Chapter 20, Problem 34P

To determine

The difference in internal energy $E_{\text{int},B}-E_{\text{int},A}$ .

Answer to Problem 34P

The difference in internal energy $E_{\text{int},B}-E_{\text{int},A}$ is $4.29\times {10}^4\text{ J}$ .

Explanation of Solution

Since the gas goes through a cyclic process, the net change in internal energy of the gas must be zero.

Write the equation for the net change in internal energy of the gas.

  $\Delta E_{\text{int}}=0$                                                                                                                   (I)

Here, $\Delta E_{\text{int}}$ is the net change in internal energy of the gas after one complete cycle.

The net change in internal energy must be equal to the sum of the change in internal energy of the gas during the different processes in the cycle.

Write the expression for $\Delta E_{\text{int}}$ .

  $\Delta E_{\text{int}}=\Delta E_{\text{int},AB}+\Delta E_{\text{int},BC}+\Delta E_{\text{int},CD}+\Delta E_{\text{int},DA}$

Here, $\Delta E_{\text{int},AB}$ is the change in internal energy of the gas during the process AB, $\Delta E_{\text{int},BC}$ is the change in internal energy of the gas during the process BC, $\Delta E_{\text{int},CD}$ is the change in internal energy of the gas during the process CD and $\Delta E_{\text{int},DA}$ is the change in internal energy of the gas during the process DA.

Put the above equation in equation (I) and rewrite it for $\Delta E_{\text{int},AB}$ .

  $\begin{array}{c}\Delta E_{\text{int},AB}+\Delta E_{\text{int},BC}+\Delta E_{\text{int},CD}+\Delta E_{\text{int},DA}=0\\ \Delta E_{\text{int},AB}=-\Delta E_{\text{int},BC}-\Delta E_{\text{int},CD}-\Delta E_{\text{int},DA}\end{array}$                    (II)

Write the expression for $\Delta E_{\text{int},AB}$ .

  $\Delta E_{\text{int},AB}=E_{\text{int},B}-E_{\text{int},A}$

Here, $E_{\text{int},B}$ is the internal energy of the gas at the point B and $E_{\text{int},A}$ is the internal energy of the gas at the point A.

Put the above equation in equation (II).

  $E_{\text{int},B}-E_{\text{int},A}=-\Delta E_{\text{int},BC}-\Delta E_{\text{int},CD}-\Delta E_{\text{int},DA}$                                                        (III)

The process BC is an isochoric process. The work done in an isochoric process is zero. This implies $W$ is $0$ for the process BC.

Write the first law of thermodynamics.

  $\Delta E_{\text{int}}=Q+W$                                                                                                          (IV)

Here, $\Delta E_{\text{int}}$ is the change in internal energy of the gas, $Q$ is the heat added to the gas and $W$ is the work done on the gas.

Use equation (IV) to write the expression for $\Delta E_{\text{int},BC}$ .

  $\Delta E_{\text{int},BC}=Q_{BC}+W_{BC}$                                                                                                (V)

Here, $Q_{BC}$ is the heat added to the gas during the process BC and $W_{BC}$ is the work done on the gas during the process BC.

The process BC is isobaric.

Write the expression for $W_{BC}$ .

  $W_{BC}=-P_B\Delta V_{BC}$

Here, $P_B$ is the pressure of the gas at the point B and $\Delta V_{BC}$ is the change in volume of the gas during the process BC.

Put the above equation in equation (V).

  $\begin{array}{c}\Delta E_{\text{int},BC}=Q_{BC}+\left(-P_B\Delta V_{BC}\right)\\ =Q_{BC}-P_B\Delta V_{BC}\end{array}$                                                                                    (VI)

The process CD is isothermal. The change in internal energy for an isothermal process is zero.

Write the expression for $\Delta E_{\text{int},CD}$ .

    $\Delta E_{\text{int},CD}=0$                                                                                                           (VII)

Use equation (IV) to write the expression for $\Delta E_{\text{int},DA}$ .

  $\Delta E_{\text{int},DA}=Q_{DA}+W_{DA}$                                                                                           (VIII)

Here, $Q_{DA}$ is the heat added to the gas during the process DA and $W_{DA}$ is the work done on the gas during the process DA.

The process DA is isobaric.

Write the expression for $W_{DA}$ .

  $W_{DA}=-P_D\Delta V_{DA}$

Here, $P_D$ is the pressure of the gas at the point D and $\Delta V_{DA}$ is the change in volume of the gas during the process DA.

Put the above equation in equation (VIII).

  $\begin{array}{c}\Delta E_{\text{int},DA}=Q_{DA}+\left(-P_D\Delta V_{DA}\right)\\ =Q_{DA}-P_D\Delta V_{DA}\end{array}$                                                                                    (IX)

Put equations (VI), (VII) and (IX) in equation (III).

  $\begin{array}{c}{E}_{\text{int},B}-E_{\text{int},A}=-\left(Q_{BC}-P_B\Delta V_{BC}\right)-0-\left(Q_{DA}-P_D\Delta V_{DA}\right)\\ =-Q_{BC}+P_B\Delta V_{BC}-Q_{DA}+P_D\Delta V_{DA}\\ =-\left(Q_{BC}+Q_{DA}\right)+P_B\Delta V_{BC}+P_D\Delta V_{DA}\end{array}$                                         (X)

Write the equation for $\Delta V_{BC}$ .

  $\Delta V_{BC}=V_C-V_B$                                                                                                    (XI)

Here, $V_C$ is the volume of the gas at the point C and $V_B$ is the volume of the gas at the point B.

Write the equation for $\Delta V_{DA}$ .

  $\Delta V_{DA}=V_A-V_D$                                                                                                       (XII)

Here, $V_A$ is the volume of the gas at the point A and $V_D$ is the volume of the gas at the point D.

Conclusion:

It is given that the heat entering the system from B to C is $345\text{ kJ}$ and the heat leaving the system from D to A is $371\text{ kJ}$ . From figure P20.34, the volume of the gas at A is $0.2{\text{ m}}^3$ , volume at B is $0.09{\text{ m}}^3$ , the volume at C is $0.4{\text{ m}}^3$ and the volume at D is $1.2{\text{ m}}^3$ , the pressure at the point B is $3\text{ atm}$ and the pressure at the point D is $1\text{ atm}$ .

Substitute $0.4{\text{ m}}^3$ for $V_C$ and $0.09{\text{ m}}^3$ for $V_B$ in equation (XI) to find $\Delta V_{BC}$ .

  $\begin{array}{c}\Delta V_{BC}=0.4{\text{ m}}^3-0.09{\text{ m}}^3\\ =0.31{\text{ m}}^3\end{array}$

Substitute $0.2{\text{ m}}^3$ for $V_A$ and $1.2{\text{ m}}^3$ for $V_D$ in equation (XII) to find $\Delta V_{DA}$ .

  $\begin{array}{c}\Delta V_{DA}=0.2{\text{ m}}^3-1.2{\text{ m}}^3\\ =-1.00{\text{ m}}^3\end{array}$

Substitute $345\text{ kJ}$ for $Q_{BC}$ , $-371\text{ kJ}$ for $Q_{DA}$ , $3\text{ atm}$ for $P_B$ , $0.31{\text{ m}}^3$ for $\Delta V_{BC}$ , $1\text{ atm}$ for $P_D$ and $-1.00{\text{ m}}^3$ for $\Delta V_{DA}$ in equation (X) to find $E_{\text{int},B}-E_{\text{int},A}$ .

$\begin{array}{c}{E}_{\text{int},B}-E_{\text{int},A}=-\left(345\text{ kJ}\frac{1000\text{ J}}{1\text{ kJ}}+\left(-371\text{ kJ}\frac{1000\text{ J}}{1\text{ kJ}}\right)\right)+\left(3\text{ atm}\frac{1.013\times {10}^5\text{ Pa}}{1\text{ atm}}\right)\left(0.31{\text{ m}}^3\right)+\left(1\text{ atm}\frac{1.013\times {10}^5\text{ Pa}}{1\text{ atm}}\right)\left(-1.00{\text{ m}}^3\right)\\ =4.29\times {10}^4\text{ J}\end{array}$

Therefore, the difference in internal energy $E_{\text{int},B}-E_{\text{int},A}$ is $4.29\times {10}^4\text{ J}$ .