Chapter 21, Problem 55AP

(a)

To determine

The initial volume of the gas.

Answer to Problem 55AP

The initial volume of the gas is $0.514{\text{m}}^3$.

Explanation of Solution

Write the expression from ideal gas law.

  $P_i{V}_i=nR{T}_i$                                                                                            (I)

Here, $P_i$ is the initial pressure of the gas, $V_i$ is the initial volume of the gas, $n$ is the number of moles, $R$ is the gas constant, and $T_i$ is the initial temperature.

Write the expression for number of moles.

  $n=\frac{m}{M}$                                                                                                  (II)

Here, $m$ is the mass and $M$ is the molar mass

Rearrange the equation (I) for $V_i$.

  $V_i=\frac{nR{T}_i}{P_i}$                                                                                              (III)

Conclusion:

Substitute $1.20\text{kg}$ for $m$ and $28.9\text{g/mol}$ for $M$ in equation (II) to find $n$.

  $\begin{array}{c}n=\frac{1.20\text{kg}}{28.9\text{g/mol}\left(\frac{0.001\text{kg}}{1\text{g}}\right)}\\ =\frac{1.20\text{kg}}{0.0289\text{kg/mol}}\\ =41.5\text{mol}\end{array}$

Substitute $41.5\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$, $25.0°\text{C}$ for $T_i$, and $2.00\times {10}^5\text{Pa}$ for $P_i$ in equation (III) to find $V_i$.

  $\begin{array}{c}{V}_i=\frac{\left(41.5\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(25.0\text{+273}\text{K}\right)}{\left(2.00\times {10}^5\text{Pa}\right)}\\ =0.514{\text{m}}^3\end{array}$

Therefore, the initial volume of the gas is $0.514{\text{m}}^3$.

(b)

To determine

The final volume of the gas.

Answer to Problem 55AP

The final volume of the gas is $2.06{\text{m}}^3$.

Explanation of Solution

Write the expression for final volume of the gas.

  $\frac{P_f}{P_i}=\frac{\sqrt{V_f}}{\sqrt{V_i}}$                                                                                         (IV)

Here, $P_f$ is the final pressure and $V_f$ is the final volume.

Rearrange the equation (IV) for $V_f$.

  $\begin{array}{c}{\left(\frac{P_f}{P_i}\right)}^2=\frac{V_f}{V_i}\\ V_f=V_i{\left(\frac{P_f}{P_i}\right)}^2\end{array}$

Conclusion:

Substitute $0.514{\text{m}}^3$ for $V_i$ and $4.00\times {10}^5\text{Pa}$ for $P_f$, and $2.00\times {10}^5\text{Pa}$ for $P_i$ in above equation to find $V_f$.

  $\begin{array}{c}{V}_f=\left(0.514{\text{m}}^3\right){\left(\frac{4.00\times {10}^5\text{Pa}}{2.00\times {10}^5\text{Pa}}\right)}^2\\ =2.06{\text{m}}^3\end{array}$

Therefore, the final volume of the gas is $2.06{\text{m}}^3$.

(c)

To determine

The final temperature of the gas.

Answer to Problem 55AP

The final temperature of the gas is $2.38\times {10}^3\text{K}$.

Explanation of Solution

Write the expression from ideal gas law.

  $P_f{V}_f=nR{T}_f$

Rearrange the above equation for $T_f$.

  $T_f=\frac{P_f{V}_f}{nR}$

Conclusion:

Substitute $2.06{\text{m}}^3$ for $V_f$ and $4.00\times {10}^5\text{Pa}$ for $P_f$, $41.5\text{mol}$ for $n$, and $8.314\text{J/mol}\cdot \text{K}$ for $R$ to find $T_f$.

  $\begin{array}{c}{T}_f=\frac{\left(4.00\times {10}^5\text{Pa}\right)\left(2.06{\text{m}}^3\right)}{\left(41.5\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)}\\ =0.0238\times {10}^5\text{K}\\ \approx \text{2}\text{.38}\times {\text{10}}^3\text{K}\end{array}$

Therefore, the final temperature of the gas is $2.38\times {10}^3\text{K}$.

(d)

To determine

The work done on the air.

Answer to Problem 55AP

The work done on the air is $-480\text{kJ}$.

Explanation of Solution

Write the expression for work done on the air.

  $W=-{\displaystyle \underset{V_i}{\overset{V_f}{\int }}PdV}$

Rearrange the above equation for $T_f$.

  $\begin{array}{c}W=-C{\displaystyle \underset{V_i}{\overset{V_f}{\int }}{V}^{1/2}dV}\\ =\left(\frac{P_i}{V_i^{1/2}}\right){\left[\frac{2V^{3/2}}{3}\right]}_{V_i}^{V_f}\\ =-\frac{2}{3}\left(\frac{P_i}{V_i^{1/2}}\right)\left(V_f^{3/2}-V_i^{3/2}\right)\end{array}$

Conclusion:

Substitute $2.06{\text{m}}^3$ for $V_f$ $0.514{\text{m}}^3$ for $V_i$ and $2.00\times {10}^5\text{Pa}$ for $P_i$ in above equation to find $W$.

  $\begin{array}{c}W=-\frac{2}{3}\left(\frac{2.00\times {10}^5\text{Pa}}{{\left(0.514{\text{m}}^3\right)}^{1/2}}\right)\left[{\left(2.06{\text{m}}^3\right)}^{3/2}-{\left(0.514{\text{m}}^3\right)}^{3/2}\right]2.96-0.37\\ =-4.792\times {10}^5\text{J}\\ =-480\times {10}^3\text{J}\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\\ \approx -480\text{kJ}\end{array}$

Therefore, the work done on the air is $-480\text{kJ}$.

(e)

To determine

The energy transferred by the heat.

Answer to Problem 55AP

The energy transferred by the heat is $2.28\text{MJ}$.

Explanation of Solution

Write the expression for internal energy of the diatomic gas.

  $\Delta E_{\mathrm{int}}=n{C}_V\Delta T$                                                                                         (V)

Here, $C_V$ is the specific heat capacity at constant volume and $\Delta T$ is the change in temperature.

Rearrange the equation (V).

  $\Delta E_{\mathrm{int}}=n\left(\frac{5}{2}R\right)\Delta T$                                                                                   (VI)

Write the expression for energy transferred by heat.

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

Here, $Q$ is the amount of heat energy transferred to the system.

Conclusion:

Substitute $41.5\text{mol}$ for $n$, $8.314\text{J/mol}\cdot \text{K}$ for $R$ and $2.38\times {10}^3\text{K}-298\text{K}$ for $\Delta T$ in equation (VI) to find $\Delta E_{\mathrm{int}}$.

  $\begin{array}{c}\Delta E_{\mathrm{int}}=\left(\frac{5}{2}\right)\left(41.5\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(2.38\times {10}^3\text{K}-298\text{K}\right)\\ =\left(\frac{5}{2}\right)\left(41.5\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(2082\text{K}\right)\\ =1.80\times {10}^6\text{J}\end{array}$

Substitute $1.80\times {10}^6\text{J}$ for $\Delta E_{\mathrm{int}}$ and $-480\times {10}^3\text{J}$ for $W$ in equation (VII) to find $Q$.

  $\begin{array}{c}Q=1.80\times {10}^6\text{J}+480\times {10}^3\text{J}\\ =1800\times {10}^3\text{J}+480\times {10}^3\text{J}\\ =2.28\times {10}^6\text{J}\left(\frac{{10}^{-6}\text{MJ}}{1\text{J}}\right)\\ =2.28\text{MJ}\end{array}$

Therefore, the energy transferred by the heat is $2.28\text{MJ}$.