Chapter 15, Problem 13P

(a)

To determine

The temperature and pressure at point $D$.

Answer to Problem 13P

The pressure at point $D$ is $8.87\text{kPa}$. The temperature at point $D$ is $1200\text{K}$.

Explanation of Solution

There is $2.00\text{mol}$ of ideal gas in the refrigerator.

The cycle of the ideal gas is shown below in figure 1.

The pressure at point $A$ and $D$ are same and is $P_2$. The pressure at point $B$ and $C$ are same and is $1.30\text{kPa}$. The volume at $A$ and $B$ is same and is $1.50{\text{m}}^{\text{3}}$. The volume at $C$ and $D$ is same and is $2.25{\text{m}}^{\text{3}}$. The temperature at point $A$ is $800.0\text{K}$.

Write the formula for the pressure at point $D$ and $A$.

$P_2=\frac{nR{T}_A}{V_A}$ (I)

Here, $P_2$ is the pressure at point $D$ and $A$, $n$ is the number of moles, $R$ is the universal gas constant, $T_A$ is the temperature at point $A$, $V_A$ is the volume at point $A$.

Write the formula for the temperature at point $D$.

$T_D=\frac{P_2{V}_D}{nR}$ (II)

Here, $T_D$ is the temperature at point $D$, $V_D$ is the volume at point $D$.

Conclusion:

Substitute $2.00\text{mol}$ for $n$ , $8.314\text{J/mol}\cdot \text{K}$ for $R$, $800.0\text{K}$ for $T_A$, $1.50{\text{m}}^{\text{3}}$ for $V_A$ in equation (I).

$\begin{array}{c}{P}_2=\frac{\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(800.0\text{K}\right)}{1.50{\text{m}}^{\text{3}}}\\ =8.87\text{kPa}\end{array}$

Substitute $8.87\text{kPa}$ for $P_2$, $2.00\text{mol}$ for $n$ , $8.314\text{J/mol}\cdot \text{K}$ for $R$, $2.25{\text{m}}^{\text{3}}$ for $V_D$ in equation (II).

$\begin{array}{c}{T}_D=\frac{\left(8.87\text{kPa}\right)\left(2.25{\text{m}}^{\text{3}}\right)}{\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)}\\ =1200\text{K}\end{array}$

The pressure at point $D$ is $8.87\text{kPa}$. The temperature at point $D$ is $1200\text{K}$.

(b)

To determine

The net work done on the gas if the gas completes 4 cycles.

Answer to Problem 13P

The net work done on the gas if the gas completes 4 cycles is $23\text{kJ}$.

Explanation of Solution

There is $2.00\text{mol}$ of ideal gas in the refrigerator.

The cycle of the ideal gas is shown below in figure 1.

The pressure at point $A$ and $D$ are same and is $P_2$. The pressure at point $B$ and $C$ are same and is $1.30\text{kPa}$. The volume at $A$ and $B$ is same and is $1.50{\text{m}}^{\text{3}}$. The volume at $C$ and $D$ is same and is $2.25{\text{m}}^{\text{3}}$. The temperature at point $A$ is $800.0\text{K}$.

Write the formula for the work done in four cycle of the gas.

$W=4\left(P_2-P_1\right)\left(V_2-V_1\right)$ (III)

Here, $P_2$ is the pressure at point $D$ and $A$, $P_1$ is pressure at point $B$ and $C$, $V_1$ is volume at  $A$ and $B$, $V_2$ is volume at $C$ and $D$.

Conclusion:

Substitute $1.50{\text{m}}^{\text{3}}$ for $V_1$, $2.25{\text{m}}^{\text{3}}$ for $V_2$, $1.30\text{kPa}$ for $P_1$, $8.87\text{kPa}$ for $P_2$ in equation (III).

$\begin{array}{c}W=4\left(8.87\text{kPa}-1.30\text{kPa}\right)\left(2.25{\text{m}}^{\text{3}}-1.50{\text{m}}^{\text{3}}\right)\\ =23\text{kJ}\end{array}$

The net work done on the gas if the gas completes 4 cycles is $23\text{kJ}$.

(c)

To determine

The internal energy of the gas at point $A$.

Answer to Problem 13P

The internal energy of the gas at point $A$ is $20.0\text{kJ}$.

Explanation of Solution

There is $2.00\text{mol}$ of ideal gas in the refrigerator.

The cycle of the ideal gas is shown below in figure 1.

The pressure at point $A$ and $D$ are same and is $P_2$. The pressure at point $B$ and $C$ are same and is $1.30\text{kPa}$. The volume at $A$ and $B$ is same and is $1.50{\text{m}}^{\text{3}}$. The volume at $C$ and $D$ is same and is $2.25{\text{m}}^{\text{3}}$. The temperature at point $A$ is $800.0\text{K}$.

Write the formula for the internal energy at point $A$.

$U=\frac{3}{2}nRT$ (IV)

Here, $U$ is the internal energy at $A$, $n$ is the number of moles, $R$ is the universal gas constant, $T$ is the temperature at point $A$.

Conclusion:

Substitute $2.00\text{mol}$ for $n$ , $8.314\text{J/mol}\cdot \text{K}$ for $R$, $800.0\text{K}$ for $T$ in equation (IV).

$\begin{array}{c}U=\frac{3}{2}\left(2.00\text{mol}\right)\left(8.314\text{J/mol}\cdot \text{K}\right)\left(800.0\text{K}\right)\\ =20.0\text{kJ}\end{array}$

The internal energy of the gas at point $A$ is $20.0\text{kJ}$.

(d)

To determine

The total change in internal energy of the gas during the four cycles.

Answer to Problem 13P

There is no change in internal energy of the gas through the complete four cycles.

Explanation of Solution

There is $2.00\text{mol}$ of ideal gas in the refrigerator.

The pressure at point $A$ and $D$ are same and is $P_2$. The pressure at point $B$ and $C$ are same and is $1.30\text{kPa}$. The volume at $A$ and $B$ is same and is $1.50{\text{m}}^{\text{3}}$. The volume at $C$ and $D$ is same and is $2.25{\text{m}}^{\text{3}}$. The temperature at point $A$ is $800.0\text{K}$.

For ideal gases, change in internal energy depends on the change in temperature only. If the temperature remains same, there is no change in internal energy.

Since in the course of the complete four cycles of the gas, there is no change in temperature of the gas, there is no change in the internal energy.

Conclusion:

There is no change in internal energy of the gas through the complete four cycles.