Chapter 21, Problem 47PQ

(a)

To determine

The temperature at B and C.

Answer to Problem 47PQ

The temperature at B is $\underset{\_}{1.20\times {10}^2\text{K}}$ and the temperature at C is $\underset{\_}{1.64\times {10}^2\text{K}}$.

Explanation of Solution

Given that the pressure at point A is $2.00\times {10}^5\text{Pa}$, and at B is $1.00\times {10}^5\text{Pa}$. The volume of as at A is $0.0550{\text{m}}^3$, and C is $0.0750{\text{m}}^3$. Also the number of moles of the gas is $5.50$.

Write the expression for the temperature at $B$ from ideal gas equation.

  $T_B=\frac{P_B{V}_B}{nR}$                                                                                                (I)

Here, $T_B$ is the temperature at $B$, $P_B$ is the pressure at $B$, $V_B$ is the volume at $B$, $n$ is the number of moles of gas, and $R$ is the universal gas constant.

Write the expression for the temperature at C from ideal gas equation.

  $T_C=\frac{P_C{V}_C}{nR}$                                                                                                   (II)

Here, $T_C$ is the temperature at $C$ , $P_C$ is the pressure at $C$ , and $V_C$ is the volume at $C$.

Conclusion:

Substitute $1.00\times {10}^5\text{Pa}$ for $P_B$, $0.0550{\text{m}}^3$ for $V_B$, $5.50\text{mol}$ for $n$, and $8.315\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (I) to find $T_B$.

  $\begin{array}{c}{T}_B=\frac{\left(1.00\times {10}^5\text{Pa}\right)\left(0.0550{\text{m}}^3\right)}{\left(5.50\text{mol}\right)\left(8.315\text{J}/\text{mol}\cdot \text{K}\right)}\\ =1.20\times {10}^2\text{K}\end{array}$

Point $B$ and $C$ are lying on the same plane. Here the pressure is constant at both $B$ and $C$.

Substitute $1.00\times {10}^5\text{Pa}$ for $P_C$, $0.0750{\text{m}}^3$ for $V_C$, $5.50\text{mol}$ for $n$, and $8.315\text{J}/\text{mol}\cdot \text{K}$ for $R$ in equation (II) to find $T_C$.

  $\begin{array}{c}{T}_C=\frac{\left(1.00\times {10}^5\text{Pa}\right)\left(0.0750{\text{m}}^3\right)}{\left(5.50\text{mol}\right)\left(8.315\text{J}/\text{mol}\cdot \text{K}\right)}\\ =1.64\times {10}^2\text{K}\end{array}$

Therefore, the temperature at B is $\underset{\_}{1.20\times {10}^2\text{K}}$ and the temperature at C is $\underset{\_}{1.64\times {10}^2\text{K}}$.

(b)

To determine

The total work done on the gas during one cycle.

Answer to Problem 47PQ

The total work done on the gas during one cycle is $\underset{\_}{1.00\times {10}^3\text{J}}$.

Explanation of Solution

The total work done is equivalent to the area under the $PV$ curve. The area enclosed is of a triangular area.

Write the expression for the area of triangle.

  $A=\frac{1}{2}bh$                                                                                               (III)

Here, $A$ is the area of the triangle, $b$ is the base of the triangle, and $h$ is the height of the triangle.

Write the expression for the base of the triangle.

  $b=V_2-V_1$                                                                                                   (IV)

Here, $V_2$ is the final volume of the gas, and $V_1$ is the initial volume of gas.

Write the expression for height of the triangle.

  $h=P_2-P_1$                                                                                                                (V)

Here, $P_2$ is the final pressure of the gas, and $P_1$ is the initial pressure.

Use equations (IV) and (V) in expression (III)

  $A=\frac{1}{2}\left(V_2-V_1\right)\left(P_2-P_1\right)$                                                                                (VI)

Conclusion:

Substitute $0.0750{\text{m}}^3$ for $V_2$, $0.0550{\text{m}}^3$ for $V_1$, $2.00\times {10}^5\text{Pa}$ for $P_2$, and $1.00\times {10}^5\text{Pa}$ for $P_1$ in equation (IV) to find $A$.

  $\begin{array}{c}A=\frac{1}{2}\left(0.0750{\text{m}}^2-0.0550{\text{m}}^3\right)\left(2.00\times {10}^5\text{Pa}-1.00\times {10}^5\text{Pa}\right)\\ =1.00\times {10}^3\text{J}\end{array}$

Therefore, the total work done on the gas during one cycle is $\underset{\_}{1.00\times {10}^3\text{J}}$.

(c)

To determine

The amount of heat transferred when the gas goes from B to C.

Answer to Problem 47PQ

The amount of heat transferred when the gas goes from B to C is $\underset{\_}{5.02\times {10}^3\text{J}}$.

Explanation of Solution

Write the expression for the change in thermal energy of the gas.

  $\Delta E_{\text{th}}=\frac{3}{2}nR\left(T_2-T_1\right)$                                                                                     (VII)

Here, $\Delta E_{\text{th}}$ is the change in thermal energy of the gas, $n$ is the number of moles of gas, $R$ is the universal gas constant, $T_2$ is the final temperature, and $T_1$ is the initial temperature.

The process done between points B and C is adiabatic in nature.

Write the expression for the work done in an isobaric process.

  $W=-P\left(V_2-V_1\right)$                                                                                           (VIII)

Here, $W$ is the work done, $P$ is the pressure, $V_2$ is the final volume, and $V_1$ is the initial volume.

Apply first law of thermodynamics. The heat flowing through $B$ and $C$ is equal to the difference between the change in thermal energy and the work done.

Write the expression for the heat flowing between $B$ and $C$.

  $Q=\Delta E_{\text{th}}-W$                                                                                             (IX)

Conclusion:

Substitute $1.00\times {10}^5\text{Pa}$ for $P$, $0.0750{\text{m}}^3$ for $V_2$, and $0.0550{\text{m}}^3$ for $V_1$ in equation (VII) to find $W$.

  $\begin{array}{c}W=-\left(1.00\times {10}^5\text{Pa}\right)\left(0.0750{\text{m}}^3-0.0550{\text{m}}^3\right)\\ =-2.00\times {10}^3\text{J}\end{array}$

Substitute $3.02\times {10}^3\text{J}$ for $\Delta E_{\text{th}}$, and $-2.00\times {10}^3\text{J}$ for $W$ in equation (IX) for $Q$.

  $\begin{array}{c}Q=3.02\times {10}^3\text{J}-\left(-2.00\times {10}^3\text{J}\right)\\ =5.02\times {10}^3\text{J}\end{array}$

Therefore, the amount of heat transferred when the gas goes from B to C is $\underset{\_}{5.02\times {10}^3\text{J}}$.