Chapter 20, Problem 48SP

Five moles of neon gas at 2.00 atm and 27.0 °C is adiabatically compressed to one-third its initial volume. Find the final pressure, final temperature, and external work done on the gas. For neon, $\gamma =1.67,c_{\upsilon }=0.148\text{ cal/g}\cdot \text{°C}$, and $M=20.18\text{ kg/kmol}$.

To determine

The final temperature, the final pressure, and the external work done on a neon gas if five moles of gas are adiabatically compressed to one-third of its initial volume.

Answer to Problem 48SP

Solution:

$1.27\text{ MPa}$, $626.8\text{ K}$, and $20.24\text{ kJ}$

Explanation of Solution

Given data:

Five moles of neon gas are initially at $2.00\text{ atm}$ and $27.0°\text{C}$

The neon gas is adiabatically compressed to one-third of its initial volume.

The specific heat of the neon gas at a constant volume is $0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$.

The specific heat ratio of the neon gas is $1.67$.

The molar mass of the neon gas is $20.18\text{ kg/kmol}$.

Formula used:

The formula for specific heat ratio is

$\gamma =\frac{c_p}{c_v}$

Here, $c_p$ is the specific heat of the gas at a constant pressure, $c_v$ is the specific heat of the gas at a constant volume, and $\gamma$ is the specific heat ratio of the gas.

The relation between specific heats and universal gas constant is written as

$c_p-c_v=\frac{R}{M}$

Here, $R$ is the universal gas constant and $M$ is the molar mass of the gas.

The ideal gas equation is written as

$PV=nRT$

Here, $n$ is the number of moles of the gas, $P$ is the pressure of the gas, $V$ is the volume of the gas, and $T$ is the temperature of the gas in Kelvin.

The number of moles are calculated by the formula:

$n=\frac{m}{M}$

Here, $m$ is the mass of the gas.

The formula for conversion of temperature of gas from Celsius scale to Kelvin scale is

$T=\left[t+273\right]\text{ K}$

Here, $T$ is the temperature in Kelvin, and $t$ is the temperature of the gas in Celsius.

The formula for conversion of temperature of gas from Kelvin scale to Celsius scale is

$t=\left[T-273\right]°\text{C}$

The equation for an adiabatic process is written as

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

Here, $P_1$ is the initial pressure before the adiabatic process, and $P_2$ is the final pressure after the adiabatic process. Here, $V_1$ is the initial volume before the adiabatic process and $V_2$ is the final volume after the adiabatic process.

The heat transferred in an adiabatic process is zero. It is expressed as

$\Delta Q=0$

Here, $\Delta Q$ is the heat interaction of the system in an adiabatic process.

The expression for the first law of thermodynamics for an adiabatic process is written as

$0=\Delta U+\Delta W$

Here, $\Delta U$ is the change in internal energy of the system and $\Delta W$ is the work done by the system.

The formula for change in internal energy of a gas is written as

$\Delta U=m{c}_v\Delta T$

Here, $\Delta T$ is the change in temperature of the gas, $m$ is the mass of the gas, and $c_v$ is the specific heat of the gas at constant volume.

Explanation:

Recall the formula for specific heat ratio for neon gas:

$\gamma =\frac{c_p}{c_v}$

Substitute $1.67$ for $\gamma$ and $0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_v$

$\begin{array}{c}1.67=\frac{c_p}{0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)}\\ c_p=\left(1.67\right)\left[0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\\ c_p=0.247\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\end{array}$

Recall the expression relating specific heats and universal gas constant:

$c_p-c_v=\frac{R}{M}$

Substitute $0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_v$, $0.247\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_p$, and $20.18\text{ kg/kmol}$ for $M$

$\begin{array}{c}0.247\text{ cal/}\left(\text{g}\cdot °\text{C}\right)-0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)=\frac{R}{20.18\text{ kg/kmol}}\\ R=0.099\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\end{array}$

Further solve as

$\begin{array}{c}R=\left[0.099\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(\frac{1\text{ g}}{0.001\text{ kg}}\right)\left[20.18\text{ kg/kmol}\right]\left(\frac{1\text{ kmol}}{\text{1000 mol}}\right)\\ =2\text{ cal/}\left(\text{mol}\cdot \text{K}\right)\end{array}$

Recall the expression for conversion of initial temperature of neon gas from Celsius to Kelvin scale:

$T_1=\left[t_1+273\right]\text{ K}$

Here, $t_1$ is the initial temperature of the neon gas in Celsius scale and $T_1$ is the initial temperature of the neon gas before compression in Kelvin scale.

Substitute $27.0°\text{C}$ for $t_1$

$\begin{array}{c}{T}_1=\left[27.0°\text{C}+273\right]\text{ K}\\ =300\text{ K}\end{array}$

Recall the ideal gas equation for initial conditions of neon gas, that is, before compression:

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

The number of moles of neon gas are five. Therefore, substitute $2\text{ cal/}\left(\text{mol}\cdot \text{K}\right)$ for $R$, $300\text{ K}$ for $T_1$, $2.00\text{ atm}$ for $P_1$, and $5\text{ moles}$ for $n$:

$\begin{array}{c}\left(2.00\text{ atm}\right)V_1=\left(5\text{ moles}\right)\left[2\text{ cal/}\left(\text{mol}\cdot \text{K}\right)\right]\left(300\text{ K}\right)\\ V_1=\frac{\left(5\text{ moles}\right)\left[2\text{ cal/}\left(\text{mol}\cdot \text{K}\right)\right]\left(300\text{ K}\right)}{\left(2.00\text{ atm}\right)}\\ V_1=\frac{\left(5\text{ moles}\right)\left[2\text{ cal/}\left(\text{mol}\cdot \text{K}\right)\right]\left(\frac{4.186\text{ J}}{1\text{ cal}}\right)\left(300\text{ K}\right)}{\left[\left(2.00\text{ atm}\right)\left(\frac{1.013\times {10}^5\text{ Pa}}{1\text{ atm}}\right)\right]}\\ =0.06198{\text{ m}}^3\end{array}$

Further solve as

$\begin{array}{c}{V}_1=\left(0.061984{\text{ m}}^3\right)\left(\frac{{10}^6{\text{ cm}}^3}{1{\text{ m}}^3}\right)\\ =61984{\text{ cm}}^3\end{array}$

According to the problem, the final volume of the gas after compression is one third of its initial volume. Therefore,

$V_2=\frac{1}{3}{V}_1$

Substitute $61984{\text{ cm}}^3$ for $V_1$

$\begin{array}{c}{V}_2=\frac{1}{3}\left(61984{\text{ cm}}^3\right)\\ =20661{\text{ cm}}^3\end{array}$

Recall the equation for pressure and volume for an adiabatic process:

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

Substitute $61984{\text{ cm}}^3$ for $V_1$, $20661{\text{ cm}}^3$ for $V_2$, $2.00\text{ atm}$ for $P_1$, and $1.67$ for $\gamma$

$\begin{array}{c}\left(2.00\text{ atm}\right){\left(61984{\text{ cm}}^3\right)}^{1.67}=P_2{\left(20661{\text{ cm}}^3\right)}^{1.67}\\ P_2=\frac{\left(2.00\text{ atm}\right){\left(61984{\text{ cm}}^3\right)}^{1.67}}{{\left(20661{\text{ cm}}^3\right)}^{1.67}}\\ =\frac{\left[\left(2.00\text{ atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{ atm}}\right)\left(\frac{{10}^{-6}\text{ MPa}}{1\text{ Pa}}\right)\right]{\left(61984{\text{ cm}}^3\right)}^{1.67}}{{\left(20661{\text{ cm}}^3\right)}^{1.67}}\\ \simeq 1.27\text{ MPa}\end{array}$

The final pressure after adiabatic compression is $1.27\text{ MPa}$.

Recall the ideal gas equation for gas after adiabatic compression:

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

Here, $T_2$ is the temperature of the gas after adiabatic compression.

Substitute $1.27\text{ MPa}$ for $P_2$, $20661{\text{ cm}}^3$ for $V_2$, $2\text{ cal/}\left(\text{mol}\cdot \text{K}\right)$ for $R$, and $5\text{ moles}$ for $n$

$\begin{array}{c}\left(1.27\text{ MPa}\right)\left(20661{\text{ cm}}^3\right)=\left(5\text{ moles}\right)\left[2\text{ cal/}\left(\text{mol}\cdot \text{K}\right)\right]T_2\\ \left[\left(1.27\text{ MPa}\right)\left(\frac{{10}^6\text{ Pa}}{1\text{ MPa}}\right)\right]\left[\left(20661{\text{ cm}}^3\right)\left(\frac{{10}^{-6}{\text{ m}}^3}{1{\text{ cm}}^3}\right)\right]=\left(5\text{ moles}\right)\left[2\text{ cal/}\left(\text{mol}\cdot \text{K}\right)\right]\left(\frac{4.186\text{ J}}{1\text{ cal}}\right)T_2\\ \left(1270000\text{ Pa}\right)\left(0.020661{\text{ m}}^3\right)=83.72\text{ J}\\ \left(1270000\right)\left(0.020661\right)=5\left(8.372\right)T_2\\ T_2=626.8\text{ K}\end{array}$

The final temperature of the gas after adiabatic compression is $626.8\text{ K}$.

Recall the expression for the final temperature of the gas from Celsius to Kelvin scale:

$T_2=\left[t_2+273\right]\text{ K}$

Here, $t_2$ is the final temperature of the gas after adiabatic compression in Celsius.

Rearrange for $t_2$

$t_2=\left[T_2-273\right]°\text{C}$

Substitute $626.8\text{ K}$ for $T_2$

$\begin{array}{c}{t}_2=\left[626.8\text{ K}-273\right]\text{ K}\\ =353.8°\text{C}\end{array}$

Calculate the change in temperature of the gas:

$\Delta T=t_2-t_1$

Substitute $353.8°\text{C}$ for $t_2$ and $27.0°\text{C}$ for $t_1$

$\begin{array}{c}\Delta T=353.8°\text{C}-27.0°\text{C}\\ =\text{326}\text{.8}°\text{C}\end{array}$

Recall the expression for number of moles of gas:

$n=\frac{m}{M}$

Substitute $20.18\text{ kg/kmol}$ for $M$ and $5\text{ moles}$ for $n$

$\begin{array}{c}5\text{ moles}=\frac{m}{20.18\text{ kg/kmol}}\\ m=5\text{ moles}\left(20.18\frac{\text{kg}}{\text{kmol}}\right)\\ m=5\text{ moles}\left[20.18\frac{\text{kg}}{\text{kmol}}\left(\frac{1\text{ kmol}}{1000\text{ moles}}\right)\right]\\ m=0.10\text{ kg}\end{array}$

Recall the formula for change in internal energy of the gas:

$\Delta U=m{c}_v\Delta T$

Substitute $\text{326}\text{.8}°\text{C}$ for $\Delta T$, $0.10\text{ kg}$ for $m$, and $0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_v$

$\begin{array}{c}\Delta U=\left(0.10\text{ kg}\right)\left[0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(\text{326}\text{.8}°\text{C}\right)\\ =\left[0.10\text{ kg}\left(\frac{1000\text{ g}}{1\text{ kg}}\right)\right]\left[0.148\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(\text{326}\text{.8}°\text{C}\right)\\ =100\left(0.148\right)\left(326.8\right)\\ =4836\text{ cal}\end{array}$

Recall the expression for the first law of thermodynamics for adiabatic process:

$\begin{array}{c}0=\Delta U+\Delta W\\ \Delta W=-\Delta U\end{array}$

Substitute $4836\text{ cal}$ for $\Delta U$

$\begin{array}{c}\Delta W=-4836\text{ cal}\\ =\left[-4836\text{ cal}\left(\frac{\text{0}\text{.004186 kJ}}{\text{1 cal}}\right)\right]\\ =-20.24\text{ kJ}\end{array}$

The negative sign indicates that the work is done on the system.

Therefore, the work done on the system is $20.24\text{ kJ}$.

Conclusion:

The final pressure is $1.27\text{ MPa}$. The final temperature is $626.8\text{ K}$. The external work done on the gas is $20.24\text{ kJ}$.