Chapter 21, Problem 21.51AP

A certain ideal gas has a molar specific heat of C_v = $\frac{7}{2}$R. A 2.00-mol sample of the gas always starts at pressure 1.00 × 10⁵ Pa and temperature 300 K. For each of the following processes, determine (a) the final pressure, (b) the final volume, (c) the final temperature, (d) the change in internal energy of the gas, (e) the energy added to the gas by heat, and (f) the work done on the gas. (i) The gas is heated at constant pressure to 400 K. (ii) The gas is heated at constant volume to 400 K. (iii) The gas is compressed at constant temperature to 1.20 × 10⁵ Pa. (iv) The gas is compressed adiabatically to 1.20 × 10⁵ Pa.

To determine

 The final pressure, final volume, final temperature, change in the internal energy, the energy added to the gas by heat and the work done on the gas.

Answer to Problem 21.51AP

 The final pressure is $100\text{kPa}$ , final volume is $66.5\text{L}$, final temperature is $400\text{K}$, change in the internal energy is $\text{5}\text{.82}\text{kJ}$ , the energy added to the gas by heat is $7.48\text{kJ}$ and the work done on the gas is $-1.66\text{kJ}$.

Explanation of Solution

Given info: The molar specific heat of the gas at constant volume is $\frac{7}{2}R$, the amount of the gas is $2.00\text{mol}$, the initial pressure of the gas is $1.00\times {10}^5\text{Pa}$ and the initial temperature of the gas is $300\text{K}$. The final temperature of the gas is $400\text{K}$.

The value of universal gas constant is $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$.

The gas is heated at constant pressure.

Write the expression for molar specific heat of the gas at constant pressure.

$C_{\text{p}}=R+C_{\text{v}}$

Here,

$R$ is the universal gas constant .

$C_{\text{v}}$ is the molar specific heat of the gas at constant volume.

Substitute $\frac{7}{2}R$ for $C_{\text{v}}$ in the above expression.

$\begin{array}{c}{C}_{\text{p}}=R+\frac{7}{2}R\\ =\frac{9}{2}R\end{array}$

The gas is heated at constant pressure, therefore the final pressure of the gas is equal to the initial pressure of the gas.

The expression for the final pressure of the gas is,

$P_2=P_1$

Here,

$P_1$ is the initial pressure of the gas.

$P_2$ is the final pressure of the gas.

Substitute $1.00\times {10}^5\text{Pa}$ for $P_1$ in the above expression.

$\begin{array}{c}{P}_2=1.00\times {10}^5\text{Pa}\\ =100\text{kPa}\end{array}$

The final pressure of the gas is $100\text{kPa}$.

The final temperature is the temperature to which the gas is heated.

The value of final temperature of the gas is,

$T_2=400\text{K}$

Write the expression for the change in the temperature of the gas.

$\Delta T=T_2-T_1$

Here,

$T_2$ is the final temperature.

$T_1$ is the initial temperature

Substitute $300\text{K}$ for $T_1$ and $400\text{K}$ for $T_2$ in the above expression.

$\begin{array}{c}\Delta T=400\text{K}-100\text{K}\\ \text{=100}\text{K}\end{array}$

The ideal gas equation is,

$PV=nRT$

Here,

$P$ is the pressure.

$V$ is the volume.

$n$ is the number of moles.

$T$ is the temperature.

Rearrange the above equation for the value of $V$.

$V=\frac{nRT}{P}$

Substitute $V_2$ for $V$, $P_2$ for $P$ and $T_2$ for $T$ in the above expression.

$V_2=\frac{nR{T}_2}{P_2}$

Substitute $1.00\times {10}^5\text{Pa}$ for $P_2$ $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $400\text{K}$ for $T_2$ in the above expression.

$\begin{array}{c}{V}_2=\frac{\left(2.00\text{mol}\right)\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(400\text{K}\right)}{\left(1.00\times {10}^5\text{Pa}\right)}\\ =66.5\text{L}\end{array}$

Thus, the final volume of the gas is $66.5\text{L}$.

The expression for the change in the internal energy of the gas is,

$\Delta E=n{C}_{\text{v}}\Delta T$

Substitute $\frac{7}{2}R$ for $C_{\text{v}}$ in the above expression.

$\Delta E=n\frac{7}{2}R\Delta T$

Substitute $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $100\text{K}$ for $\Delta T$ in the above expression.

$\begin{array}{c}\Delta E=\left(2.00\text{mol}\right)\frac{7}{2}\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(100\text{K}\right)\\ =5.82\times {10}^3\text{J}\\ =\text{5}\text{.82}\text{kJ}\end{array}$

Thus the change in the internal energy of the gas is $\text{5}\text{.82}\text{kJ}$.

The expression for the energy added to the gas by heat is,

$Q=n{C}_{\text{p}}\Delta T$

Substitute $\frac{9}{2}R$ for $C_{\text{p}}$ in the above expression.

$Q=n\frac{9}{2}R\Delta T$

Substitute $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $100\text{K}$ for $\Delta T$ in the above expression.

$\begin{array}{c}Q=\left(2.00\text{mol}\right)\frac{9}{2}\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(100\text{K}\right)\\ =7.48\times {10}^3\text{J}\times \frac{{10}^3\text{kJ}}{1\text{J}}\\ =7.48\text{kJ}\end{array}$

Thus the energy added to the gas by heat is $7.48\text{kJ}$.

The formula for the work done on the gas is,

$W=\Delta E-Q$

Substitute $\text{5}\text{.82}\text{kJ}$ for $\Delta E$ and $7.48\text{kJ}$ for $Q$ in the above expression.

$\begin{array}{c}W=\text{5}\text{.82}\text{kJ}-7.48\text{kJ}\\ =-1.66\text{kJ}\end{array}$

Thus the work done on the gas is $-1.66\text{kJ}$.

Conclusion:

Therefore, for the gas, the final pressure is $1.00\times {10}^5\text{Pa}$ , final volume is $66.5\text{L}$, final temperature is $400\text{K}$, change in the internal energy is $\text{5}\text{.82}\text{kJ}$ , the energy added to the gas by heat is $7.48\text{kJ}$ and the work done on the gas is $-1.66\text{kJ}$.

To determine

 The final pressure, final volume, final temperature, change in the internal energy, the energy added to the gas by heat and the work done on the gas.

Answer to Problem 21.51AP

 For the gas, the final pressure is $133\text{kPa}$ , final volume is $49.9\text{L}$, final temperature is $400\text{K}$, change in the internal energy is $\text{5}\text{.82}\text{kJ}$ , the energy added to the gas by heat is $\text{5}\text{.82}\text{kJ}$ and the work done on the gas is $0$.

Explanation of Solution

Given info: The molar specific heat of the gas at constant volume is $\frac{7}{2}R$, the amount of the gas is $2.00\text{mol}$, the initial pressure of the gas is $1.00\times {10}^5\text{Pa}$ and the initial temperature of the gas is $300\text{K}$. The final temperature of the gas is $400\text{K}$.

The value of universal gas constant is $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$.

The gas is heated at constant volume.

The gas is heated at constant volume, therefore the final volume of the gas is equal to the initial pressure of the gas.

The formula or ideal gas is,

$PV=nRT$

Rearrange the above equation for the value of $V$.

$V=\frac{nRT}{P}$

Substitute $V_1$ for $V$, $P_1$ for $P$ and $T_1$ for $T$ in the above expression.

$V_1=\frac{nR{T}_1}{P_1}$

Write the expression for the final volume of the gas.

$V_2=V_1$

Substitute $\frac{nR{T}_1}{P_1}$ for $V_1$ in the above expression.

$V_2=\frac{nR{T}_1}{P_1}$

Substitute $1.00\times {10}^5\text{Pa}$ for $P_1$ $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $300\text{K}$ for $T_1$ in the above expression.

$\begin{array}{c}{V}_2=\frac{\left(2.00\text{mol}\right)\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(300\text{K}\right)}{\left(1.00\times {10}^5\text{Pa}\right)}\\ =49.87\text{L}\\ \approx 49.9\text{L}\end{array}$

Thus, the final volume of the gas is $49.9\text{L}$.

The final temperature is the temperature to which the gas is heated.

The value of final temperature of the gas is,

$T_2=400\text{K}$

Write the expression for the change in the temperature of the gas.

$\Delta T=T_2-T_1$

Here,

$T_2$ is the final temperature.

$T_1$ is the initial temperature

Substitute $300\text{K}$ for $T_1$ and $400\text{K}$ for $T_2$ in the above expression.

$\begin{array}{c}\Delta T=400\text{K}-100\text{K}\\ \text{=100}\text{K}\end{array}$

Write the expression for ideal gas.

$PV=nRT$

Rearrange the above equation for the value of $P$.

$P=\frac{nRT}{V}$

Substitute $V_1$ for $V$, $P_1$ for $P$ and $T_1$ for $T$ in the above expression.

$P_1=\frac{nR{T}_1}{V_1}$ (1)

Substitute $V_2$ for $V$, $P_2$ for $P$ and $T_2$ for $T$ in the rearranged expression of the ideal gas for value of $P$

$P_2=\frac{nR{T}_2}{V_2}$ (2)

Divide equation (2) from equation (1).

$\begin{array}{c}\frac{P_2}{P_1}=\frac{\frac{nR{T}_2}{V_2}}{\frac{nR{T}_1}{V_1}}\\ \frac{P_2}{P_1}=\frac{T_2}{T_1}\\ P_2=\frac{P_1{T}_2}{T_1}\end{array}$

Substitute $1.00\times {10}^5\text{Pa}$ for $P_1$, $300\text{K}$ for $T_1$ and $400\text{K}$ for $T_2$ in the above expression.

$\begin{array}{c}{P}_2=\frac{\left(1.00\times {10}^5\text{Pa}\right)\left(400\text{K}\right)}{300\text{K}}\\ =133.33\text{kPa}\\ \approx 133\text{kPa}\end{array}$

The expression for the change in the internal energy of the gas is,

$\Delta E=n{C}_{\text{v}}\Delta T$

Substitute $\frac{7}{2}R$ for $C_{\text{v}}$ in the above expression.

$\Delta E=n\frac{7}{2}R\Delta T$

Substitute $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $100\text{K}$ for $\Delta T$ in the above expression.

$\begin{array}{c}\Delta E=\left(2.00\text{mol}\right)\frac{7}{2}\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(100\text{K}\right)\\ =5.82\times {10}^3\text{J}\times \frac{{10}^{-3}\text{J}}{1\text{kJ}}\\ =\text{5}\text{.82}\text{kJ}\end{array}$

Thus the change in the internal energy of the gas is $\text{5}\text{.82}\text{kJ}$.

The formula for the energy added to the gas by heat for constant volume is,

$Q=n{C}_{\text{v}}\Delta T$

Substitute $\frac{7}{2}R$ for $C_{\text{v}}$ in the above expression.

$Q=n\frac{7}{2}R\Delta T$

Substitute $\Delta E$ for $n\frac{7}{2}R\Delta T$ in the above expression.

$Q=\Delta E$

Substitute $\text{5}\text{.82}\text{kJ}$ for $\Delta E$ in the above expression.

$Q=\text{5}\text{.82}\text{kJ}$

Thus the energy added to the gas by heat is $\text{5}\text{.82}\text{kJ}$.

The forrmula for the work done on the gas is,

$W=\Delta E-Q$

Substitute $\text{5}\text{.82}\text{kJ}$ for $\Delta E$ and $\text{5}\text{.82}\text{kJ}$ for $Q$ in the above expression.

$\begin{array}{c}W=\text{5}\text{.82}\text{kJ}-\text{5}\text{.82}\text{kJ}\\ =\text{0}\end{array}$

Thus the work done on the gas is $0$.

Conclusion:

Therefore, for the gas, the final pressure is $133\text{kPa}$ , final volume is $49.9\text{L}$, final temperature is $400\text{K}$, change in the internal energy is $\text{5}\text{.82}\text{kJ}$ , the energy added to the gas by heat is $\text{5}\text{.82}\text{kJ}$ and the work done on the gas is $0$.

To determine

 The final pressure, final volume, final temperature, change in the internal energy, the energy added to the gas by heat and the work done on the gas.

Answer to Problem 21.51AP

 For the gas, the final pressure is $120\text{kPa}$ , final volume is $41.6\text{L}$, final temperature is $300\text{K}$, change in the internal energy is $\text{0}$ , the energy added to the gas by heat is $-909\text{J}$ and the work done on the gas is $+909\text{J}$.

Explanation of Solution

Given info: The molar specific heat of the gas at constant volume is $\frac{7}{2}R$, the amount of the gas is $2.00\text{mol}$, the initial pressure of the gas is $1.00\times {10}^5\text{Pa}$ and the initial temperature of the gas is $300\text{K}$. The final pressure of the gas is $1.20\times {10}^5\text{Pa}$.

The value of universal gas constant is $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$.

The gas is compressed at constant temperature.

The gas is heated at constant volume, therefore the final volume of the gas is equal to the initial pressure of the gas.

The formula for ideal gas is,

$PV=nRT$

Rearrange the above equation for the value of $V$.

$V=\frac{nRT}{P}$

Substitute $V_1$ for $V$, $P_1$ for $P$ and $T_1$ for $T$ in the above expression.

$V_1=\frac{nR{T}_1}{P_1}$

Substitute $1.00\times {10}^5\text{Pa}$ for $P_1$ $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $300\text{K}$ for $T_1$ in the above expression.

$\begin{array}{c}{V}_1=\frac{\left(2.00\text{mol}\right)\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(300\text{K}\right)}{\left(1.00\times {10}^5\text{Pa}\right)}\\ =49.87\text{L}\\ \approx 49.9\text{L}\end{array}$

Thus, the initial volume of the gas is $49.9\text{L}$.

Substitute $V_2$ for $V$, $P_2$ for $P$ and $T_2$ for $T$ in the above expression.

$V_2=\frac{nR{T}_2}{P_2}$

Substitute $1.20\times {10}^5\text{Pa}$ for $P_2$ $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $300\text{K}$ for $T_2$ in the above expression.

$\begin{array}{c}{V}_2=\frac{\left(2.00\text{mol}\right)\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(300\text{K}\right)}{\left(1.20\times {10}^5\text{Pa}\right)}\\ =41.58\text{L}\\ \approx 41.6\text{L}\end{array}$

Thus, the final volume of the gas is $41.6\text{L}$.

The expression for the final temperature of the gas is,

$T_2=T_1$

Substitute $300\text{K}$ for $T_1$ in the above expression.

$T_2=300\text{K}$

The final temperature of the gas is $300\text{K}$.

The expression for the change in the temperature of the gas is,

$\Delta T=T_2-T_1$

Substitute $300\text{K}$ for $T_1$ and $300\text{K}$ for $T_2$ in the above expression.

$\begin{array}{c}\Delta T=300\text{K}-300\text{K}\\ \text{=0}\end{array}$

The final pressure of the gas is the pressure at which the gas is compressed.

The value of final; pressure of the gas is,

$\begin{array}{c}{P}_2=1.20\times {10}^5\text{Pa}\times \frac{1\text{kPa}}{{10}^3\text{Pa}}\\ =120\text{kPa}\end{array}$

The expression for the change in the internal energy of the gas is,

$\Delta E=n{C}_{\text{v}}\Delta T$

Substitute $\frac{7}{2}R$ for $C_{\text{v}}$ in the above expression.

$\Delta E=n\frac{7}{2}R\Delta T$

Substitute $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $0$ for $\Delta T$ in the above expression.

$\begin{array}{c}\Delta E=\left(2.00\text{mol}\right)\frac{7}{2}\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(0\right)\\ =0\end{array}$

Thus the change in the internal energy of the gas is $\text{0}$.

The expression for the work done on the gas is,

$W=-P{\displaystyle \underset{{\text{V}}_1}{\overset{{\text{V}}_2}{\int }}d\text{V}}$

Divide and multiply the above equation by $V$

$W=-PV{\displaystyle \underset{V_1}{\overset{V_2}{\int }}\frac{d\text{V}}{V}}$

Substitute $nrT$ for $PV$ in the above expression.

$\begin{array}{c}W=-nrT{\displaystyle \underset{V_1}{\overset{V_2}{\int }}\frac{dV}{V}}\\ =-nrT\ln \left(\frac{V_2}{V_1}\right)\end{array}$

Substitute $49.9\text{L}$ for ${\text{V}}_1$, $\text{41}\text{.6}\text{L}$ for ${\text{V}}_2$ and $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $300\text{K}$ for $T$ in the above expression.

$\begin{array}{c}W=-\left(2.00\text{mol}\right)\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(300\text{K}\right)\ln \left(\frac{\text{41}\text{.6}\text{L}}{49.9\text{L}}\right)\\ =907.5\text{J}\\ \approx 909\text{J}\end{array}$

Thus, the work done on the system is $909\text{J}$.

The expression for the energy added to the gas by heat for constant volume is,

$Q=\Delta E-W$

Substitute $909\text{J}$ for $W$ and $0$ for $\Delta E$ in the above expression.

$\begin{array}{c}Q=0-909\text{J}\\ =-909\text{J}\end{array}$

Thus the energy added to the gas by heat is $-909\text{J}$.

Conclusion:

Therefore, for the gas, the final pressure is $120\text{kPa}$ , final volume is $41.6\text{L}$, final temperature is $300\text{K}$, change in the internal energy is $\text{0}$ , the energy added to the gas by heat is $-909\text{J}$ and the work done on the gas is $+909\text{J}$.

To determine

 The final pressure, final volume, final temperature, change in the internal energy, the energy added to the gas by heat and the work done on the gas.

Answer to Problem 21.51AP

 For the gas, the final pressure is $120\text{kPa}$ , final volume is $43.3\text{L}$, final temperature is $312\text{K}$, change in the internal energy is $\text{+722}\text{J}$ , the energy added to the gas by heat is $0$ and the work done on the gas is $\text{+722}\text{J}$.

Explanation of Solution

Given info: The molar specific heat of the gas at constant volume is $\frac{7}{2}R$, the amount of the gas is $2.00\text{mol}$, the initial pressure of the gas is $1.00\times {10}^5\text{Pa}$ and the initial temperature of the gas is $300\text{K}$. The final pressure of the gas is $1.20\times {10}^5\text{Pa}$.

The value of universal gas constant is $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$.

The gas is compressed adiabatically to the final pressure.

The value of the final pressure is,

$\begin{array}{c}{P}_2=1.20\times {10}^5\text{Pa}\times \frac{1\text{kPa}}{{10}^3\text{Pa}}\\ =120\text{kPa}\end{array}$

Thus, the value of the final pressure is $120\text{kPa}$.

The expression for the ratio of the specific heats is,

$\gamma =\frac{C_{\text{P}}}{C_{\text{V}}}$

Substitute $\frac{7}{2}R$ for $C_{\text{V}}$ and $\frac{9}{2}R$ for $C_{\text{P}}$ in the above expression.

$\begin{array}{c}\gamma =\frac{\frac{9}{2}R}{\frac{7}{2}R}\\ =\frac{9}{7}\end{array}$

The expression for an adiabatic process for the initial condition of gas is,

$P_1{V}_1^{\gamma }=\text{k}$ (3)

Here,

$\text{k}$ is constant value.

The expression for an adiabatic process for the final condition of gas is,

$P_2{V}_2^{\gamma }=\text{k}$ (4)

Divide equation (4) by equation (3).

$\frac{P_2{V}_2^{\gamma }}{P_1{V}_1^{\gamma }}=\frac{\text{k}}{\text{k}}$

Rearrange the above equation for the value of $V_2$.

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

Substitute $49.9\text{L}$ for $V_1$, $1.00\times {10}^5\text{Pa}$ for $P_1$, $1.20\times {10}^5\text{Pa}$ for $P_2$ and $\frac{9}{7}$ for $\gamma$ in the above expression.

$\begin{array}{c}{V}_2=\left(49.9\text{L}\right){\left(\frac{1.00\times {10}^5\text{Pa}}{1.20\times {10}^5\text{Pa}}\right)}^{\frac{1}{\frac{9}{7}}}\\ =43.3\text{L}\end{array}$

Thus, the final value of the volume is $43.3\text{L}$.

The expression for ideal gas is,

$PV=nRT$

Rearrange the above equation .

$\frac{VP}{T}=nR$ (5)

Substitute $V_1$ for $V$, $P_1$ for $P$ and $T_1$ for $T$ in the above expression.

$\frac{P_1{V}_1}{T_1}=nR$ (6)

Substitute $V_2$ for $V$, $P_2$ for $P$ and $T_2$ for $T$ in the equation (5).

$\frac{P_2{V}_2}{T_2}=nR$ (7)

Divide equation (7) by equation (6).

$\frac{\frac{P_2{V}_2}{T_2}}{\frac{P_1{V}_1}{T_1}}=\frac{nR}{nR}$

Rearrange the above expression for the value of $T_2$.

$T_2=T_1\left(\frac{P_2{V}_2}{P_1{V}_1}\right)$

Substitute $1.00\times {10}^5\text{Pa}$ for $P_1$, $1.20\times {10}^5\text{Pa}$ for $P_2$ $49.9\text{L}$ for ${\text{V}}_1$, $\text{43}\text{.3}\text{L}$ for ${\text{V}}_2$ and $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $300\text{K}$ for $T_1$ in the above expression.

$\begin{array}{c}{T}_2=\left(300\text{K}\right)\left(\frac{\left(1.20\times {10}^5\text{Pa}\right)\left(43.3\text{L}\right)}{\left(1.00\times {10}^5\text{Pa}\right)\left(49.9\text{L}\right)}\right)\\ =312.38\text{K}\\ \approx 312\text{K}\end{array}$

The expression for the change in the temperature of the gas is,

$\Delta T=T_2-T_1$

Here,

$T_2$ is the final temperature.

$T_1$ is the initial temperature

Substitute $300\text{K}$ for $T_1$ and $312.4\text{K}$ for $T_2$ in the above expression.

$\begin{array}{c}\Delta T=312.4\text{K}-100\text{K}\\ \text{=12}\text{.4}\text{K}\end{array}$

Thus the change in the temperature is $12\text{K}$.

The expression for the change in the internal energy of the gas is,

$\Delta E=n{C}_{\text{v}}\Delta T$

Substitute $\frac{7}{2}R$ for $C_{\text{v}}$ in the above expression.

$\Delta E=n\frac{7}{2}R\Delta T$

Substitute $2.00\text{mol}$ for $n$, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $12.4\text{K}$ for $\Delta T$ in the above expression.

$\begin{array}{c}\Delta E=\left(2.00\text{mol}\right)\frac{7}{2}\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(12.4\text{K}\right)\\ =721.7\text{J}\\ \approx \text{722}\text{J}\end{array}$

Thus the change in the internal energy of the gas is $\text{722}\text{J}$.

The adiabatic process is insulated to heat supplied externally.

The expression for the energy added to the gas by heat is,

$Q=0$

Thus the energy added to the gas by heat is $0$.

The expression for the work done on the gas is,

$W=\Delta E-Q$

Substitute $\text{722}\text{J}$ for $\Delta E$ and $0$ for $Q$ in the above expression.

$\begin{array}{c}W=\text{722}\text{J}-0\\ =\text{722}\text{J}\end{array}$

Thus the work done on the gas is $\text{722}\text{J}$.

Conclusion:

Therefore, for the gas, the final pressure is $120\text{kPa}$ , final volume is $43.3\text{L}$, final temperature is $312\text{K}$, change in the internal energy is $\text{+722}\text{J}$ , the energy added to the gas by heat is $0$ and the work done on the gas is $\text{+722}\text{J}$.