Chapter 16, Problem 121MP

Consider a sample containing 5.00 moles of a monatomic ideal gas that is taken from state A to state B by the following two pathways:

For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, ∆E and ∆H for each step in kJ, and calculate overall values for each pathway. Explain how the overall values for the two pathways illustrate that ∆E and ∆H are state functions, whereas q and w are path functions. Hint: In a more rigorous study of thermochemistry, it can be shown that for an ideal gas:

$\Delta E={\text{nC}}_{\text{v}}\Delta T\text{ and}$

$\Delta H={\text{nC}}_{\text{p}}\Delta T$

where C_v is the molar heat capacity at constant volume and C_p is the molar heat capacity at constant pressure. In addition, for a monatomic ideal gas, ${\text{C}}_{\text{v}}=\frac{3}{2}R$ and ${\text{C}}_{\text{p}}=\frac{5}{2}R$ , where R = 8.3145 J/K ·mol.

Interpretation Introduction

Interpretation: The values of $q,w,\Delta E$ and $\Delta H$ for each of the given steps in the two pathways in the given state change is to be calculated.

Concept Introduction: The internal energy change for the monoatomic ideal gas is calculated by using the formula,

$\Delta E=\frac{3}{2}nR\left(T_2-T_1\right)$

The enthalpy change of the monoatomic ideal gas is calculated by using the formula,

$\Delta H=\frac{5}{2}nR\left(T_2-T_1\right)$

To determine: The values of $q,w,\Delta E$ and $\Delta H$ for each of the given steps.

Answer to Problem 121MP

The values of $q,w,\Delta E$ and $\Delta H$ for each of the given steps has been rightfully stated.

Explanation of Solution

Given

The pathway one for the change in state from A to B is,

$\begin{array}{l}\begin{array}{l}{P}_{\text{A}}=3.00\text{atm}\hfill \\ V_{\text{A}}=15.0\text{L}\hfill \end{array}\stackrel{1}{\to }\begin{array}{l}{P}_{\text{C}}=3.00\text{atm}\hfill \\ V_{\text{C}}=55.0\text{L}\hfill \end{array}\\ \begin{array}{l}{P}_{\text{C}}=3.00\text{atm}\hfill \\ V_{\text{C}}=55.0\text{L}\hfill \end{array}\stackrel{2}{\to }\begin{array}{l}{P}_{\text{B}}=6.00\text{atm}\hfill \\ V_{\text{B}}=20.0\text{L}\hfill \end{array}\end{array}$

The pathway two for the change in state from A to B is,

$\begin{array}{l}\begin{array}{l}{P}_{\text{A}}=3.00\text{atm}\hfill \\ V_{\text{A}}=15.0\text{L}\hfill \end{array}\stackrel{3}{\to }\begin{array}{l}{P}_{\text{D}}=6.00\text{atm}\hfill \\ V_{\text{D}}=15.0\text{L}\hfill \end{array}\\ \begin{array}{l}{P}_{\text{D}}=6.00\text{atm}\hfill \\ V_{\text{D}}=15.0\text{L}\hfill \end{array}\stackrel{4}{\to }\begin{array}{l}{P}_{\text{B}}=6.00\text{atm}\hfill \\ V_{\text{B}}=20.0\text{L}\hfill \end{array}\end{array}$

The number of moles of monoatomic gas in the sample is $5.00\text{mol}$.

The temperature corresponding to each of the given states is calculated by using the ideal gas equation.

$\begin{array}{c}PV=nRT\\ T=\frac{PV}{nR}\end{array}$ (1)

Where,

• $n$ is the number of moles of the gas.
• $P$ is the absolute pressure of the gas.
• $V$ is the volume of the gas.
• $T$ is the absolute temperature of the gas.
• $R$ is the gas constant.

Substitute the given values of $P,V,n$ and $R$ for state A in the equation (1).

$\begin{array}{c}{T}_{\text{A}}=\frac{3.00\text{atm}\times 15.00\text{L}}{5.00\text{mol}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}}\\ =109.68\text{K}\end{array}$

Substitute the given values of $P,V,n$ and $R$ for state B in the equation (1).

$\begin{array}{c}{T}_{\text{B}}=\frac{6.00\text{atm}\times 20.00\text{L}}{5.00\text{mol}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}}\\ =292.47\text{K}\end{array}$

Substitute the given values of $P,V,n$ and $R$ for state C in the equation (1).

$\begin{array}{c}{T}_{\text{C}}=\frac{3.00\text{atm}\times 55.00\text{L}}{5.00\text{mol}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}}\\ =402.14\text{K}\end{array}$

Substitute the given values of $P,V,n$ and $R$ for state D in the equation (1).

$\begin{array}{c}{T}_{\text{D}}=\frac{6.00\text{atm}\times 15.00\text{L}}{5.00\text{mol}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}}\\ =219.35\text{K}\end{array}$

The internal energy change for each of the given steps for the monoatomic gas is calculated by using the formula,

$\Delta E=\frac{3}{2}nR\left(T_2-T_1\right)$ (2)

Where,

• $n$ is the number of moles of the gas.
• $R$ is the gas constant.
• $T_2$ is the final temperature.
• $T_1$ is the initial temperature.
• $\Delta E$ is the internal energy change.

Substitute the values of $n$, $R$, $T_2$ and $T_1$ for the step 1 in the equation (2).

$\begin{array}{c}\Delta E=\frac{3}{2}\times 5.00\text{mol}\times 8.3145\text{J/mol}\cdot \text{K}\times \left(402.14-109.68\right)\text{K}\\ =18237.4\text{J}\left(=18.237\text{kJ}\right)\end{array}$

Substitute the values of $n$, $R$, $T_2$ and $T_1$ for the step 2 in the equation (2).

$\begin{array}{c}\Delta E=\frac{3}{2}\times 5.00\text{mol}\times 8.3145\text{J/mol}\cdot \text{K}\times \left(292.47-402.14\right)\text{K}\\ =-6838.9\text{J}\left(=-6.84\text{kJ}\right)\end{array}$

Substitute the values of $n$, $R$, $T_2$ and $T_1$ for the step 3 in the equation (2).

$\begin{array}{c}\Delta E=\frac{3}{2}\times 5.00\text{mol}\times 8.3145\text{J/mol}\cdot \text{K}\times \left(219.35-109.68\right)\text{K}\\ =6838.9\text{J}\left(=6.84\text{kJ}\right)\end{array}$

Substitute the values of $n$, $R$, $T_2$ and $T_1$ for the step 4 in the equation (2).

$\begin{array}{c}\Delta E=\frac{3}{2}\times 5.00\text{mol}\times 8.3145\text{J/mol}\cdot \text{K}\times \left(292.47-219.35\right)\text{K}\\ =4559.4\text{J}\left(=4.56\text{kJ}\right)\end{array}$

The enthalpy change for each of the given steps for the monoatomic gas is calculated by using the formula,

$\Delta H=\frac{5}{2}nR\left(T_2-T_1\right)$ (3)

Where,

• $n$ is the number of moles of the gas.
• $R$ is the gas constant.
• $T_2$ is the final temperature.
• $T_1$ is the initial temperature.
• $\Delta H$ is the enthalpy change.

Substitute the values of $n$, $R$, $T_2$ and $T_1$ for the step 1 in the equation (3).

$\begin{array}{c}\Delta H=\frac{5}{2}\times 5.00\text{mol}\times 8.3145\text{J/mol}\cdot \text{K}\times \left(402.14-109.68\right)\text{K}\\ =30395.67\text{J}\left(=30.396\text{kJ}\right)\end{array}$

Substitute the values of $n$, $R$, $T_2$ and $T_1$ for the step 2 in the equation (3).

$\begin{array}{c}\Delta H=\frac{5}{2}\times 5.00\text{mol}\times 8.3145\text{J/mol}\cdot \text{K}\times \left(292.47-402.14\right)\text{K}\\ =-11398.14\text{J}\left(=-11.39\text{kJ}\right)\end{array}$

Substitute the values of $n$, $R$, $T_2$ and $T_1$ for the step 3 in the equation (3).

$\begin{array}{c}\Delta H=\frac{5}{2}\times 5.00\text{mol}\times 8.3145\text{J/mol}\cdot \text{K}\times \left(219.35-109.68\right)\text{K}\\ =11397.4\text{J}\left(=11.39\text{kJ}\right)\end{array}$

Substitute the values of $n$, $R$, $T_2$ and $T_1$ for the step 4 in the equation (3).

$\begin{array}{c}\Delta H=\frac{5}{2}\times 5.00\text{mol}\times 8.3145\text{J/mol}\cdot \text{K}\times \left(292.47-219.35\right)\text{K}\\ =7598\text{J}\left(=7.598\text{kJ}\right)\end{array}$

Since, the enthalpy change of the process is equal to the heat transferred in the cycle. Therefore, the $q=\Delta H$ is true for all the given steps in the state change.

The work done for the each of the given steps is calculated by using the formula,

$\begin{array}{c}\Delta E=q+w\\ w=\Delta E-q\end{array}$ (4)

Where,

• $\Delta E$ is the internal energy change.
• $q$ is the heat transferred.
• $w$ is the work done.

Substitute the values of $\Delta E$ and $q$ for step 1 in the equation (4).

$\begin{array}{c}w=18.237\text{kJ}-30.396\text{kJ}\\ =-12.159\text{kJ}\end{array}$

Substitute the values of $\Delta E$ and $q$ for step 2 in the equation (4).

$\begin{array}{c}w=\left(-6.84\text{kJ}\right)-\left(-11.39\text{kJ}\right)\\ =4.55\text{kJ}\end{array}$

Substitute the values of $\Delta E$ and $q$ for step 3. in the equation (4).

$\begin{array}{c}w=6.84\text{kJ}-11.39\text{kJ}\\ =-4.55\text{kJ}\end{array}$

Substitute the values of $\Delta E$ and $q$ for step 4 in the equation (4).

$\begin{array}{c}w=4.56\text{kJ}-7.598\text{kJ}\\ =-3.038\text{kJ}\end{array}$

Conclusion

The internal energy change of step 1 is $18.237kJ$.

The enthalpy change of step 1 is $30.396kJ$.

The heat transferred for the step 1 is $30.396kJ$.

The work done in the step 1 is $-12.159kJ$.

The internal energy change of step 2 is $-6.84kJ$.

The enthalpy change of step 2 is $-11.39kJ$.

The heat transferred for the step 2 is $-11.39kJ$.

The work done in the step 2 is $4.55kJ$.

The internal energy change of step 3 is $6.84kJ$.

The enthalpy change of step 3 is $11.39kJ$.

The heat transferred for the step 3 is $11.39kJ$.

The work done in the step 3 is $-4.55kJ$.

The internal energy change of step 4 is $4.56kJ$.

The enthalpy change of step 4 is $7.598kJ$.

The heat transferred for the step 4 is $7.598kJ$.

The work done in the step 4 is $-3.038kJ$