Chapter 21, Problem 21.55AP

Model air as a diatomic ideal gas with M = 28.9 g/mol. A cylinder with a piston contains 1.20 kg of air at 25.0°C and 2.00 × 10⁵ Pa. Energy is transferred by heat into the system as it is permitted to expand, with the pressure rising to 4.00 × 10⁵ Pa. Throughout the expansion, the relationship between pressure and volume is given by P = CV^1/2 where C is a constant. Find (a) the initial volume, (b) the final volume, (c) the final temperature, (d) the work done on the air, and (e) the energy transferred by heat.

To determine

The initial volume.

Answer to Problem 21.55AP

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

Explanation of Solution

Given info: Molar mass is $28.9\text{g}/\text{mol}$, mass of gas is $1.20\text{kg}$, temperature is $25.0°\text{C}$, initial pressure is $2.00\times {10}^5\text{Pa}$ and final pressure is $4.00\times {10}^5\text{Pa}$.

Write the expression for ideal gas equation.

$P_1{V}_1=nR{T}_1$

Here,

$P_1$ is initial pressure.

$V_1$ is initial volume.

$n$ is no. of molecule of gas.

$R$ is universal gas constant.

$T_{\text{1}}$ is initial temperature.

Reaarange above equation for $V_{\text{1}}$.

$V_1=\frac{nR{T}_1}{P_1}$ (2

Write the expression for number of moles of gas.

$n=\frac{m_{\text{g}}}{M}$ (3)

Here,

$m_{\text{g}}$ is mass of gas.

$M$ is molar mass.

Substitute $1.20\text{kg}$ for $m_{\text{g}}$ and $28.9\text{g}/\text{mol}$ for $M$ in equation (3).

$\begin{array}{c}n=\frac{1.20\text{kg}}{28.9\text{g}/\text{mol}\times \frac{1\text{kg}}{1000\text{g}}}\\ =41.5\text{mol}\end{array}$

Thus, the number of moles of gas is $41.5\text{mol}$.

Substitute $41.5\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$, $25.0°\text{C}$ for $T_1$ and $2.00\times {10}^5\text{Pa}$ in equation (2).

$\begin{array}{c}{V}_1=\frac{\left(41.5\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(25.0+273\text{K}\right)}{2.00\times {10}^5\text{Pa}}\\ =0.514\left(\frac{\text{J}\times \frac{1\text{N}\cdot \text{m}}{1\text{J}}}{\text{Pa}\times \frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}}\right)\\ =0.514{\text{m}}^3\end{array}$

Conclusion:

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

To determine

The final volume.

Answer to Problem 21.55AP

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

Explanation of Solution

Given info: Molar mass is $28.9\text{g}/\text{mol}$, mass of gas is $1.20\text{kg}$, temperature is $25.0°\text{C}$, initial pressure is $2.00\times {10}^5\text{Pa}$ and final pressure is $4.00\times {10}^5\text{Pa}$.

Write the expression for the relationship between pressure and volume for adiabatic process for an ideal gas:

$\frac{P_1}{\sqrt{V_1}}=\frac{P_2}{\sqrt{V_2}}$

Rearrange equation (4) for $V_2$.

$\frac{\sqrt{V_2}}{\sqrt{V_1}}=\frac{P_2}{P_1}$

Square both side in above equation.

$\begin{array}{c}{\left(\frac{\sqrt{V_2}}{\sqrt{V_1}}\right)}^2={\left(\frac{P_2}{P_1}\right)}^2\\ V_2=V_1{\left(\frac{P_2}{P_1}\right)}^2\end{array}$ (4)

Substitute $0.514{\text{m}}^3$ for $V_1$, $2.00\times {10}^5\text{Pa}$ for $P_1$ and $4.00\times {10}^5\text{Pa}$ in equation (4).

$\begin{array}{c}{V}_2=\left(0.514{\text{m}}^3\right){\left(\frac{4.00\times {10}^5\text{Pa}}{2.00\times {10}^5\text{Pa}}\right)}^2\\ =2.06{\text{m}}^3\end{array}$

Conclusion:

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

To determine

The final temperature.

Answer to Problem 21.55AP

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

Explanation of Solution

Given info: Molar mass is $28.9\text{g}/\text{mol}$, mass of gas is $1.20\text{kg}$, temperature is $25.0°\text{C}$, initial pressure is $2.00\times {10}^5\text{Pa}$ and final pressure is $4.00\times {10}^5\text{Pa}$.

Write the expression for ideal gas equation.

$P_2{V}_2=nR{T}_2$

Here,

$P_2$ is final pressure.

$V_2$ is final volume.

$n$ is no. of molecule of gas.

$R$ is universal gas constant.

$T_2$ is final temperature.

Rearrange above equation for $T_2$.

$T_2=\frac{P_2{V}_2}{nR}$ (5)

Substitute $4.00\times {10}^5\text{Pa}$ for $P_2$, $2.06{\text{m}}^3$ for $V_2$ as calculated in above part, $41.5\text{mol}$ for $n$ as calculated in above part and $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (5).

$\begin{array}{c}{T}_2=\frac{\left(4.00\times {10}^5\text{Pa}\times \frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\left(2.06{\text{m}}^3\right)}{\left(41.5\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{k}\times \frac{1\text{N}\cdot \text{m}}{1\text{J}}\right)}\\ =2.38\times {10}^3\text{K}\end{array}$

Conclusion:

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

To determine

The workdone on the air.

Answer to Problem 21.55AP

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

Explanation of Solution

Given info: Molar mass is $28.9\text{g}/\text{mol}$, mass of gas is $1.20\text{kg}$, temperature is $25.0°\text{C}$, initial pressure is $2.00\times {10}^5\text{Pa}$ and final pressure is $4.00\times {10}^5\text{Pa}$.

Write the expression for work done.

$W={\displaystyle {\int }_{V_2}^{V_1}PdV}$ (6)

Here,

$W$ is work done.

$P$ is pressure.

$V$ is volume.

Write the expression for relationship between pressure and volume:

$P=C{V}^{\frac{1}{2}}$

Here,

$C$ is constant.

Rewrite above expression for $C$.

$C=\frac{P}{V^{\frac{1}{2}}}$

Substitute $C{V}^{1/2}$ for $P$ in equation (6).

$\begin{array}{c}W={\displaystyle {\int }_{V_2}^{V_1}C{V}^{\frac{1}{2}}dV}\\ =C{\left|\frac{V^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right|}_{V_2}^{V_1}\\ =\frac{2C}{3}\left(V_1^{\frac{3}{2}}-V_2^{\frac{3}{2}}\right)\end{array}$

Substitute $\frac{P}{V^{\frac{1}{2}}}$ for $C$ in above expression.

$W=\frac{2}{3}\left(\frac{P_1}{V_1^{\frac{1}{2}}}\right)\left(V_1^{\frac{3}{2}}-V_2^{\frac{3}{2}}\right)$ (7)

Substitute $2.00\times {10}^5\text{Pa}$ for $P_1$, $0.514{\text{m}}^3$ for $V_1$ and $2.06{\text{m}}^3$ for $V_2$ in above equation.

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

Conclusion:

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

To determine

The energy transferred by heat.

Answer to Problem 21.55AP

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

Explanation of Solution

Given info: Molar mass is $28.9\text{g}/\text{mol}$, mass of gas is $1.20\text{kg}$, temperature is $25.0°\text{C}$, initial pressure is $2.00\times {10}^5\text{Pa}$ and final pressure is $4.00\times {10}^5\text{Pa}$.

The molar specific heat at constant pressure is,

$c_{\text{P}}=\frac{5R}{2}$

Write the expression for change in internal energy.

$\Delta E_{\text{i}}=n{c}_{\text{p}}\Delta T$

Here,

$\Delta E_{\text{i}}$ is change in internal energy.

$n$ is no. of molecule of gas.

$c_{\text{p}}$ is molar specific heat at constant pressure.

$\Delta T$ is change in temperature.

Substitute $\frac{5R}{2}$ for $c_{\text{p}}$ and $\left(T_2-T_1\right)$ in above equation.

$\Delta E_{\text{i}}=n\frac{5R}{2}\left(T_2-T_1\right)$ (8)

Here,

$R$ is universal gas constant.

$T_2$ is final temperature.

$T_1$ is initial temperature.

Substitute $41.5\text{mol}$ for $n$ as calculated in above part, $8.14\text{J}/\text{mol}\cdot \text{K}$ for $R$, $2.38\times {10}^3\text{K}$ for $T_2$ and $298\text{K}$ for $T_1$ in equation (8).

$\begin{array}{c}\Delta E_{\text{i}}=\frac{\left(41.5\text{mol}\right)\left(5\times 8.14\text{J}/\text{mol}\cdot \text{K}\right)\left(2.38\times {10}^3\text{K}-298\text{K}\right)}{2}\\ =1.80\times {10}^6\text{J}\end{array}$

Thus, the change in internal energy is $1.80\times {10}^6\text{J}$.

Write the expression for the energy transferred as heat.

$Q=\Delta E_{\text{i}}-W$ (9)

Here,

$Q$ is energy transferred as heat.

$\Delta E_{\text{i}}$ is change in internal energy.

$W$ is work done.

Substitute $1.80\times {10}^6\text{J}$ for $\Delta E_{\text{i}}$ and $-480\text{kJ}$ for $W$ as calculated in above parts in equation (9).

$\begin{array}{c}Q=1.80\times {10}^6\text{J}-\left(-480\text{kJ}\times \frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ =2.276\times {10}^6\text{J}\times \frac{1\text{MJ}}{{10}^6\text{J}}\\ =2.28\text{MJ}\end{array}$

Conclusion:

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