Chapter 17, Problem 52P

(a)

To determine

The work done on the gas in an isothermal process.

Answer to Problem 52P

Thework done on the gas in an isothermal process is $\underset{\_}{28.0\text{kJ}}$.

Explanation of Solution

Write the expression for the work done on the gas,

    $W_{ab}=-{\displaystyle \underset{V_a}{\overset{V_b}{\int }}PdV}$        (I)

Here, $W_{ab}$ is the work done on the gas, $P$ is the pressure of the gas and $dV$ is the volume change.

For the isothermal process, the work done on the gas,

    $\begin{array}{c}{W}_{ab\text{'}}=-nR{T}_a{\displaystyle \underset{V_a}{\overset{V_b}{\int }}\left(\frac{1}{V}\right)dV}\\ =-nR{T}_a\ln \left(\frac{V_{b\text{'}}}{V_a}\right)\end{array}$

    $W_{ab\text{'}}=nR{T}_a\ln \left(\frac{V_a}{V_{b\text{'}}}\right)$        (II)

Here, $W_{ab\text{'}}$ is the work done on the gas in an isothermal process, $T_a$ is the temperature at point $a$ , $V_{b\text{'}}$ is the volume at point $b\text{'}$ , $V_b$ is the volume at point $b$  and $V_a$ is the volume at $a$ .

Conclusion:

Substitute $5.00\text{mol}$ for $n$ , $8.314\text{J/mol}\text{.K}$ for $R$, $293\text{K}$ for $T_a$ and $10.0$ for $\frac{V_a}{V_{b\text{'}}}$ in the above equation to find $W_{ab\text{'}}$,

    $\begin{array}{c}{W}_{ab\text{'}}=\left(5.00\text{mol}\right)\left(8.314\text{J/mol}\text{.K}\right)\left(293\text{K}\right)\ln \left(10.0\right)\\ =28.0\text{kJ}\end{array}$

Therefore, the work done on the gas in an isothermal process is $\underset{\_}{28.0\text{kJ}}$.

(b)

To determine

Thework done on the gas in an adiabatic process.

Answer to Problem 52P

Thework done on the gas in an adiabatic process is $\underset{\_}{46.0\text{kJ}}$.

Explanation of Solution

Write the expression for the temperature of the system at point $b$ in an adiabatic process,

    $T_b=T_a{\left(\frac{V_a}{V_b}\right)}^{\gamma -1}$        (III)

Here, $T_b$ is the temperature of the system at point $b$, $T_a$ is the temperature of the system at point $a$, $V_a$ is the volume of the gas at point $a$ , $V_b$ is the volume of the gas at point $b$ and $\gamma$ is the ratio of specific heat.

Write the expression for the molar specific heat at constant volume,

    $C_{V,air}=\frac{5R}{2}$        (IV)

Here, $C_{V,air}$ is the molar specific heat at constant volume and $R$ is the universal gas constant.

Write the expression for the work done on the gas in an adiabatic process,

    $\begin{array}{c}{W}_{ab}={\left(-Q+\Delta E_{\mathrm{int}}\right)}_{ab}\\ ={\left(0+n{C}_V\Delta T\right)}_{ab}\\ =n{C}_V\left(T_b-T_a\right)\end{array}$        (V)

Here, $W_{ab}$ is the work done on the gas in an adiabatic process, $Q$ is the heat energy transferred in the system at $ab$ , $\Delta E_{\mathrm{int}}$ is the internal energy change in the gas at $ab$ , $T_b$ is the temperature at point $b$ and $T_a$ is the temperature at point $a$ .

Conclusion:

Substitute $293\text{K}$ for $T_a$, $0.400$ for $\gamma$ and $10.0$ for $\frac{V_a}{V_b}$ in (III) to find $T_b$,

    $\begin{array}{c}{T}_b=\left(293\text{K}\right){\left(10.0\right)}^{0.400}\\ =736\text{K}\end{array}$

Substitute $8.314\text{J/mol}\text{.K}$ for $R$ in (IV) to find $C_{V,air}$,

    $\begin{array}{c}{C}_{V,air}=\frac{5\left(8.314\text{J/mol}\text{.K}\right)}{2}\\ =20.8\text{J/mol}\text{.K}\end{array}$

Substitute $5.00\text{mol}$ for $n$ , $20.8\text{J/mol}\text{.K}$ for $C_V$ , $736\text{K}$ for $T_b$ and $293\text{K}$ for $T_a$ in (V) to find $W_{ab}$ ,

    $\begin{array}{c}{W}_{ab}=\left(5.00\text{mol}\right)\left(20.8\text{J/mol}\text{.K}\right)\left(736\text{K}-293\text{K}\right)\\ =46.0\text{kJ}\end{array}$

Therefore, the work done on the gas in an adiabatic process is $\underset{\_}{46.0\text{kJ}}$ .

(c)

To determine

The final pressure of the gas in an isothermal process.

Answer to Problem 52P

The final pressure of the gas in an isothermal process is $\underset{\_}{10.0\text{atm}}$.

Explanation of Solution

Write the expression for the ideal gas law,

    $P_B{V}_B=nR{T}_B$        (VI)

For the isothermal process, $P_{b\text{'}}{V}_b^{\gamma }=P_a{V}_a^{\gamma }$.

    $P_{b\text{'}}=P_a\left(\frac{V_a}{V_{b\text{'}}}\right)$        (VII)

Here, $P_{b\text{'}}$ is the final pressure of the gas in an isothermal process and $P_a$ is the pressure of the gas at point $a$ .

Conclusion:

Substitute $1.00\text{atm}$ for $P_a$ and $10.0$ for $\frac{V_a}{V_{b\text{'}}}$ in (VII),

    $\begin{array}{c}{P}_{b\text{'}}=\left(1.00\text{atm}\right)\left(10.0\right)\\ =10.0\text{atm}\end{array}$

Therefore, the final pressure of the gas in an isothermal process is $\underset{\_}{10.0\text{atm}}$.

(d)

To determine

The final pressure of the gas in an adiabatic process.

Answer to Problem 52P

The final pressure of the gas in an adiabatic process is $\underset{\_}{25.1\text{atm}}$.

Explanation of Solution

For the adiabatic process,

Write the expression for the final pressure of the gas,

    $P_b=P_a{\left(\frac{V_a}{V_b}\right)}^{\gamma }$        (VIII)

Conclusion:

Substitute $1.00\text{atm}$ for $P_a$ , $10.0$ for $\frac{V_a}{V_b}$ and $1.40$ for $\gamma$ in (VIII),

    $\begin{array}{c}{P}_b=\left(1.00\text{atm}\right){\left(10.0\right)}^{1.40}\\ =25.1\text{atm}\end{array}$

Therefore, the temperature at the end of the cycle is $\underset{\_}{T_i}$.