}^2 = {{3RT} \over M}$

$ \Rightarrow {V_{ms}} \propto T$

$ \Rightarrow {V_{ms}} \propto T$

i.e. mean squared velocity is directly proportional to temperature.

$ \Rightarrow {V_{ms}} = {\left( {{{3R} \over M}} \right)^T}$

by comparing it with $y=mx$ (straight line)

Hence, option 3 is correct.

We know, efficiency of a carnot engine, $\eta$ or $e = 1 - {{{T_C}} \over {{T_H}}}$

where, T$_C$ = temperature of cold sink

T$_H$ = temperature of hot source

So, ${\eta _{12}} = 1 - {{273} \over {473}} = {{200} \over {473}} = 0.423$

${\eta _1} = 1 - {{373} \over {473}} = {{100} \over {473}} = 0.211$

${\eta _2} = 1 - {{273} \over {373}} = {{100} \over {373}} = 0.268$

Here, ${\eta _1} + {\eta _2} = 0.211 + 0.268 = 0.479 > 0.423$

So, ${\eta _{12}} < {\eta _1} + {\eta _2}$

$\begin{aligned} & \frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{t}}=\mathrm{K}\left[\mathrm{~T}_{\text {avg }}-\mathrm{T}_{\mathrm{s}}\right] \\ & \mathrm{T}_1=24^{\circ} \mathrm{C} ; \mathrm{T}_2=40^{\circ} \mathrm{C}, \mathrm{t}=4, \mathrm{~T}_{\mathrm{s}}=16^{\circ} \mathrm{C} \\ & \frac{40-24}{4}=\mathrm{K}[32-16] \\ & \mathrm{K}=\frac{4}{16}=\frac{1}{4} \\ & \text { Now } \frac{24-\mathrm{T}}{4}=\mathrm{K}\left[\frac{\mathrm{~T}+24}{2}-16\right] \\ & 24-\mathrm{T}=\frac{\mathrm{T}-16}{2}+16 \\ & \frac{3 \mathrm{~T}}{2}=28 \\ & \mathrm{~T}=\frac{56}{3} \mathrm{C} \end{aligned}$

Step 1: Find the Work Done ($\mathrm{W}$) in the Cycle

The work done in a cyclic process shown in a pressure-volume (P-V) graph is equal to the area enclosed by the cycle. For a circle, this area is $\frac{1}{2} \pi R^2$, where $R$ is the radius of the circle.

Step 2: Put in the Given Values

Radius $R$ on the graph is $ R = \frac{200}{2} \times 10^3 \text{ (for Pressure, P)} \\ R = \frac{200}{2} \times 10^{-6} \text{ (for Volume, V)} $

Step 3: Calculate the Area (Work Done)

$ \mathrm{W} = \frac{1}{2} \times \pi \times \left(\frac{200}{2} \times 10^3\right) \times \left(\frac{200}{2} \times 10^{-6}\right) $

First, calculate $\frac{200}{2} = 100$, so:

$ \mathrm{W} = \frac{1}{2} \times \pi \times 100 \times 10^3 \times 100 \times 10^{-6} $

Multiplying $100 \times 100 = 10,000$, and $10^3 \times 10^{-6} = 10^{-3}$, so:

$ \mathrm{W} = \frac{1}{2} \times \pi \times 10,000 \times 10^{-3} = \frac{1}{2} \times \pi \times 10 = 5\pi~\mathrm{J} $

Step 4: Relate to Heat Exchanged

In a complete cycle, the change in internal energy is zero, so total heat exchanged is equal to the total work done.

So, the magnitude of heat exchanged is $5\pi~\mathrm{J}$.

Let's analyze both statements step by step.

Assertion (A): "In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases."

 • In an adiabatic process, no heat is exchanged with the surroundings.

 • For a reversible adiabatic (compression) process, the relationship between temperature and volume is given by

  $T V^{\gamma-1} = \text{constant},$

  where $\gamma$ is the heat capacity ratio (greater than 1).

 • If the volume is reduced from $V$ to $V/2$, then

  $T_{\text{final}} = T_{\text{initial}} \left(\frac{V}{V/2}\right)^{\gamma-1} = T_{\text{initial}}\,(2)^{\gamma-1}.$

 • Since $(2)^{\gamma-1} > 1,$ the final temperature is higher than the initial temperature.

 • Therefore, the assertion that the temperature decreases is incorrect.

Reason (R): "Free expansion of an ideal gas is an irreversible and an adiabatic process."

 • In a free expansion, the gas expands into a vacuum without doing any work on the surroundings.

 • Since the container is insulated, no heat is exchanged, making it adiabatic.

 • However, the process is inherently irreversible due to the spontaneous expansion and mixing of states.

 • Thus, this statement is true.

Since Assertion (A) is false and Reason (R) is true (but the reason does not explain the false assertion about temperature change during compression), the correct answer is:

Option C

(A) is false but (R) is true.

$\frac{C}{5}=\frac{F-32}{9} \Rightarrow C=\frac{5 F}{9}-\frac{160}{9}$

For an ideal gas, the equation of state is given by

$ pV = nRT. $

If during a process the pressure increases linearly with temperature (i.e., $p \propto T$), then from the ideal gas law (with a fixed amount of gas, $n$, and with the gas constant, $R$), it follows that

$ \frac{p}{T} = \frac{nR}{V}. $

Since $nR$ is constant, a linear relationship between $p$ and $T$ implies that $\frac{p}{T}$ remains constant throughout the process. This can only be true if the volume $V$ is constant. Hence, the process is isochoric.

For an isochoric (constant volume) process:

The work done by the gas is given by

$ W = \int p\, dV, $

and since $dV = 0$, we have $W = 0.$ (Statement A is true.)

The internal energy change for an ideal gas depends only on temperature:

$ \Delta U = nC_V\Delta T. $

If the temperature increases, then $\Delta U > 0.$ (Statement D is true.)

For an isochoric process, no expansion work is done, so any heat added goes entirely into increasing the internal energy. This means the heat added is not different from the change in internal energy (in fact, $Q = \Delta U$). (Statement B is false.)

Since the process is isochoric, the volume does not change (and certainly is not increased). (Statement C is false, and Statement E stating it is isochoric is true.)

Thus, the correct true statements are A, D, and E.

To determine the change in entropy when water is heated from temperature $T_1$ to $T_2$ (here, $T_\gamma$ is equivalent to $T_2$), we use the definition of entropy change for a reversible process:

$ \Delta S = \int_{T_1}^{T_2} \frac{dQ}{T} $

Since the water is being heated slowly, the process is reversible, and the heat added is given by:

$ dQ = mc\,dT $

where:

• $m$ is the mass of the water,

• $c$ is the specific heat capacity (given as $1\,\mathrm{J\,kg^{-1}\,K^{-1}}$).

Substitute $dQ$ into the integral:

$ \Delta S = \int_{T_1}^{T_2} \frac{mc\,dT}{T} = mc\, \int_{T_1}^{T_2} \frac{dT}{T}. $

Since $c = 1$, the equation simplifies to:

$ \Delta S = m \int_{T_1}^{T_2} \frac{dT}{T}. $

Evaluating the integral, we have:

$ \int_{T_1}^{T_2} \frac{dT}{T} = \ln{T}\Big|_{T_1}^{T_2} = \ln\left(\frac{T_2}{T_1}\right). $

Thus, the change in entropy is:

$ \Delta S = m \ln\left(\frac{T_2}{T_1}\right). $

Comparing with the options provided, the correct answer is:

Option D: $m \ln\left(\frac{T_2}{T_1}\right).$

To find the work done by an ideal gas along the path $ABCD$ on the given $P-V$ diagram, we calculate the work done on each segment individually and then sum them up.

Segment $AB$: The work is given by $\text{w}_{AB} = P_0 \times V_0$, contributing $P_0 V_0$.

Segment $BC$: No volume change occurs along this path, resulting in zero work: $\text{w}_{BC} = 0$.

Segment $CD$: The gas compresses, performing negative work: $\text{w}_{CD} = -(2P_0 \times 2V_0) = -4P_0 V_0$.

Adding the work done in each segment:

$ \begin{align*} \text{w}_{ABCD} &= \text{w}_{AB} + \text{w}_{BC} + \text{w}_{CD} \\ &= P_0 V_0 + 0 - 4P_0 V_0 \\ &= -3P_0 V_0 \end{align*} $

Thus, the total work done by the gas along the path $ABCD$ is $-3P_0 V_0$.

Let's analyze each statement step by step:

For (A): “Pressure varies inversely with volume of an ideal gas.”

At constant temperature (isothermal process), Boyle's law applies:

$PV = \text{constant}.$

Thus, pressure and volume have an inverse relationship.

Hence, (A) corresponds to the Isothermal process, which is (III).

For (B): “Heat absorbed goes partly to increase internal energy and partly to do work.”

In a constant pressure (isobaric) process, when heat is supplied, part of it is used to do work (expansion, for example) and the rest increases the internal energy.

Therefore, (B) matches with the Isobaric process, which is (IV).

For (C): “Heat is neither absorbed nor released by a system.”

This implies that the process is adiabatic (no heat exchange).

Thus, (C) matches with the Adiabatic process, which is (I).

For (D): “No work is done on or by a gas.”

Work done by a gas is given by $W = P\Delta V.$

If the volume does not change (constant volume), then no work is done.

So, (D) identifies with the Isochoric process, which is (II).

Putting it all together:

(A) → (III) Isothermal process

(B) → (IV) Isobaric process

(C) → (I) Adiabatic process

(D) → (II) Isochoric process

Thus, the correct answer is:

Option D

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

Let's determine the mass step by step:

The bullet is heated from 300 K to 600 K, so the increase in temperature is:

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

The heat required to raise the temperature of the bullet (using the specific heat capacity of lead) is:

$Q_1 = m \times c \times \Delta T = m \times 125\, \text{J/kg K} \times 300\, \text{K} = 37500\, m \, \text{J},$

where $ m $ is the mass in kilograms.

After reaching 600 K, the bullet melts. The heat required for the phase change (melting) is given by:

$Q_2 = m \times L = m \times 2.5 \times 10^4\, \text{J/kg} = 25000\, m \, \text{J}.$

The total heat required for the process is:

$Q = Q_1 + Q_2 = 37500\, m + 25000\, m = 62500\, m \, \text{J}.$

We are given that the total heat is 625 J, so:

$62500\, m = 625.$

Solving for $ m $:

$m = \frac{625}{62500} = 0.01\, \text{kg}.$

Finally, converting the mass from kilograms to grams:

$0.01\, \text{kg} = 0.01 \times 1000\, \text{g} = 10\, \text{g}.$

Thus, the mass of the bullet is 10 grams. This corresponds to Option D.

Check Statement (A):

“Internal energy (U), volume (V), and mass (M) are extensive variables.”

Internal energy is extensive.

Volume is extensive.

Mass is extensive.

Hence, statement (A) is correct.

Check Statement (B):

“Pressure (P), temperature (T), and density (ρ) are intensive variables.”

Pressure is intensive.

Temperature is intensive.

Density is intensive.

Hence, statement (B) is correct.

Check Statement (C):

“Volume (V), temperature (T), and density (ρ) are intensive variables.”

Volume is not intensive; it is an extensive property.

(T) and (ρ) are indeed intensive.

Since volume is miscategorized here, (C) is incorrect.

Check Statement (D):

“Mass (M), temperature (T), and internal energy are extensive variables.”

Mass is extensive.

Temperature is not extensive (it is intensive).

Internal energy is extensive.

Because temperature is miscategorized, (D) is incorrect.

Only (A) and (B) are correct statements.

Therefore, the correct choice is:

Option B: (A) and (B) Only.

Let's analyze the problem step by step:

For a rigid diatomic molecule (without vibrational modes):

Degrees of freedom:

Translation: 3

Rotation: 2

Total: 5

The molar specific heat at constant volume is given by:

$C_v = \frac{5}{2}R$

At constant pressure, it is:

$C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$

Therefore, the ratio is:

$\gamma_1 = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \approx 1.4$

For the diatomic molecule that also has vibrational modes (assuming the vibrational mode is fully excited):

Additional vibrational mode contributions:

Each vibrational mode contributes 2 degrees of freedom (one kinetic and one potential).

Therefore, the degrees of freedom become:

$3\ (\text{translation}) + 2\ (\text{rotation}) + 2\ (\text{vibration}) = 7$

The molar specific heat at constant volume now is:

$C_v = \frac{7}{2}R$

And at constant pressure:

$C_p = C_v + R = \frac{7}{2}R + R = \frac{9}{2}R$

Thus, the ratio is:

$\gamma_2 = \frac{C_p}{C_v} = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} \approx 1.2857$

Comparison:

For the rigid molecule, $\gamma_1 \approx 1.4$.

For the molecule with vibrational modes, $\gamma_2 \approx 1.2857$.

Clearly, $\gamma_2 < \gamma_1$.

Therefore, the correct option is:

Option A: $\gamma_2 < \gamma_1$

To determine the energy radiated by two spherical bodies, both made of the same material but with different radii and temperatures, we can use the Stefan-Boltzmann law. This law states that the energy radiated by a black body per unit time is proportional to the product of the surface area $ A $ and the fourth power of its temperature $ T $.

The formula for power $ P $ is given by:

$ P = \sigma e A T^4 $

Where:

$ \sigma $ is the Stefan-Boltzmann constant

$ e $ is the emissivity of the material

$ A $ is the surface area of the sphere

$ T $ is the temperature in Kelvin

Expressing this relationship as a proportionality:

$ P \propto A T^4 $

For a sphere, the surface area $ A $ is:

$ A = 4\pi R^2 $

Hence, the power radiated is proportional to:

$ P \propto R^2 T^4 $

For two bodies with radii $ R_1 $ and $ R_2 $, and temperatures $ T_1 $ and $ T_2 $, the ratio of their powers is:

$ \frac{P_1}{P_2} = \left( \frac{R_1}{R_2} \right)^2 \left( \frac{T_1}{T_2} \right)^4 $

Given:

$ R_1 = 0.2 \, \text{m} $, $ T_1 = 800 \, \text{K} $

$ R_2 = 0.8 \, \text{m} $, $ T_2 = 400 \, \text{K} $

Substituting into the formula:

$ \frac{E}{E'} = \left( \frac{0.2}{0.8} \right)^2 \left( \frac{800}{400} \right)^4 $

Calculating each part:

$\left( \frac{0.2}{0.8} \right)^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16}$

$\left( \frac{800}{400} \right)^4 = 2^4 = 16$

Combining these calculations:

$ \frac{E}{E'} = \frac{1}{16} \times 16 = 1 $

This implies that the energy radiated by the bigger body $ E' $ equals the energy radiated by the smaller body $ E $. Thus, $ E' = E $.

$\Delta {Q_1} = m{s_i}\Delta {T_2},\,\Delta {Q_2} = m{L_i}$

$\Delta {Q_3} = m{S_w}\Delta {T_3},\,\Delta {Q_4} = m{L_s},\,\Delta {Q_5} = m{S_s}\Delta {T_5}$

$\Delta {Q_{net}} = m({S_i}\Delta {T_1} + {L_i} + {S_w}\Delta {T_3} + {L_s} + {S_s}\Delta {T_s})$

$ = {10^{ - 3}}(2100 \times 10 + 335000 + 4180 \times 100 + 2250000 + 1920 \times 10)$

$ = {10^{ - 3}}(3043200) = 3043.2\,J$

So, $\Delta {Q_{net}} \approx 3043\,J$

For $A$ to $B$ :

Given $\mathrm{PV}^3=\mathrm{RT} \Rightarrow \mathrm{P}=\frac{\mathrm{RT}}{\mathbf{V}^3}$

Work done $\mathrm{W}_{\mathrm{AB}}=\int \mathrm{PdV}$

$\begin{aligned} & =\int \frac{R T}{V^3} d V=R T\left[\frac{V^{-2}}{-2}\right]_{V_1=2}^{V_B=4} \\\\ & =-\frac{R T}{2}\left[\frac{1}{V^2}\right]=-\frac{R T}{2}\left[\frac{1}{16}-\frac{1}{4}\right] \\\\ & =-\frac{R T}{2}\left(\frac{-3}{16}\right)=\frac{3 R T}{32}\end{aligned}$

$\begin{aligned} & =\frac{3}{32} \times 8 \times 300=225 \mathrm{~J} \\\\ & \mathrm{~W}_{\mathrm{B} . \mathrm{C}}(\text { Isobaric process ) } \\\\ & \mathrm{W}_{\mathrm{BC}}=\mathrm{P} \Delta \mathrm{V}=10(2-4)=-20 \mathrm{~J} \\\\ & \mathrm{~W}_{\mathrm{B} . \mathrm{C}}(\text { Isochoric process })=0 \\\\ & \mathrm{~W}_{\text {nct }}=\mathrm{W_{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{C A}=225-20+0=205 \mathrm{~J}\end{aligned}$

The average translational kinetic energy ($K_{avg}$) of a molecule is directly proportional to the absolute temperature (T) of the gas, as described by the equation:

$K_{avg} = \frac{3}{2}kT$

Where:

• $K_{avg}$ is the average kinetic energy,
• $k$ is the Boltzmann's constant, and
• $T$ is the temperature in Kelvin.

From the given problem, if the temperature of the gas is $-78^{\circ}C$, which in Kelvin is $T_1 = -78 + 273 = 195K$, and the average translational kinetic energy is $K$, when the energy becomes $2K$, we need to find the new temperature $(T_2)$.

Using the direct proportionality relation, we can set up the following equation:

$K \propto T$

$2K = K_2 = \frac{3}{2}kT_2$

Given that at $T_1$, the kinetic energy is $K$, and at $T_2$, it's $2K$, we can use the ratio as follows:

$\frac{K_2}{K_1} = \frac{2K}{K} = 2 = \frac{T_2}{T_1}$

Thus, we have:

$T_2 = 2T_1 = 2 \times 195K = 390K$

To find the temperature in Celsius, we convert $390K$ back to Celsius:

$T_{2(Celsius)} = 390 - 273 = 117^{\circ}C$

Therefore, the temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2K$ is $117^{\circ}C$. The correct answer is:

Option D $117^{\circ} \mathrm{C}$

To find the ratio of the initial pressure to the final pressure of an ideal gas undergoing an adiabatic process, we can use the adiabatic equation for an ideal gas, which relates pressure (P) and volume (V) as follows:

$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$

Here, $P_{1}$ and $V_{1}$ are the initial pressure and volume, respectively, $P_{2}$ and $V_{2}$ are the final pressure and volume, respectively, and $\gamma$ is the heat capacity ratio of the gas.

Given in the question, $\gamma = 1.5$, $V_{1} = 5$ litres and $V_{2} = 4$ litres. We need to find the ratio $\frac{P_{1}}{P_{2}}$.

Rearranging the adiabatic equation for the ratio $\frac{P_{1}}{P_{2}}$, we get:

$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma} \Rightarrow \frac{P_{1}}{P_{2}} = \left(\frac{V_{2}}{V_{1}}\right)^{\gamma}$

Substituting the given values:

$\frac{P_{1}}{P_{2}} = \left(\frac{V_{2}}{V_{1}}\right)^{\gamma} = \left(\frac{4}{5}\right)^{1.5}$

To calculate the value:

$\frac{P_{1}}{P_{2}} = \left(\frac{4}{5}\right)^{1.5} = \left(\frac{4}{5}\right)^{\frac{3}{2}}$

Simplifying further:

$\frac{P_{1}}{P_{2}} = \left(\frac{2^2}{5}\right)^{\frac{3}{2}} = \left(\frac{2^3}{5^{\frac{3}{2}}}\right) = \frac{8}{5\sqrt{5}}$

Therefore, the ratio of the initial pressure to the final pressure is $\frac{8}{5\sqrt{5}}$, which corresponds to Option B.

For an adiabatic process, the work done by the gas can be found using the formula:

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

Given that the volume is doubled ($V_2 = 2V_1$) and the adiabatic constant $\gamma = \frac{3}{2}$, we can manipulate the ideal gas law and the adiabatic process relationship to find an expression for work done in terms of the initial conditions and $\gamma$.

Recall, for an adiabatic process, $PV^{\gamma} = \text{constant}$, so we can write:

$ P_1V_1^{\gamma} = P_2V_2^{\gamma} $

Using the fact that $V_2 = 2V_1$, we can express $P_2$ in terms of $P_1$ and $V_1$ as follows:

$ P_1V_1^{\gamma} = P_2(2V_1)^{\gamma} $

$ P_2 = P_1 \left( \frac{V_1}{2V_1} \right)^{\gamma} $

$ P_2 = P_1 \left( \frac{1}{2} \right)^{\gamma} $

The work done then becomes:

$ W = \frac{P_1 V_1 - P_1 \left( \frac{1}{2} \right)^{\gamma} \cdot 2V_1}{\gamma - 1} $

$ W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\gamma} \cdot 2}{\gamma - 1} $

Plugging in $\gamma = \frac{3}{2}$, we get:

$ W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\frac{3}{2}} \cdot 2}{\frac{3}{2} - 1} $

$ W = P_1V_1 \frac{1 - \sqrt{\frac{1}{2}} \cdot 2}{\frac{1}{2}} $

$ W = 2P_1V_1 (1 - \sqrt{\frac{1}{2}}) $

Since $P_1V_1 = nRT$ for 1 mole ($n = 1$) of gas at temperature $T$, we can further simplify:

$ W = 2RT(1 - \sqrt{\frac{1}{2}}) $

$ W = 2RT(1 - \frac{1}{\sqrt{2}}) $

$ W = RT(2 - \sqrt{2}) $

Therefore, the correct option is:

Option B $\mathrm{RT}[2-\sqrt{2}]$.

To find the heat given to a diatomic gas during an isobaric (constant pressure) expansion, we can use the formula that relates the work done by the gas, the heat added to the system, and the change in the internal energy of the system. The first law of thermodynamics states that:

$\Delta Q = \Delta U + W$

where:

• $\Delta Q$ is the heat added to the system,
• $\Delta U$ is the change in internal energy of the system, and
• $W$ is the work done by the gas.

For an isobaric process, the work done $W$ is given by:

$W = P \Delta V$

We're given that $W = 100 \ \mathrm{J}$ for this process, so:

$W = 100 \ \mathrm{J}$

The change in internal energy $\Delta U$ for an ideal gas can also be related to the temperature change and the specific heat capacity at constant volume $C_v$. Using the equation:

$\Delta U = nC_v\Delta T$

However, without direct values for $n$, $\Delta T$, or $C_v$, we need to rely on the relation between the provided work and the heat capacity ratio $\gamma$ to find the heat added. For a diatomic gas, $\gamma = C_p/C_v$. The heat added at constant pressure can also be described as:

$\Delta Q = nC_p\Delta T$

Since we know that for an ideal gas, the work done on the gas during an isobaric process is related to the heat added by the ratio of the specific heats ($\gamma$), we can use the fact that $C_p = \gamma C_v$, and relating that to the work done, we get:

$\Delta Q = \frac{\gamma}{\gamma - 1} W$

Substituting the known values, with $\gamma = 1.4$, and $W = 100 \ \mathrm{J}$, we find:

$\Delta Q = \frac{1.4}{1.4 - 1} \times 100 \ \mathrm{J} = \frac{1.4}{0.4} \times 100 \ \mathrm{J} = 3.5 \times 100 \ \mathrm{J} = 350 \ \mathrm{J}$

Therefore, the heat given to the gas is $\Delta Q = 350 \ \mathrm{J}$, so the correct option is:

Option C

350 J

Let's analyze the given statements one by one in detail:

Statement (I): The mean free path of gas molecules is inversely proportional to the square of molecular diameter.

The mean free path ($ \lambda $) of gas molecules is the average distance a molecule travels before colliding with another molecule. The formula for the mean free path in terms of molecular diameter ($ d $) is given by:

$\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$

Here, $ k_B $ is the Boltzmann constant, $ T $ is the temperature, and $ P $ is the pressure. From this equation, it is evident that the mean free path ($ \lambda $) is inversely proportional to the square of the molecular diameter ($ d^2 $). Hence, Statement I is true.

Statement (II): Average kinetic energy of gas molecules is directly proportional to absolute temperature of gas.

The average kinetic energy ($ E_{\text{avg}} $) of gas molecules is given by:

$E_{\text{avg}} = \frac{3}{2} k_B T$

where $ k_B $ is the Boltzmann constant and $ T $ is the absolute temperature. This shows that the average kinetic energy is directly proportional to the absolute temperature of the gas. Hence, Statement II is true.

Considering the analysis above, the correct answer is:

Option B
Both Statement I and Statement II are true

To find the ratio of specific heats at constant volume ($C_V$) of the gases, we need to understand the degrees of freedom each type of gas molecule has, as this determines their specific heat capacity at constant volume. Degrees of freedom refer to the number of independent ways in which a molecule can store energy.

A monoatomic gas molecule has 3 translational degrees of freedom. This is because it can move in three dimensions: x, y, and z. A diatomic gas, if we assume it to be rigid for this context, has 5 degrees of freedom: 3 translational like the monoatomic gas and 2 rotational since it can rotate around two axes perpendicular to the bond axis connecting the two atoms. Vibrational modes are not considered at room temperature for a rigid diatomic gas, as these require higher energy to become accessible.

The specific heat capacity at constant volume for a monoatomic gas is given by:

$C_{V, mono} = \frac{3}{2} R$

And for a diatomic gas, it's:

$C_{V, diatomic} = \frac{5}{2} R$

Where $R$ is the ideal gas constant.

To find the ratio of their specific heats at constant volume, we divide the specific heat of the monoatomic gas by that of the diatomic gas:

$\frac{C_{V, mono}}{C_{V, diatomic}} = \frac{\frac{3}{2} R}{\frac{5}{2} R} = \frac{3}{2} \times \frac{2}{5} = \frac{3}{5}$

Thus, the correct ratio of the specific heats at constant volume of the monoatomic to diatomic (rigid) gases is given by Option B: $\frac{3}{5}$.

$\begin{aligned} \text { (1) } P_a V_a & =P_b V_b \quad \text{.... (i)}\\ P_c V_c & =P_d V_d \quad \text{.... (ii)} \end{aligned}$

$\begin{aligned} \text { (2) } P_a V_a^{\gamma-1} & =P_d V_d^{\gamma-1} \quad \text{.... (iii)}\\ P_b V_b^{\gamma-1} & =P_c V_c^{\gamma-1} \quad \text{.... (iv)} \end{aligned}$

$\begin{aligned} & \text { (i) } \div \text { (iii) } \Rightarrow \frac{V_a}{V_a^{\gamma-1}}=\frac{P_b V_b}{P_d V_d{ }^{\gamma-1}} \text { (v) } \\ & \text { (ii) } \div \text { (iv) } \Rightarrow \frac{V_c}{V_c^{\gamma-1}}=\frac{P_d V_d}{P_b V_b{ }^{\gamma-1}} \text { (vi) } \\ & \text { (v) } \times \text { (vi) } \Rightarrow \frac{V_a}{V_a^{\gamma-1}} \frac{V_c}{V_c{ }^{\gamma-1}}=\frac{P_b V_b}{P_d V_d^{\gamma-1}} \times \frac{P_d V_d}{P_b V_b^{\gamma-1}} \\ & \Rightarrow V_a^{\gamma-2} V_c{ }^{\gamma-2}=V_d^{\gamma-1} V_b{ }^{\gamma-2} \\ & \Rightarrow \frac{V_a}{V_d}=\frac{V_b}{V_c} \end{aligned}$

To evaluate the veracity of the given statements, we need to understand the physical quantities involved and their dimensional formulas. Specifically, we're looking at the dimensions of specific heat and gas constant.

Specific Heat:

Specific heat (c) is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin, since the increment is the same in both scales). Its formula is $q = mc\Delta T$, where $q$ is the heat added, $m$ is the mass, $c$ is the specific heat, and $\Delta T$ is the change in temperature.

From the formula, we can deduce the dimensions of specific heat as follows:

$[q] = [m][c][\Delta T]$

Knowing that the dimension of heat (q) is equivalent to energy, which is $[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]$, the mass (m) is $[\mathrm{M}]$, and temperature ($\Delta T$) is $[\mathrm{K}]$, we can solve for $[c]$:

$[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}] = [\mathrm{M}][c][\mathrm{K}]$

So, $[c] = [\mathrm{L}^2 \mathrm{T}^{-2} \mathrm{K}^{-1}]$

This reveals that Statement (I) is correct.

Gas Constant:

The gas constant (R) appears in the ideal gas law, represented as $PV = nRT$, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature. The dimensions of the gas constant can be derived from this relation.

Pressure (P) has dimensions $[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}]$, volume (V) has dimensions $[\mathrm{L}^3]$, and temperature (T) has dimensions $[\mathrm{K}]$.

Therefore, $[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}][\mathrm{L}^3] = [n][R][\mathrm{K}]$

Considering that the mole (n) is a dimensionless quantity, we can deduce the dimensions of R as:

$[R] = \frac{[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]}{[\mathrm{K}]} = [\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2} \mathrm{K}^{-1}]$

This indicates that Statement (II) has stated the dimensions incorrectly, presenting them as $[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-1} \mathrm{K}^{-1}]$ when it should be $[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2} \mathrm{K}^{-1}]$, making Statement (II) incorrect.

Based on the analysis:

Option A, "Statement (I) is incorrect but statement (II) is correct," is wrong because Statement (I) is correct.

Option B, "Both statement (I) and statement (II) are incorrect," is wrong because Statement (I) is correct.

Option C, "Both statement (I) and statement (II) are correct," is wrong because Statement (II) is incorrect.

Option D, "Statement (I) is correct but statement (II) is incorrect," is the correct choice, reflecting the true nature of the statements provided.

The energy of a diatomic molecule depends on the degrees of freedom it has. For a non-rigid diatomic molecule, there are more degrees of freedom compared to a rigid diatomic molecule. Specifically, a non-rigid diatomic molecule has translational, rotational, and vibrational degrees of freedom. The translational and rotational degrees of freedom are the same for both rigid and non-rigid diatomic molecules, which include:

• 3 translational degrees of freedom,
• 2 rotational degrees of freedom (since linear molecules do not have the third rotational freedom because it requires rotating around the bond axis, which doesn't change the energy for linear molecules).

Additionally, non-rigid diatomic molecules have vibrational degrees of freedom. For a simple diatomic molecule, there is 1 vibrational degree of freedom (since vibration along the bond axis is possible). However, considering the energy distribution across these vibrations requires accounting for both the potential and kinetic energy associated with vibrations, effectively doubling the vibrational degrees of freedom for energy calculations to 2 (one for kinetic and one for potential energy).

Thus, the total degrees of freedom for a non-rigid diatomic molecule are:

• 3 (translational) + 2 (rotational) + 2 (vibrational) = 7 degrees of freedom.

The formula for the average energy of a molecule in terms of degrees of freedom at temperature T is given by:

$ E = \frac{f}{2}k_\mathrm{B}T $

where $f$ is the total number of degrees of freedom, $k_\mathrm{B}$ is the Boltzmann constant, and $T$ is the temperature.

Plugging in the values for a non-rigid diatomic molecule:

$ E = \frac{7}{2}k_\mathrm{B}T $

Therefore, the total energy for 10 such molecules would be:

$ 10 \times \frac{7}{2}k_\mathrm{B}T = 35k_\mathrm{B}T $

So, the correct answer is:

Option D: 35 $k_\mathrm{B}T$.

To solve this problem, we can use the first law of thermodynamics, which states:

$\Delta Q = \Delta U + W$

where:

• $\Delta Q$ is the heat transferred to the system, in this case, $48 \mathrm{~J}$.
• $\Delta U$ is the change in internal energy of the system.
• $W$ is the work done by the system.

For one mole of an ideal gas, the change in internal energy can be calculated using the equation:

$\Delta U = nC_{v}\Delta T$

where:

• $n$ is the number of moles, which is $1$ in this case.
• $C_{v}$ is the molar specific heat capacity at constant volume.
• $\Delta T$ is the change in temperature, in Kelvins.

Given that we are dealing with helium, a monatomic ideal gas, we know that:

$C_{v} = \frac{3}{2}R$

Substituting the given values, we get:

$\Delta U = (1) \frac{3}{2}(8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1})(2^{\circ} \mathrm{C})$

Since a change in temperature of $1^{\circ} \mathrm{C}$ is equivalent to a change in temperature of $1 \mathrm{~K}$, we can directly substitute $2^{\circ} \mathrm{C}$ as a $2 \mathrm{~K}$ change without needing to convert Celsius to Kelvin as both scales have the same magnitude of degree.

Therefore,

$\Delta U = 1 \times \frac{3}{2} \times 8.3 \times 2 = 24.9 \mathrm{~J}$

Now, substituting $\Delta U$ and $\Delta Q$ into the first law of thermodynamics:

$48 = 24.9 + W$

Solving for $W$ we find:

$W = 48 - 24.9 = 23.1 \mathrm{~J}$

Thus, the work done by the gas is $23.1 \mathrm{~J}$, which corresponds to Option A.

$\begin{aligned} & \because \quad P V^2=R T \\ & P(2 v d v)+V^2(d P)=R d T \end{aligned}$

at $P=$ const.

$P d v=\frac{R d T}{2 V} \quad \text{... (i)}$

Now, for $n=1$

$\begin{aligned} & d \theta=d v+d w \\ & C_P d T=C_v d T+P d v \quad \text{... (ii)} \end{aligned}$

from (i) and (ii)

$C_P=C_V+\frac{R}{2 V}$

$\begin{aligned} & v=\sqrt{\frac{3 R T}{M}} \\ & \Rightarrow \frac{v_{\mathrm{H}_{\mathrm{e}}}}{v_{\mathrm{O}_2}}=\sqrt{\frac{M_{\mathrm{o}_2}}{M_{\mathrm{H}_e}}}=\frac{2 \sqrt{2 }}{1} \end{aligned}$