Heat and Thermodynamics

$ \begin{aligned} &\text { Increase in internal energy, }\\ &\begin{aligned} \Delta U & =m s \Delta T \\ & =5 \times 4200(30-20) \\ & =210000 \mathrm{~J} \\ & =210 \times 10^3 \mathrm{~J}=210 \mathrm{~kJ} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Work done by Carnot engine }\\ &W=\left(Q_1-Q_2\right) \propto\left(T_1-T_2\right) \end{aligned} $

Now for engine $A$

$ W_1=\left(Q_1-Q_2\right) \propto(600-T) $

and for engine $B$

$ W_2=\left(Q_1-Q_2\right) \propto(T-400) $

When work done by both engine is same than

$ \begin{aligned} & 600-T=T-400 \\ & 2 T=1000 \Rightarrow T=500 \mathrm{~K} \end{aligned} $

At constant pressure, the heat supplied goes into both increasing the internal energy and doing work.

For diatomic gas, $C_p=\frac{7}{2} R$ and $C_V=\frac{5}{2} R$

$ \frac{(\Delta U)_V}{(\Delta U)_p}=\frac{{ }^n C_V \Delta T}{{ }^n C_p \Delta T}=\frac{C_V}{C_p}=\frac{\frac{5}{2}}{\frac{7}{2}}=\frac{5}{7} $

$ \begin{aligned} &\text { Energy per molecule, }\\ &\begin{aligned} E=\frac{f}{2} & k T=\frac{6}{2} \times 1.38 \times 10^{-23} \times(273+47) \\ & =3 \times 1.38 \times 10^{-23} \times 320 \\ & =13.28 \times 10^{-21} \mathrm{~J} \\ & =\frac{13.28 \times 10^{-21}}{1.6 \times 10^{-19}} \mathrm{eV} \\ & =830.6 \times 10^{-4} \mathrm{eV} \approx 828 \times 10^{-4} \mathrm{eV} \end{aligned} \end{aligned} $

$F: C=13: 5$

$ \Rightarrow \quad \frac{F}{C}=\frac{13}{5} \Rightarrow F=\frac{13}{5} C $

We know that,

$ \begin{aligned} & \frac{C}{5}=\frac{F-32}{9} \\ \Rightarrow & \frac{C}{5}=\frac{\frac{13}{5} C-32}{9} \\ \Rightarrow & \frac{9 C}{5}=\frac{13 C}{5}-32 \\ \Rightarrow & \frac{13 C}{5}-\frac{9 C}{5}=32 \\ \Rightarrow & \frac{4 C}{5}=32 \Rightarrow C=40^{\circ} \mathrm{C} \end{aligned} $

$ \begin{aligned} &\text { Efficiency of Carnot engine, }\\ &\eta=1-\frac{Q_2}{Q_1}=1-\frac{400}{600}=\frac{1}{3}\\ &\begin{aligned} & \text { but } \eta=1-\frac{T_2}{T_1} \\ & \Rightarrow \quad \frac{1}{3}=1-\frac{T_2}{(273+127)} \\ & \Rightarrow \quad \frac{1}{3}=1-\frac{T_2}{400} \\ & \Rightarrow \frac{T_2}{400}=1-\frac{1}{3}=\frac{2}{3} \\ & \therefore \quad T_2=\frac{2}{3} \times 400=266.7 \mathrm{~K} \end{aligned} \end{aligned} $

We know that in adiabatic expansion, there is no heat exchange. So, $\Delta Q = 0$.

From the first law of thermodynamics:

$ \Delta Q = \Delta U + W $ Since $\Delta Q = 0$, this means: $ 0 = \Delta U + W $ So, $ W = -\Delta U $

The change in internal energy for $n$ moles of a diatomic gas is: $ \Delta U = n C_V \Delta T $ where $C_V$ for a diatomic gas is $\frac{5}{2}R$.

Here, $n = 3$ moles and the temperature change $\Delta T = -50^\circ \mathrm{C}$ (or $-50$ in Kelvin, since the size of the degree is the same).

So, $ W = -\Delta U = -n C_V \Delta T = -3 \times \frac{5}{2}R \times (-50) $

Multiplying out: $ -3 \times \frac{5}{2} R \times (-50) = 7.5 R \times 50 = 375R $

The work done by the gas is $375R$.

The fundamental limitation to the coefficient of performance of a refrigerator is given by second law of thermodynamics.

$ \begin{aligned} &\text { As we know that, }\\ &\begin{aligned} \gamma & =1+\frac{2}{f} \\ \Rightarrow \quad \frac{2}{f} & =\gamma-1 \Rightarrow f=\frac{2}{\gamma-1} \end{aligned} \end{aligned} $

Given:

• Initial volume $ V_1 = 400 \text{ cc} $

• Final volume $ V_2 = 100 \text{ cc} $

• Initial pressure $ P_1 = p $

• The process is sudden (adiabatic)

Thus, for an adiabatic process:

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

Given:

$ \gamma = 1.5 $

Step 1: Write the equation for $ P_2 $

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

Step 2: Substitute the values

$ P_2 = p \left( \frac{400}{100} \right)^{1.5} = p (4)^{1.5} $

Step 3: Simplify

$ (4)^{1.5} = 4^{1} \times 4^{0.5} = 4 \times 2 = 8 $

$ P_2 = 8p $

✅ Final Answer:

$ \boxed{P_2 = 8p} $

Correct Option: (B) $8p$

$ \begin{aligned} &\text { Efficiency of Carnot's engine is }\\ &\begin{aligned} \eta_1 & =1-\frac{T_2}{T_1} \\ 0.6 & =1-\frac{T_2}{600} \\ \Rightarrow \quad T_2 & =240 \mathrm{~K} \\ \text { Again, } \eta_2 & =80 \%=0.8 \\ \therefore \quad \eta_2 & =1-\frac{T_2^{\prime}}{T_1^{\prime}} \\ 0.8 & =1-\frac{240}{T_1^{\prime}} \quad\left[\because T_2^{\prime}=T_2=240 \mathrm{~K}\right] \\ \Rightarrow T_1^{\prime}= & 1200 \mathrm{~K} \end{aligned} \end{aligned} $

Total internal energy of mixture of the gases $=$ Total internal energy of $\mathrm{O}_2+$ Total internal energy of Ar

$ \begin{aligned} & =2 \times \frac{5}{2} R T+4 \times \frac{3}{2} R T \\ & =11 R T \end{aligned} $

Given:

• Temperature $ T = 127^\circ \text{C} $

• Boltzmann constant $ k = 1.38 \times 10^{-23} \, \text{J/K} $

We know that:

$ \text{Average translational kinetic energy per molecule} = \frac{3}{2} k T $

Step 1: Convert temperature to kelvins

$ T = 127 + 273 = 400 \, \text{K} $

Step 2: Substitute the values

$ E = \frac{3}{2} k T = \frac{3}{2} (1.38 \times 10^{-23}) (400) $

Step 3: Simplify

$ E = 1.5 \times 1.38 \times 400 \times 10^{-23} $

$ E = 2.07 \times 400 \times 10^{-23} $

$ E = 8.28 \times 10^{-21} \, \text{J} $

✅ Final Answer:

$ \boxed{8.28 \times 10^{-21} \, \text{J}} $

Correct Option: C

Power consumed, $P=V I=220 \times 4$

$ =880 \mathrm{~W} $

Heat energy required

$ \begin{aligned} H & =m s \Delta T \\ & =1000 \times 1 \times(100-34) \times 4.2 \\ & =2.772 \times 10^5 \mathrm{~J} \end{aligned} $

$ \begin{aligned} & \therefore \text { Time taken }=\frac{H}{P}=\frac{2.772 \times 10^5}{880} \\ & \quad=315 \mathrm{~s} \\ & \quad=\frac{315}{60} \text { minutes } \\ & \quad=5.25 \text { minutes } \end{aligned} $

According to Zeroth law of thermodynamics, temperature is same for two bodies in thermal equilibrium.

Coefficient of performance of refrigerator

$ \begin{aligned} \mathrm{COP} & =\frac{T_c}{T_h-T_c} \\ \quad 5 & =\frac{(-13+273)}{T_h-(-13+273)} \\ \Rightarrow \quad 5 & =\frac{260}{T_h-260} \\ \Rightarrow 5 T_h & -1300=260 \end{aligned} $

$ \begin{aligned} \Rightarrow T_h & =\frac{1560}{5} \\ & =312^{\circ} \mathrm{K}=(312-273)^{\circ} \mathrm{C} \\ & =39^{\circ} \mathrm{C} \end{aligned} $

Process $I$ is isochoric because volume is constant.

In process II, no heat exchange takes place.

$\therefore$ Process is adiabatic.

In process III, temperature is constant. So it is isothermal process.

In process IV, pressure is constant. Thus, it is isobaric process.

Total initial internal energy

$ \begin{aligned} U^{\prime}=U_1 & +U_2=\frac{3}{2} n_1 R T_1+\frac{3}{2} n_2 R T_2 \\ & =\frac{3}{2} R\left(n_1 T_1+n_2 T_2\right) \\ & =\frac{3}{2} R(2 \times 300+4 \times 600) \\ & =4500 R \end{aligned} $

Total number of moles after mixing

$ \begin{aligned} n_{\text {total }} & =n_1+n_2=2+4=6 \\ \therefore \quad U_{\text {final }} & =\frac{3}{2} n_{\text {total }} R T_{\text {final }} \\ U_{\text {final }} & =\frac{3}{2} \times 6 \times R T_{\text {final }} \end{aligned} $

Since, $U^{\prime}=U_{\text {final }}$

$ \begin{aligned} & \Rightarrow 4500 R=\frac{3}{2} \times 6 \times R \times T_{\text {final }} \\ & T_{\text {final }}=500 \mathrm{~K}=(500-273)^{\circ} \mathrm{C}=227^{\circ} \mathrm{C} \end{aligned} $

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)

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

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

To begin with, we're told that during an adiabatic process, the pressure $P$ of a gas is directly proportional to the cube of its absolute temperature $T$, that is, $P \propto T^3$. From this, we can express the relation as $P = kT^3$, where $k$ is a constant.

In an adiabatic process, $PV^\gamma = \text{constant}$, where $\gamma = \frac{C_P}{C_V}$, $P$ is the pressure, $V$ is the volume, $C_P$ is the heat capacity at constant pressure, and $C_V$ is the heat capacity at constant volume.

Also, the ideal gas equation is $PV = nRT$, where $n$ is the amount of substance, $R$ is the ideal gas constant, and $T$ is the absolute temperature. This can be rearranged to express $P$ in terms of $T$ and $V$, resulting in $P = \frac{nRT}{V}$.

Given that $P = kT^3$, we can equate this to the expression we got from the ideal gas law: $\frac{nRT}{V} = kT^3$. Simplifying, we get $V = \frac{nR}{kT^2}$, which shows that $V$ is inversely proportional to $T^2$.

Now, let's recall the adiabatic condition $PV^\gamma = \text{constant}$. Substituting the proportionalities $P \propto T^3$ and $V \propto T^{-2}$ into this equation, it becomes $T^3(T^{-2})^\gamma = \text{constant}$, simplifying to $T^{3-2\gamma} = \text{constant}$.

For the expression $T^{3-2\gamma} = \text{constant}$ to hold true in an adiabatic process, where the only variable is temperature $T$, the exponent must equal to zero (as the quantity of $T$ to some power equals to a constant suggests that the change in $T$ does not alter the value of the expression). This implies $3 - 2\gamma = 0$, solving for $\gamma$:

$3 - 2\gamma = 0 \implies 2\gamma = 3 \implies \gamma = \frac{3}{2}$

Therefore, the ratio of $\frac{C_P}{C_V}$ for the gas is $\frac{3}{2}$, which matches with Option B $\frac{3}{2}$.

The mean free path $\lambda$ is the average distance covered by a molecule between two successive collisions. It is given by the formula:

$\lambda = \frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^2}$

Where:

• $n$ is the number density of the molecules, i.e., the number of molecules per unit volume,
• $d$ is the diameter of the molecules, and
• $\sqrt{2}$ arises from considering the relative motion of a pair of particles in a gas, taking into account that either particle can be moving towards the other, which effectively doubles the cross-sectional area through which they can collide.

This formula shows that the mean free path is inversely proportional to the number density $n$ of the molecules and the square of the diameter $d$ of the molecules. It also takes into account that collisions are more likely as the cross-sectional area of the molecules increases, or as the density of the molecules increases.

Therefore, the correct answer is:

Option A: $\frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^2}$

$\begin{aligned} \Delta Q & =\Delta U+w \\ & =\pi(140)^2 \times 10^3 \times 10^{-6} \\ & =61.6 \mathrm{~J} \end{aligned}$

The collision frequency ($Z$) of gas molecules is proportional to the square root of the absolute temperature ($T$) of the system. Mathematically, it can be represented as $Z \propto \sqrt{T}$. This implies that when the temperature changes, the collision frequency changes as well according to the formula:

$Z_1 = Z_0 \sqrt{\frac{T_1}{T_0}}$

where:

• $Z_1$ is the collision frequency at temperature $T_1$,
• $Z_0$ is the initial collision frequency at temperature $T_0$,
• $T_1$ is the final absolute temperature, and
• $T_0$ is the initial absolute temperature.

To solve the given problem, we first need to convert the provided temperatures from Celsius to Kelvin (since absolute temperature in Kelvin should be used):

• $T_0 = 27^{\circ}C = 27 + 273 = 300$ K
• $T_1 = 127^{\circ}C = 127 + 273 = 400$ K

Using the formula for collision frequency change and substituting the given values:

$Z_1 = Z_0 \sqrt{\frac{400}{300}} = Z_0 \sqrt{\frac{4}{3}}$

This can be simplified to:

$Z_1 = Z_0 \times \frac{2}{\sqrt{3}}$

Thus, the collision frequency of the system at $127^{\circ}C$ is $\frac{2}{\sqrt{3}}$ times the collision frequency at $27^{\circ}C$, making Option B the correct answer:

$\frac{2}{\sqrt{3}} \mathrm{Z}$

$\begin{aligned} & w=\frac{-n R}{\gamma-1}(\Delta T) \\ &=\frac{-R}{1 / 2}\left(\frac{T}{\sqrt{2}}-T\right) \\ &=2 R\left(\frac{\sqrt{2} T-T}{\sqrt{2}}\right) \\ &=R T(2-\sqrt{2}) \\ & \therefore \quad T V_\gamma^{-1}=\text { cons. } \\ & T V_\gamma^{-1}=T_f(2 V)^{\gamma-1} \\ & T_f=\frac{T}{\sqrt{2}} \end{aligned}$

For non-linear polyatomic molecules, both translational and rotational degree of freedom have same value and is equal to 3.

$\begin{aligned} & P M=\rho R T \\ & \Rightarrow \frac{P}{T}=\left(\frac{R}{m}\right) \rho=\text { slope } \end{aligned}$

So from given curve, $\rho_1>\rho_2>\rho_3$

The resistance of a platinum resistance thermometer varies linearly with temperature. The relation can be given by:

$R_t = R_0(1 + \alpha t)$

where:

• $R_t$ is the resistance at temperature $t$,


• $R_0$ is the resistance at 0°C (ice point),


• $\alpha$ is the temperature coefficient of resistance, and


• $t$ is the temperature in degrees Celsius.

In this question, we are given:

• Resistance at the ice point ($R_{0} = 8 \Omega$),


• Resistance at the steam point ($R_{100} = 10 \Omega$).

To find the temperature coefficient of resistance ($\alpha$), we use the resistance values at the ice and steam points:

$\alpha = \frac{R_{100} - R_0}{R_0 \times 100} = \frac{10\Omega - 8\Omega}{8\Omega \times 100} = \frac{2\Omega}{800\Omega} = \frac{1}{400} \text{ per } ^{\circ}C$

Now, to find the resistance $R_t$ at $400 ^{\circ}C$, we substitute $\alpha$, $R_0$, and $t = 400$ into the formula:

$R_t = R_0(1 + \alpha t) = 8\Omega(1 + \frac{1}{400} \times 400) = 8\Omega(1 + 1) = 8\Omega \times 2 = 16\Omega$

Hence, the resistance of the platinum wire at $400^{\circ}C$ is $16 \Omega$. The correct option is:

Option B: 16 $\Omega$

To find the increase in temperature on the Fahrenheit scale, we use the relationship between the Celsius and Fahrenheit temperature scales. The formula to convert Celsius to Fahrenheit is:

$F = \frac{9}{5}C + 32$

However, since we are interested in the increase in temperature, we can ignore the "+ 32" part of the formula, because this constant does not affect the change in temperature, only the absolute temperatures. Thus, to find the increase in temperature on the Fahrenheit scale, we can use:

$\Delta F = \frac{9}{5}\Delta C$

Given that the increase in temperature is $40^{\circ}C$, we can substitute this value into the equation:

$\Delta F = \frac{9}{5} \times 40^{\circ}C$

$\Delta F = 9 \times 8$

$\Delta F = 72^{\circ}F$

Therefore, the increase in temperature on the Fahrenheit scale is $72^{\circ}F$. So, the correct answer is:

Option C $72^{\circ} \mathrm{F}$

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To calculate the root mean square (rms) velocity of gas molecules, we can use the formula:

$ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} $

where:

• $ v_{\text{rms}} $ is the root mean square velocity of the gas molecules,
• $ k $ is the Boltzmann constant,
• $ T $ is the absolute temperature in Kelvin, and
• $ m $ is the mass of one molecule of the gas.

Since the temperature and pressure are the same for hydrogen and oxygen, we can ignore the constant and temperature parts of the equation because they will cancel out in the comparison between the two gases.

Now, we need to compare the mass of one molecule of hydrogen to that of one molecule of oxygen. The molecular mass of hydrogen (H₂) is approximately 2 g/mol, while the molecular mass of oxygen (O₂) is approximately 32 g/mol.

We know the rms speed of hydrogen is 2 km/s, so let's find the mass ratio and then determine the speed of oxygen molecules. Using the rms velocity formula and considering the ratio of masses:

$ \frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{m_{\text{O₂}}}{m_{\text{H₂}}}} $

We know the mass m is proportional to the molecular weight for each gas, so we can substitute:

$ \frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{M_{\text{O₂}}}{M_{\text{H₂}}}} $

Now we plug in the values:

$ \frac{2 \text{ km/s}}{v_{\text{rms, O₂}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 $

Solving for $ v_{\text{rms, O₂}} $:

$ v_{\text{rms, O₂}} = \frac{2 \text{ km/s}}{4} = 0.5 \text{ km/s} $

Therefore, the rms velocity of oxygen molecules at the same temperature and pressure conditions as that of hydrogen with a rms velocity of 2 km/s is 0.5 km/s.

The correct answer is Option D : 0.5.

To calculate the molar specific heat at constant volume of a gas mixture, we can use a weighted average based on the molar specific heats of the individual gases and their respective amounts (moles).

Let's call $C_{V,m}$ the molar specific heat at constant volume of the mixture, $n_1$ the number of moles of the monoatomic gas, $n_2$ the number of moles of the diatomic gas, $C_{V,m1}$ the molar specific heat at constant volume of the monoatomic gas, and $C_{V,m2}$ the molar specific heat at constant volume of the diatomic gas.

For a monoatomic ideal gas, the molar specific heat at constant volume is:

$ C_{V,m1} = \frac{3}{2}R $

For a diatomic ideal gas, if we assume the gas is rigid and does not exhibit vibrational modes, the molar specific heat at constant volume is:

$ C_{V,m2} = \frac{5}{2}R $

The weighted average of the molar specific heat for the mixture is:

$ C_{V,m} = \frac{(n_1 \cdot C_{V,m1} + n_2 \cdot C_{V,m2})}{n_1 + n_2} $

Putting the values into the equation, we get:

$ C_{V,m} = \frac{(2 \cdot \frac{3}{2}R + 6 \cdot \frac{5}{2}R)}{2 + 6} $

$ C_{V,m} = \frac{(3R + 15R)}{8} $

$ C_{V,m} = \frac{18R}{8} $

$ C_{V,m} = \frac{9}{4}R $

Therefore, the molar specific heat of the mixture at constant volume is $\frac{9}{4}R$. The correct answer is:

Option D

$ \frac{9}{4}R $

To find the work done by the gas when it goes from state A to state B, we can look at the definition of work done on or by a gas in a thermodynamic process. For a quasi-static process, the work done $ W $ is given by the integral of the pressure $ P $ with respect to the volume $ V $:

$ W = \int_{V_1}^{V_2} P dV $

Given the relationship $ PV^{\frac{3}{2}} = K $, we can solve for $ P $:

$ P = \frac{K}{V^{\frac{3}{2}}} $

Now, substitute $ P $ into the work integral and evaluate it:

$ W = \int_{V_1}^{V_2} \frac{K}{V^{\frac{3}{2}}} dV = K \int_{V_1}^{V_2} V^{-\frac{3}{2}} dV $

To integrate this, we'll use the power rule for integration. The integral of $ V^{-\frac{3}{2}} $ is $ -2V^{-\frac{1}{2}} $, so the work done is:

$ W = K \left[-2V^{-\frac{1}{2}}\right]_{V_1}^{V_2} = K \left(-2V_2^{-\frac{1}{2}} + 2V_1^{-\frac{1}{2}} \right) $

We can rewrite $ V^{-\frac{1}{2}} $ as $ \frac{1}{\sqrt{V}} $:

$ W = K \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right) $

Since $ P_2V_2^{\frac{3}{2}} = K $ and $ P_1V_1^{\frac{3}{2}} = K $, we can express $ K $ in terms of $ P_1 $ and $ V_1 $ (or $ P_2 $ and $ V_2 $, but we’ll use $ P_1 $ and $ V_1 $ for now):

$ W = P_1V_1^{\frac{3}{2}} \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right) $

$ W = -2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_2}} + 2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_1}} $

$ W = -2P_1V_1\frac{\sqrt{V_1}}{\sqrt{V_2}} + 2P_1V_1 $

Since $ \frac{\sqrt{V_1}}{\sqrt{V_2}} = \sqrt{\frac{V_1}{V_2}} $, and recalling $ P_2V_2^{\frac{3}{2}} = K $ once more:

$ W = -2P_1V_1\sqrt{\frac{V_1}{V_2}} + 2P_1V_1 $

$ W = -2 \frac{P_1V_1^{\frac{3}{2}}}{\sqrt{V_2}} + 2P_1V_1 $

Putting $ P_1V_1^{\frac{3}{2}} $ back as $ K $:

$ W = -2\frac{K}{\sqrt{V_2}} + 2P_1V_1 $

$ W = -2P_2V_2 \sqrt{V_2}\frac{1}{\sqrt{V_2}} + 2P_1V_1 $

$ W = -2P_2V_2 + 2P_1V_1 $

So the work done is:

$ W = 2P_1V_1 - 2P_2V_2 $

Therefore, the correct answer matching the given options is:

Option D: $ 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right) $

Shortcut :

The speed of sound in a gas at standard temperature and pressure (STP) can be calculated using the following formula derived from the ideal gas law and the speed of sound relation in a gas:

$ v = \sqrt{\gamma \frac{R T}{M}} $

Where:

• $ v $ = speed of sound in the gas
• $ \gamma $ = adiabatic index (ratio of specific heats, $ C_p/C_v $)
• $ R $ = universal gas constant
• $ T $ = temperature in Kelvin
• $ M $ = molar mass of the gas in kilograms per mole (kg/mol)

Given:

• $ R = 8.3 \, J \cdot mol^{-1} \cdot K^{-1} $


• $ \gamma = 1.4 $ (For diatomic gases such as oxygen)


• $ T = 273.15 \, K $ (Standard temperature, 0°C in Kelvin)


• Molar mass of Oxygen ($ O_2 $) $ M = 32 \times 10^{-3} \, kg/mol $ (32 g/mol converted to kg/mol)

Plugging these values into the formula:

$ v = \sqrt{1.4 \times \frac{8.3 \times 273.15}{32 \times 10^{-3}}} $

Calculating the values inside the square root:

$ v = \sqrt{1.4 \times \frac{2268.745}{0.032}} $

$ v = \sqrt{1.4 \times 70896.40625} $

$ v = \sqrt{99304.96875} $

$ v \approx 315 \, m/s $

To determine the total internal energy of the gas mixture, we adhere to the equipartition theorem, which dictates that each degree of freedom contributes $\frac{1}{2} RT$ to the internal energy per mole, where $R$ is the gas constant and $T$ is the temperature.

An argon atom, being a noble gas, is monoatomic, with 3 translational degrees of freedom. Since we neglect all vibrational modes, and monoatomic gases have no rotational or vibrational degrees of freedom that contribute to energy at our specified temperature, each mole of argon has $3 \times \frac{1}{2} RT = \frac{3}{2} RT$ of energy.

Oxygen, on the other hand, is a diatomic molecule. This implies that under normal conditions, it has 3 translational and 2 rotational degrees of freedom. So, each mole of oxygen has $5 \times \frac{1}{2} RT = \frac{5}{2} RT$ of energy as we are neglecting vibrational modes which typically become relevant only at higher temperatures.

Hence, the total internal energy ($U$) of the gas mixture is:

$ U = (\text{Energy per mole of Ar} \times \text{Number of moles of Ar}) + (\text{Energy per mole of O}_{2} \times \text{Number of moles of O}_{2}) $

$ U = \left(\frac{3}{2} RT \times 8\right) + \left(\frac{5}{2} RT \times 6\right) $

$ U = \left(12 RT + 15 RT\right) $

$ U = 27 RT $

Thus, the total internal energy of the system is $27 RT$.

The two isobaric processes depicted in the figure show changes in the volume of an ideal gas with temperature under constant pressure conditions.

Isobaric processes follow the equation $PV = nRT$, where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is temperature.

When we rearrange the equation in terms of V (Volume), we get $V = \left(\frac{nR}{P}\right)T$. The slope of the volume-temperature graph in such a scenario is given by $\frac{nR}{P}$. This implies that the slope is inversely proportional to the pressure (i.e., slope $\propto \frac{1}{P}$).

From this relationship, if one process has a higher slope compared to another, its corresponding pressure must be lower. Observing the given figure and applying this understanding, if Process 2 has a greater slope than Process 1, it indicates that $P_2 < P_1$.

$\mathrm{KE}=\frac{\mathrm{f}}{2} \mathrm{kT}$

$\begin{array}{ll} \mathrm{f}_1=3, & \mathrm{f}_2=5 \\ \mathrm{n}_1=3, & \mathrm{n}_2=2 \end{array}$

$\begin{aligned} & \mathrm{f}_{\text {mixture }}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}=\frac{9+10}{\mathrm{f}}=\frac{19}{5} \\ & \gamma_{\text {mixture }}=1+\frac{2 \times 5}{19}=\frac{29}{19}=1.52 \end{aligned}$