Chapter 17, Problem 43P

(a)

To determine

The specific heat of air at constant volume.

Answer to Problem 43P

The specific heat of air at constant volume is $719.25\text{J}/\text{kg}°\text{C}$

Explanation of Solution

Given Info: Temperature of air is $300\text{ K}$, Initial pressure is $2.00\times {10}^5\text{ Pa}$, Initial volume is $0.350{\text{ m}}^3$, molar mass of air is $28.9\text{g}/\text{mol}$ and specific heat at constant volume is $\frac{5}{2}R$ .

Formula for calculating the gas constant is,

$\overline{R}=\frac{R}{M}$

Here,

$\overline{R}$ is the gas constant in $\text{J}/\text{kg}°\text{C}$

$R$ is the universal gas constant

$M$ is the molar mass of the gas

Substitute $8.314\text{J}/\text{molK}$ for $R$ and $28.9\text{g}/\text{mol}$ for $M$ to get $\overline{R}$,

$\begin{array}{c}\overline{R}=\frac{\left(8.314\text{J}/\text{molK}\right)}{\left(28.9\text{g}/\text{mol}\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)}\\ =287.7\text{J}/\text{kg}°\text{C}\end{array}$

The specific heat at constant volume in $\text{J}/\text{kg}$ can be given as,

$C_{\text{V}}=\frac{5}{2}\overline{R}$

Substitute $287.7\text{J}/\text{kg}°\text{C}$ for $\overline{R}$ in the above equation,

$\begin{array}{c}{C}_{\text{V}}=\frac{5}{2}\left(287.7\text{J}/\text{kg}°\text{C}\right)\\ =719.25\text{J}/\text{kg}°\text{C}\end{array}$

Thus, the specific heat at constant volume is $719.25\text{J}/\text{kg}°\text{C}$

Conclusion:

Therefore, the specific heat at constant volume is $719.25\text{J}/\text{kg}°\text{C}$

(b)

To determine

The mass of the air in the cylinder.

Answer to Problem 43P

The mass of the air in the cylinder is $809.2\text{g}$

Explanation of Solution

Given Info Temperature of air is $300\text{ K}$, Initial pressure is $2.00\times {10}^5\text{ Pa}$, Initial volume is $0.350{\text{ m}}^3$, molar mass of air is $28.9\text{g}/\text{mol}$ and specific heat at constant volume is $\frac{5}{2}R$ .

According to the ideal gas equation,

$PV=nRT$

Here,

$P$ is the pressure of the gas

$V$ is the volume of the vessel

$n$ is the number of moles

$R$ is the gas constant

$T$ is the temperature of the gas

Rearrange the above equation for $n,$

$n=\frac{PV}{RT}$

Substitute $2.00\times {10}^5\text{ Pa}$ for $P$, $0.350{\text{ m}}^3$ for $V$, $8.314\text{J}/\text{molK}$ for $R$ and $300\text{ K}$ for $T$ in the above equation to get $n$

$\begin{array}{c}n=\frac{\left(2.00\times {10}^5\text{ Pa}\right)\left(0.350{\text{ m}}^3\right)}{\left(8.314\text{J}/\text{molK}\right)\left(300\text{ K}\right)}\\ =28\text{ mol}\end{array}$

The number of moles of a gas is,

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

Here,

$m$ is the mass of the air

$M$ is the molar mass of the air

Substitute $28\text{ mol}$ for $n$ and $28.9\text{g}/\text{mol}$ for $M$ in the above equation,

$\begin{array}{c}\left(28\text{ mol}\right)=\frac{m}{\left(28.9\text{g}/\text{mol}\right)}\\ m=809.2\text{g}\end{array}$

Thus, the mass of the air is $809.2\text{ g}$

Conclusion:

Therefore, the mass of the air will be $809.2\text{ g}$

(c)

To determine

The energy input required to raise the temperature to $700\text{K}$

Answer to Problem 43P

The energy input required to raise the temperature to $700\text{K}$ is $\text{232}\text{.8}\text{kJ}$

Explanation of Solution

Given Info: Temperature of air is $300\text{ K}$, Initial pressure is $2.00\times {10}^5\text{ Pa}$, Initial volume is $0.350{\text{ m}}^3$, molar mass of air is $28.9\text{g}/\text{mol}$ and specific heat at constant volume is $\frac{5}{2}R$, final temperature of the air is $700\text{K}$ and piston is held fixed.

As the piston is held fixed, therefore, the process will be of constant volume.

The energy input for the given condition is given as,

$\begin{array}{c}Q=m{C}_{\text{V}}\Delta T\\ =m{C}_{\text{V}}\left(T_{\text{f}}-T_{\text{i}}\right)\end{array}$

Here,

$Q$ is the energy input

$m$ is the mass of the air

$C_{\text{V}}$ is the specific heat at constant volume

$\Delta T$ is the temperature difference

$T_{\text{f}}$ is the final temperature of the air

$T_{\text{i}}$ is the initial temperature of the air

Substitute $719.25\text{J}/\text{kgK}$ for $C_{\text{V}}$, $809.2\text{ g}$ for $m$, $700\text{K}$ for $T_{\text{f}}$ and $300\text{K}$ for $T_{\text{i}}$ to get $Q$ in the above equation,

$\begin{array}{c}Q=\left[\left(809.2\text{ g}\right)\left(\frac{1\text{kg}}{1000\text{g}}\right)\right]\left(719.25\text{J}/\text{kgK}\right)\left(700-300\right)\text{K}\\ \text{=232806}\text{.84}\text{J}\\ \text{=232}\text{.8}\text{kJ}\end{array}$

Thus, the energy input required is $\text{232}\text{.8}\text{kJ}$.

Conclusion:

Therefore, the energy input required will be $\text{232}\text{.8}\text{kJ}$.

(d)

To determine

The energy input required to raise the temperature to $700\text{K}$

Answer to Problem 43P

The energy input required to raise the temperature to $700\text{K}$ is $\text{325}\text{.93}\text{kJ}$

Explanation of Solution

Given Info: Temperature of air is $300\text{ K}$, Initial pressure is $2.00\times {10}^5\text{ Pa}$, Initial volume is $0.350{\text{ m}}^3$, molar mass of air is $28.9\text{g}/\text{mol}$ and specific heat at constant volume is $\frac{5}{2}R$, final temperature of the air is $700\text{K}$ and piston is free to move.

As the piston is free to move, therefore, the process will be of constant pressure.

The energy input for the given condition is given as,

$\begin{array}{c}Q=m{C}_P\Delta T\\ =m{C}_P\left(T_{\text{f}}-T_{\text{i}}\right)\end{array}$ (1)

Here,

$Q$ is the energy input

$m$ is the mass of the air

$C_P$ is the specific heat at constant pressure

$\Delta T$ is the temperature difference

$T_{\text{f}}$ is the final temperature of the air

$T_{\text{i}}$ is the initial temperature of the air

The expression for gas constant for any ideal gas can be given as,

$C_P-C_{\text{V}}=\overline{R}$

Substitute $719.25\text{J}/\text{kgK}$ for $C_{\text{V}}$ and $287.7\text{J}/\text{kgK}$ for $\overline{R}$ in the above equation,

$\begin{array}{c}{C}_P-\left(719.25\text{J}/\text{kgK}\right)=\left(287.7\text{J}/\text{kgK}\right)\\ C_P=1006.95\text{J}/\text{kgK}\end{array}$

Substitute $1006.95\text{J}/\text{kgK}$ for $C_P$, $809.2\text{ g}$ for $m$, $700\text{K}$ for $T_{\text{f}}$ and $300\text{K}$ for $T_{\text{i}}$ to get $Q$ in the equation (1),

$\begin{array}{c}Q=\left[\left(809.2\text{ g}\right)\left(\frac{1\text{kg}}{1000\text{g}}\right)\right]\left(1006.95\text{J}/\text{kgK}\right)\left(700-300\right)\text{K}\\ \text{=325929}\text{.576}\text{J}\\ \text{=325}\text{.93}\text{kJ}\end{array}$

Thus, the energy input required is $\text{325}\text{.93}\text{kJ}$.

Conclusion:

Therefore, the energy input required will be $\text{325}\text{.93}\text{kJ}$.