Chapter 21, Problem 42PQ

(a)

To determine

To draw a PV diagram for the gas.

Answer to Problem 42PQ

The PV diagram for the gas is given below.

Explanation of Solution

It is given that pressure of the gas is $2.50\times {10}^5\text{Pa}$ , temperature of the gas is $295\text{K}$ and the gas isothermally expands from $1.25{\text{m}}^3$ to $2.75{\text{m}}^3$ .

The following figure gives the PV diagram for the gas.

In above figure, point 1 represents the initial state. The gas follows isothermal expansion from $1.25{\text{m}}^3$ to $2.75{\text{m}}^3$ which is shown by the curve 12 labeled isothermal. Thereafter gas undergoes adiabatic compression which is shown by curve 23 labeled Isobaric and curve31 is the isochoric process in which volume is constant.

Conclusion:

Thus, the PV diagram for the gas is given below.

(b)

To determine

The change in thermal energy.

Answer to Problem 42PQ

The change in thermal energy of the gas is zero.

Explanation of Solution

From figure its clear that the gas undergoes a cyclic process. A cyclic process is a thermodynamic process in which the system returns to its original state after exchange of energy as a result of heat and work. The system variables including thermal energy must return to original values. Therefore, there is no change in system’s thermal energy:

In this problem it is given that the system return’s to its original state. Thus following a cyclic process, total change in thermal energy of the gas is zero.

Conclusion:

Therefore, the change in thermal energy of the gas is zero.

(c)

To determine

The work done by environment on the gas.

Answer to Problem 42PQ

The work done by environment on the gas is $-\text{7}\text{.56}\times {\text{10}}^4\text{J}$ .

Explanation of Solution

The net work done by the gas is the area inside the curve. In isothermal process volume of the system is raised from $1.25{\text{m}}^3$ to $2.75{\text{m}}^3$ at constant temperature. For this isothermal part, the gas performed work on the environment.

Write the expression for the work done during isothermal expansion.

  $W_{\text{isothermal}}=N{k}_{\text{B}}T\ln \frac{V_1}{V_2}$                                                                                              (I)

Here, $W_{\text{isothermal}}$ is the work done by the gas during isothermal process, $N$ is the number of gas molecules, $k_{\text{B}}$ is the Boltzmann constant, $T$ is the temperature, $V_1$ is the volume before isothermal process and $V_2$ is the volume reached after isothermal process.

Write ideal gas equation.

  $PV=N{k}_{\text{B}}T$

Here, $P$ is the pressure of gas, $V$ is the volume of the gas .

Use above equation for the gas at initial state (state 1 in figure).

  $P_1{V}_1=N{k}_{\text{B}}T$                                                                                                             (II)

Here, $P_1$ is the initial pressure and $V_1$ is the initial volume.

Substitute $P_1{V}_1$ for $N{k}_{\text{B}}T$ in equation (I) to get expression of $W_{\text{isothermal}}$ .

  $W_{\text{isothermal}}=P_1{V}_1\ln \frac{V_1}{V_2}$                                                                                                (III)

For isothermal process, $PV=\text{constant}$.

Apply above equation for the isothermal process of the gas.

  $P_1{V}_1=P_2{V}_2$

Here, $P_2$ is the pressure reached after isothermal process and $V_2$ is the volume of the gas at the end of the isothermal process.

Rearrange above equation to get $P_2$ .

  $P_2=\frac{V_1}{V_2}{P}_1$                                                                                                                (IV)

In figure1, the straight line represents isobaric process where pressure is constant. During this process the system is compressed to original volume. Therefore, work is done on the system.

Write the expression for the work done in isobaric compression.

  $W_{\text{isobaric}}=-P\Delta V$

Here, $W_{\text{isobaric}}$ is the isobaric pressure, $P$ is the pressure of the gas and $\Delta V$ is the change in volume of the gas.

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

During isobaric process, the gas is at pressure $P_2$ and volume is decreased to $V_3$ from $V_2$.

Use above equation to write work done by gas during isobaric compression shown in figure1.

  $W_{\text{isothermal}}=-P_2\left(V_3-V_2\right)$                                                                                          (V)

Here, $P_2$ is the pressure after isothermal process, $V_3$ is the pressure after isobaric compression.

In an isochoric process total work done is zero.

Write the expression for the total work done by the gas.

  $W_{\text{tot}}=W_{\text{isothermal}}+W_{\text{isobaric}}$                                                                                           (VI)

Here, $W_{\text{tot}}$ is the total work done by environment on the gas.

Conclusion:

Substitute $2.50\times {10}^5\text{Pa}$ for $P_1$ , $1.25{\text{m}}^3$ for $V_1$ and $2.75{\text{m}}^3$ for $V_2$ in equation (III) to get $W_{\text{isothermal}}$.

  $\begin{array}{c}{W}_{\text{isothermal}}=\left(2.50\times {10}^5\text{Pa}\right)\left(1.25{\text{m}}^3\right)\ln \frac{1.25{\text{m}}^3}{2.75{\text{m}}^3}\\ =-2.46\times {10}^5\text{J}\end{array}$

Substitute $2.50\times {10}^5\text{Pa}$ for $P_1$ , $1.25{\text{m}}^3$ for $V_1$ and $2.75{\text{m}}^3$ for $V_2$ in equation (IV) to find $P_2$ .

  $\begin{array}{c}{P}_2=\frac{1.25{\text{m}}^3}{2.75{\text{m}}^3}\left(2.50\times {10}^5\text{Pa}\right)\\ =1.14\times {10}^5\text{Pa}\end{array}$

Substitute $1.14\times {10}^5\text{Pa}$ for $P_2$ , $2.75{\text{m}}^3$ for $V_2$ and $1.25{\text{m}}^3$ for $V_3$ in equation(V) to get $W_{\text{isobaric}}$.

  $\begin{array}{c}{W}_{\text{isobaric}}=-\left(1.14\times {10}^5\text{Pa}\right)\left(1.25{\text{m}}^3-2.75{\text{m}}^3\right)\\ =1.70\times {10}^5\text{J}\end{array}$

Substitute $-2.46\times {10}^5\text{J}$ for $W_{\text{isothermal}}$ and $1.70\times {10}^5\text{J}$ for $W_{\text{isobaric}}$ in equation (VI) to get $W_{\text{tot}}$ .

  $\begin{array}{c}{W}_{\text{tot}}=-2.46\times {10}^5\text{J}+1.70\times {10}^5\text{J}\\ \text{=}-\text{7}\text{.56}\times {\text{10}}^4\text{J}\end{array}$

Therefore, the work done by environment on the gas is $-\text{7}\text{.56}\times {\text{10}}^4\text{J}$ .

(d)

To determine

The heat that flows into the gas.

Answer to Problem 42PQ

The heat that flows into the gas is $7.56\times {10}^4\text{J}$ .

Explanation of Solution

Write the first law of thermodynamics.

  $Q+W_{\text{tot}}=\Delta E_{\text{th}}$

Here, $Q$ is the heat transferred and $\Delta E_{\text{th}}$ is the change in thermal energy of the gas.

Since the process is cyclic, $\Delta E_{\text{th}}$ is zero.

Substitute $0$ for $\Delta E_{\text{th}}$ in above equation to get $Q$ .

  $\begin{array}{c}Q+W_{\text{tot}}=0\\ Q=-W_{\text{tot}}\end{array}$                                                                                                      (VII)

Conclusion:

Substitute $-\text{7}\text{.56}\times {\text{10}}^4\text{J}$ for $W_{\text{tot}}$ in equation (VII) to get $Q$ .

  $\begin{array}{c}Q=-\left(-\text{7}\text{.56}\times {\text{10}}^4\text{J}\right)\\ =\text{7}\text{.56}\times {\text{10}}^4\text{J}\end{array}$

Therefore, the heat that flows into the gas is $7.56\times {10}^4\text{J}$ .