Chapter 12.6, Problem 40P

(a)

To determine

To drive an expression of $\Delta u$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process.

Answer to Problem 40P

The expression of $\Delta u$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process is $\Delta u=0$.

Explanation of Solution

Write the equation of state of the given gas.

$P\left(v-a\right)=RT$ (I)

Here, the temperature is $T$, the pressure is $P$, the specific volume is $v$, the universal gas constant is $R$ and the constant is $a$.

Write the general expression for change in internal energy $\left(\Delta u\right)$.

$\Delta u=u_2-u_1={\displaystyle {\int }_{T_1}^{T_2}{c}_{v}dT}+{\displaystyle {\int }_{v_1}^{v_2}\left(T{\left(\frac{\partial P}{\partial T}\right)}_v-P\right)dv}$ (II)

Here, the internal energy at state 1, 2 is $u_1,u_2$ and the specific heat at constant volume is $c_v$; the symbol $\partial$ indicates the partial derivative of variables.

Rearrange the Equation (I) to obtain $P$.

$P=\frac{RT}{v-a}$ (III)

Conclusion:

Partially differentiate the Equation (III) with respect to temperature by keeping the specific volume as constant.

$\begin{array}{c}{\left(\frac{\partial P}{\partial T}\right)}_v=\frac{\partial }{\partial T}{\left(\frac{RT}{v-a}\right)}_v\\ =\frac{R}{v-a}{\left(\frac{\partial T}{\partial T}\right)}_v\\ =\frac{R}{v-a}\left(1\right)\\ =\frac{R}{v-a}\end{array}$

Substitute $\frac{R}{v-a}$ for ${\left(\frac{\partial P}{\partial T}\right)}_v$ in Equation (II).

$\begin{array}{c}\Delta u={\displaystyle {\int }_{T_1}^{T_2}{c}_{v}dT}+{\displaystyle {\int }_{v_1}^{v_2}\left(T\left(\frac{R}{v-a}\right)-P\right)dv}\\ ={\displaystyle {\int }_{T_1}^{T_2}{c}_{v}dT}+{\displaystyle {\int }_{v_1}^{v_2}\left(\frac{RT}{v-a}-P\right)dv}\\ ={\displaystyle {\int }_{T_1}^{T_2}{c}_{v}dT}+{\displaystyle {\int }_{v_1}^{v_2}\left(P-P\right)dv}\\ ={\displaystyle {\int }_{T_1}^{T_2}{c}_{v}dT}+{\displaystyle {\int }_{v_1}^{v_2}\left(0\right)dv}\end{array}$

$={\displaystyle {\int }_{T_1}^{T_2}{c}_{v}dT}$ (IV)

For an isothermal process, the temperature is kept constant.

$T=\text{constant}$

The differential temperature or change in temperature becomes zero.

$dT=0$

Substitute $0$ for $dT$ in Equation (IV).

$\begin{array}{c}\Delta u={\displaystyle {\int }_{T_1}^{T_2}{c}_v\left(0\right)}\\ =0\end{array}$

Thus, the expression of $\Delta u$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process is $\Delta u=0$.

(b)

To determine

To drive an expression of $\Delta h$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process.

Answer to Problem 40P

The expression of $\Delta h$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process is $\Delta h=a\left(P_2-P_1\right)$.

Explanation of Solution

Write the general expression for change in enthalpy $\left(\Delta h\right)$.

$\Delta h=h_2-h_1={\displaystyle {\int }_{T_1}^{T_2}{c}_{p}dT}+{\displaystyle {\int }_{P_1}^{P_2}\left(v-T{\left(\frac{\partial v}{\partial T}\right)}_P\right)dP}$ (V)

Here, the enthalpy at state 1, 2 is $h_1,h_2$ and the specific heat at constant pressure is $c_p$; the symbol $\partial$ indicates the partial derivative of variables.

Rearrange the Equation (V) to obtain $v$.

$v=\frac{RT}{P}+a$ (VI)

Conclusion:

Partially differentiate the Equation (VI) with respect to temperature by keeping the pressure as constant.

$\begin{array}{c}{\left(\frac{\partial v}{\partial T}\right)}_P=\frac{\partial }{\partial T}{\left(\frac{RT}{P}+a\right)}_P\\ =\frac{\partial }{\partial T}{\left(\frac{RT}{P}\right)}_P+\frac{\partial }{\partial T}{\left(a\right)}_P\\ =\frac{R}{P}{\left(\frac{\partial T}{\partial T}\right)}_P+0\\ =\frac{R}{P}\left(1\right)\end{array}$

$=\frac{R}{P}$

Substitute $\frac{R}{P}$ for ${\left(\frac{\partial v}{\partial T}\right)}_P$ in Equation (V).

$\begin{array}{c}\Delta h={\displaystyle {\int }_{T_1}^{T_2}{c}_{p}dT}+{\displaystyle {\int }_{P_1}^{P_2}\left(v-T\left(\frac{R}{P}\right)\right)dP}\\ ={\displaystyle {\int }_{T_1}^{T_2}{c}_{p}dT}+{\displaystyle {\int }_{P_1}^{P_2}\left(v-\frac{RT}{P}\right)dP}\\ ={\displaystyle {\int }_{T_1}^{T_2}{c}_{p}dT}+{\displaystyle {\int }_{P_1}^{P_2}\left[v-\left(v-a\right)\right]dP}\\ ={\displaystyle {\int }_{T_1}^{T_2}{c}_{p}dT}+{\displaystyle {\int }_{P_1}^{P_2}\left[v-v+a\right]dP}\end{array}$

$\begin{array}{c}={\displaystyle {\int }_{T_1}^{T_2}{c}_{p}dT}+{\displaystyle {\int }_{P_1}^{P_2}adP}\\ ={\displaystyle {\int }_{T_1}^{T_2}{c}_{p}dT}+a\left(P_2-P_1\right)\end{array}$ (VI)

For an isothermal process, the temperature is kept constant.

$T=\text{constant}$

The differential temperature or change in temperature becomes zero.

$dT=0$

Substitute $0$ for $dT$ in Equation (VI).

$\begin{array}{c}\Delta h={\displaystyle {\int }_{T_1}^{T_2}{c}_p\left(0\right)}+a\left(P_2-P_1\right)\\ =0+a\left(P_2-P_1\right)\\ =a\left(P_2-P_1\right)\end{array}$

Thus, the expression of $\Delta h$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process is $\Delta h=a\left(P_2-P_1\right)$.

(c)

To determine

To drive an expression of $\Delta s$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process.

Answer to Problem 40P

The expression of $\Delta s$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process is $\Delta s=-R\left(\ln \frac{P_2}{P_1}\right)$.

Explanation of Solution

Write the general expression for change in entropy $\left(\Delta s\right)$.

$\Delta s=s_2-s_1={\displaystyle {\int }_{T_1}^{T_2}\frac{c_p}{T}dT-}{\displaystyle {\int }_{P_1}^{P_2}{\left(\frac{\partial v}{\partial T}\right)}_{P}dP}$ (VII)

Here, the entropy at state 1, and 2 are $s_1$ and $s_2$.

Conclusion:

Substitute $\frac{R}{P}$ for ${\left(\frac{\partial v}{\partial T}\right)}_P$ in Equation (VII).

$\begin{array}{c}\Delta s={\displaystyle {\int }_{T_1}^{T_2}\frac{c_p}{T}dT}-{\displaystyle {\int }_{P_1}^{P_2}\left(\frac{R}{P}\right)dP}\\ ={\displaystyle {\int }_{T_1}^{T_2}\frac{c_p}{T}dT}-R{\displaystyle {\int }_{P_1}^{P_2}\left(\frac{1}{P}\right)dP}\\ ={\displaystyle {\int }_{T_1}^{T_2}\frac{c_p}{T}dT}-R\left(\ln \frac{P_2}{P_1}\right)\end{array}$ (VIII)

For an isothermal process, the temperature is kept constant.

$T=\text{constant}$

The differential temperature or change in temperature becomes zero.

$dT=0$

Substitute $0$ for $dT$ in Equation (VIII).

$\begin{array}{c}\Delta s={\displaystyle {\int }_{T_1}^{T_2}\frac{c_p}{T}\left(0\right)}-R\left(\ln \frac{P_2}{P_1}\right)\\ =0-R\left(\ln \frac{P_2}{P_1}\right)\\ =-R\left(\ln \frac{P_2}{P_1}\right)\end{array}$

Thus, the expression of $\Delta s$ for a gas whose equation of state is $P\left(v-a\right)=RT$ for an isothermal process is $\Delta s=-R\left(\ln \frac{P_2}{P_1}\right)$.