Chapter 21, Problem 51AP

(i)

To determine

Find the final pressure, final volume, final temperature, change in internal energy, energy added to the gas, and work done on the gas, if the gas is heated at constant pressure at $400\text{K}$.

Answer to Problem 51AP

1. (a) The final pressure is $100\text{kPa}$.
2. (b) The final volume is $66.5\text{L}$.
3. (c) The final temperature is $400\text{K}$.
4. (d) The change in internal energy of the gas is $5.82\text{kJ}$.
5. (e) The energy added to the gas is $7.48\text{kJ}$.
6. (f) The work done on the gas is $-1.66\text{kJ}$.

Explanation of Solution

(a)

Since an isobaric process occurs at constant pressure. an isobaric expansion of a gas requires heat transfers to keep the pressure constant. So that the final pressure of the gas is equal to the initial pressure.

Find the final pressure of the gas.

  $\begin{array}{c}{P}_f=1.00\times {10}^5\text{Pa}\\ =100\times {10}^3\text{Pa}\left(\frac{{10}^{-3}\text{kPa}}{1\text{Pa}}\right)\\ =100\text{kPa}\end{array}$

(b)

Find the final volume of the gas, by using ideal gas equation.

  $V_f=\frac{nR{T}_f}{P_f}$                                                                                              (I)

Here, $V_f$ is the final volume of the gas, $P_f$ is the final pressure of the gas, $n$ is the number of moles, $T_f$ is the final temperature, and $R$ is the gas constant.

(c)

Since the final temperature of the gas heated at constant pressure is $400\text{K}$.

(d)

Find the internal energy of the gas.

  $\Delta E_{\mathrm{int}}=\frac{7}{2}nR\left(T_f-T_i\right)$                                                                              (II)

Here, $\Delta E_{\mathrm{int}}$ is the internal energy of the gas and $T_i$ is the initial temperature of the gas.

(e)

Find the energy added to the gas by heat.

  $Q=n{C}_P\left(T_f-T_i\right)$

Here, $Q$ is the amount of heat added to the system and $C_P$ is the specific heat capacity at constant pressure.

Rearrange the above equation for constant pressure.

  $Q=\frac{9}{2}nR\left(T_f-T_i\right)$                                                                                     (III)

(f)

Find the work done by the gas.

  $W=-P\Delta V$

Here, $\Delta V$ is the change in volume.

Replace $nR\Delta T$ for $P\Delta V$ in the above equation.

  $\begin{array}{c}W=-nR\Delta T\\ =-nR\left(T_f-T_i\right)\end{array}$                                                                                      (IV)

Conclusion:

Substitute $2.00\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $400\text{K}$ for $T_f$, and $100\times {10}^3\text{Pa}$ for $P_f$ in equation (I) to find $V_f$.

  $\begin{array}{c}{V}_f=\frac{\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(400\text{K}\right)}{\left(100\times {10}^3\text{Pa}\right)}\\ =66.5\times {10}^{-3}{\text{m}}^3\left(\frac{{10}^3\text{L}}{1{\text{m}}^3}\right)\\ =66.5\text{L}\end{array}$

Substitute $2.00\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $400\text{K}$ for $T_f$, and $300\text{K}$ for $T_i$ in equation (II) to find $\Delta E_{\mathrm{int}}$.

  $\begin{array}{c}\Delta E_{\mathrm{int}}=\frac{7}{2}\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(400\text{K}-300\text{K}\right)\\ =5.82\times {10}^3\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =5.82\text{kJ}\end{array}$

Substitute $2.00\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $400\text{K}$ for $T_f$, and $300\text{K}$ for $T_i$ in equation (III) to find $Q$.

  $\begin{array}{c}Q=\frac{9}{2}\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(400\text{K}-300\text{K}\right)\\ =7.48\times {10}^3\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =7.48\text{kJ}\end{array}$

Substitute $2.00\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $400\text{K}$ for $T_f$, and $300\text{K}$ for $T_i$ in equation (IV) to find $W$.

  $\begin{array}{c}W=-\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(400\text{K}-300\text{K}\right)\\ =-1.66\times {10}^3\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =-1.66\text{kJ}\end{array}$

Therefore,

1. (a) The final pressure is $100\text{kPa}$.
2. (b) The final volume is $66.5\text{L}$.
3. (c) The final temperature is $400\text{K}$.
4. (d) The change in internal energy of the gas is $5.82\text{kJ}$.
5. (e) The energy added to the gas is $7.48\text{kJ}$.
6. (f) The work done on the gas is $-1.66\text{kJ}$.

(ii)

To determine

Find the final pressure, final volume, final temperature, change in internal energy, energy added to the gas, and work done on the gas, if the gas is heated at constant volume at $400\text{K}$.

Answer to Problem 51AP

1. (a) The final pressure is $133\text{kPa}$.
2. (b) The final volume is $49.9\text{L}$.
3. (c) The final temperature is $400\text{K}$.
4. (d) The change in internal energy of the gas is $5.82\text{kJ}$.
5. (e) The energy added to the gas is $5.82\text{kJ}$.
6. (f) The work done on the gas is $0$.

Explanation of Solution

(a)

Find the final pressure of the gas for an iso-volumetric process.

  $\frac{P_f}{T_f}=\frac{P_i}{T_i}$

Rearrange the above equation for $P_f$.

  $P_f=P_i\left(\frac{T_f}{T_i}\right)$                                                                                          (V)

(b)

Find the final volume of the gas, by using ideal gas equation.

  $V_f=\frac{nR{T}_i}{P_i}$                                                                                             (VI)

(c)

Since the final temperature of the gas heated at constant pressure is $400\text{K}$.

(d)

Find the internal energy of the gas.

  $\Delta E_{\mathrm{int}}=\frac{7}{2}nR\left(T_f-T_i\right)$                                                                            (VII)

(e)

Find the energy added to the gas by heat.

  $Q=n{C}_V\left(T_f-T_i\right)$

Here, $C_V$ is the specific heat capacity at constant volume.

Rearrange the above equation for constant volume.

  $Q=\frac{7}{2}nR\left(T_f-T_i\right)$                                                                                    (VIII)

(f)

Find the work done by the gas.

  $W=-{\displaystyle \int P\Delta V}$

Since, volume is constant so that work done on the gas is zero.

Conclusion:

Substitute $300\text{K}$ for $T_i$, $1.00\times {10}^5\text{Pa}$ for $P_i$, and $400\text{K}$ for $T_f$ in equation (V) to find $P_f$.

  $\begin{array}{c}{P}_f=\left(1.00\times {10}^5\text{Pa}\right)\left(\frac{400\text{K}}{300\text{K}}\right)\\ =133\times {10}^3\text{Pa}\left(\frac{{10}^{-3}\text{kPa}}{1\text{Pa}}\right)\\ =133\text{kPa}\end{array}$

Substitute $300\text{K}$ for $T_i$, $1.00\times {10}^5\text{Pa}$ for $P_i$, $2.00\text{mol}$ for $n$, and $8.314\text{J/mol}\cdot \text{K}$ for $R$ in equation (VI) to find $V_f$.

  $\begin{array}{c}{V}_f=\frac{\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(300\text{K}\right)}{\left(1.00\times {10}^5\text{Pa}\right)}\\ =49.9\times {10}^{-3}{\text{m}}^3\left(\frac{{10}^3\text{L}}{1{\text{m}}^3}\right)\\ =49.9\text{L}\end{array}$

Substitute $2.00\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $400\text{K}$ for $T_f$, and $300\text{K}$ for $T_i$ in equation (VII) to find $\Delta E_{\mathrm{int}}$.

  $\begin{array}{c}\Delta E_{\mathrm{int}}=\frac{7}{2}\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(400\text{K}-300\text{K}\right)\\ =5.82\times {10}^3\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =5.82\text{kJ}\end{array}$

Substitute $2.00\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $400\text{K}$ for $T_f$, and $300\text{K}$ for $T_i$ in equation (VIII) to find $Q$.

  $\begin{array}{c}Q=\frac{7}{2}\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(400\text{K}-300\text{K}\right)\\ =5.82\times {10}^3\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ =5.82\text{kJ}\end{array}$

Therefore,

1. (a) The final pressure is $133\text{kPa}$.
2. (b) The final volume is $49.9\text{L}$.
3. (c) The final temperature is $400\text{K}$.
4. (d) The change in internal energy of the gas is $5.82\text{kJ}$.
5. (e) The energy added to the gas is $5.82\text{kJ}$.
6. (f) The work done on the gas is $0$.

(iii)

To determine

Find the final pressure, final volume, final temperature, change in internal energy, energy added to the gas, and work done on the gas, if the gas is compressed at constant temperature to $1.20\times {10}^5\text{Pa}$.

Answer to Problem 51AP

1. (a) The final pressure is $120\text{kPa}$.
2. (b) The final volume is $41.6\text{L}$.
3. (c) The final temperature is $300\text{K}$.
4. (d) The change in internal energy of the gas is $0$.
5. (e) The energy added to the gas is $-\text{909J}$.
6. (f) The work done on the gas is $909\text{J}$.

Explanation of Solution

(a)

Since an isobaric process occurs at constant pressure. An isobaric expansion of a gas requires heat transfers to keep the pressure constant. So that the final pressure of the gas is equal to the initial pressure.

Find the final pressure of the gas.

  $\begin{array}{c}{P}_f=1.20\times {10}^5\text{Pa}\\ =120\times {10}^3\text{Pa}\left(\frac{{10}^{-3}\text{kPa}}{1\text{Pa}}\right)\\ =120\text{kPa}\end{array}$

(b)

Find the final volume of the gas, by using ideal gas equation.

  $V_f=V_i\left(\frac{P_i}{P_f}\right)$                                                                                              (IX)

(c)

Since the final temperature of the gas heated at constant temperature is $300\text{K}$.

(d)

Find the internal energy of the gas.

  $\Delta E_{\mathrm{int}}=\frac{7}{2}nR\left(T_f-T_i\right)$

Since the temperature is constant for the compressed gas, so that internal energy of the gas is zero.

(e)

Find the energy added to the gas by heat.

  $Q=\Delta E_{\mathrm{int}}-W$                                                                                              (X)

(f)

Find the work done by the gas.

  $W=-{\displaystyle \int P\Delta V}$

Integrate the above equation.

  $\begin{array}{c}W=-nR{T}_i{\displaystyle \underset{V_i}{\overset{V_f}{\int }}\frac{dV}{V}}\\ =-nR{T}_i\ln \left(\frac{V_f}{V_i}\right)\\ =-nR{T}_i\ln \left(\frac{P_i}{P_f}\right)\end{array}$                                                                                      (XI)

Conclusion:

Substitute $120\text{kPa}$ for $P_f$, $100\text{kPa}$ for $P_i$, and $49.9\text{L}$ for $V_i$ in equation (IX) to find $V_f$.

  $\begin{array}{c}{V}_f=\left(49.9\text{L}\right)\left(\frac{100\text{kPa}}{120\text{kPa}}\right)\\ =41.6\text{L}\end{array}$

Substitute $0$ for $\Delta E_{\mathrm{int}}$ and $909\text{J}$ for $W$ in equation (X) to find $Q$.

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

Substitute $120\text{kPa}$ for $P_f$, $100\text{kPa}$ for $P_i$, $2.00\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, and $300\text{K}$ for $T_i$ in equation (XI) to find $W$.

  $\begin{array}{c}W=-\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(300\text{K}\right)\ln \left(\frac{100\text{kPa}}{120\text{kPa}}\right)\\ =909\text{J}\end{array}$

Therefore,

1. (a) The final pressure is $120\text{kPa}$.
2. (b) The final volume is $41.6\text{L}$.
3. (c) The final temperature is $300\text{K}$.
4. (d) The change in internal energy of the gas is $0$.
5. (e) The energy added to the gas is $-\text{909J}$.
6. (f) The work done on the gas is $909\text{J}$.

(iv)

To determine

Find the final pressure, final volume, final temperature, change in internal energy, energy added to the gas, and work done on the gas, if the gas is compressed adiabatically to $1.20\times {10}^5\text{Pa}$.

Answer to Problem 51AP

1. (a) The final pressure is $120\text{kPa}$.
2. (b) The final volume is $43.3\text{L}$.
3. (c) The final temperature is $312\text{K}$.
4. (d) The change in internal energy of the gas is $722\text{J}$.
5. (e) The energy added to the gas is $0$.
6. (f) The work done on the gas is $722\text{J}$.

Explanation of Solution

(a)

Since an isobaric process occurs without transfer of heat between a thermodynamic system.

Find the final pressure of the gas.

  $\begin{array}{c}{P}_f=1.20\times {10}^5\text{Pa}\\ =120\times {10}^3\text{Pa}\left(\frac{{10}^{-3}\text{kPa}}{1\text{Pa}}\right)\\ =120\text{kPa}\end{array}$

(b)

From the adiabatic compression ratio.

  $\gamma =\frac{C_P}{C_V}$

Rewrite the above expression.

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

Find the final volume of the gas, by using ideal gas equation.

  $\begin{array}{c}{P}_f{V}_f^{\gamma }=P_i{V}_i^{\gamma }\\ V_f=V_i{\left(\frac{P_i}{P_f}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\gamma $}\right.}\end{array}$                                                                                     (XII)

(c)

Find the final temperature of the gas, by using ideal gas equation.

  $\begin{array}{c}\frac{T_i}{P_i{V}_i}=\frac{T_f}{P_f{V}_f}\\ T_f=T_i\left(\frac{P_f{V}_f}{P_i{V}_i}\right)\end{array}$                                                                                        (XIII)

(d)

Find the internal energy of the gas.

  $\Delta E_{\mathrm{int}}=\frac{7}{2}nR\left(T_f-T_i\right)$                                                                                      (XIV)

(e)

Since the energy added to the gas by heat is zero.

(f)

Find the work done by the gas.

  $W=-Q+\Delta E_{\mathrm{int}}$                                                                                                 (XV)

Conclusion:

Substitute $120\text{kPa}$ for $P_f$, $100\text{kPa}$ for $P_i$, and $49.9\text{L}$ for $V_i$ in equation (XII) to find $V_f$.

  $\begin{array}{c}{V}_f=\left(49.9\text{L}\right){\left(\frac{100\text{kPa}}{120\text{kPa}}\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$9$}\right.}\\ =43.3\text{L}\end{array}$

Substitute $120\text{kPa}$ for $P_f$, $100\text{kPa}$ for $P_i$, $300\text{K}$ for $T_i$, $43.3\text{L}$ for $V_f$, and $49.9\text{L}$ for $V_i$ in equation (XIII) to find $T_f$.

  $\begin{array}{c}{T}_f=\left(300\text{K}\right)\left[\frac{\left(120\text{kPa}\right)\left(43.3\text{L}\right)}{\left(100\text{kPa}\right)\left(49.9\text{L}\right)}\right]\\ =312.4\text{K}\\ \approx \text{312}\text{K}\end{array}$

Substitute $312.4\text{K}$ for $T_f$, $300\text{K}$ for $T_i$, $2.00\text{mol}$ for $n$, and $8.314\text{J/mol}\cdot \text{K}$ for $R$ in equation (XIV) to find $\Delta E_{\mathrm{int}}$.

  $\begin{array}{c}\Delta E_{\mathrm{int}}=\frac{7}{2}\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(312.4\text{K}-300\text{K}\right)\\ =722\text{J}\end{array}$

Substitute $0$ for $Q$ and $722\text{J}$ for $\Delta E_{\mathrm{int}}$ in equation (XV) to find $W$.

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

Therefore,

1. (a) The final pressure is $120\text{kPa}$.
2. (b) The final volume is $43.3\text{L}$.
3. (c) The final temperature is $312\text{K}$.
4. (d) The change in internal energy of the gas is $722\text{J}$.
5. (e) The energy added to the gas is $0$.
6. (f) The work done on the gas is $722\text{J}$.