Chapter 19, Problem 22P

In Figure P19.22, the change in internal energy of a gas that is taken from A to C along the blue path is +800 J. The work done on the gas along the red path ABC is −500 J. (a) How much energy must be added to the system by heat as it goes from A through B to C? (b) If the pressure at point A is five times that of point C, what is the work done on the system in going from C to D?

Figure P19.22

(c) What is the energy exchanged with the surroundings by heat as the gas goes from C to A along the green path? (d) If the change in internal energy in going from point D to point A is +500 J, how much energy must be added to the system by heat as it goes from point C to point D?

To determine

The energy which must be added to the system by heat which goes from A through B to C.

Answer to Problem 22P

The energy which must be added to the system by heat which goes from A through B to C is $1300\text{J}$.

Explanation of Solution

Given info: The change in internal energy of a gas which is taken along the blue path from A to C is $+800\text{J}$, the work done on the gas along the red path ABC is $-500\text{J}$.

Write the equation of first law of thermodynamics.

$\begin{array}{c}{\left(\Delta E_{\mathrm{int}}\right)}_{ABC}=Q_{ABC}+W_{ABC}\\ Q_{ABC}={\left(\Delta E_{\mathrm{int}}\right)}_{ABC}-W_{ABC}\end{array}$ (1)

Here,

${\left(\Delta E_{\mathrm{int}}\right)}_{ABC}$ is the change in internal energy along the red path ABC.

$Q_{ABC}$ is the energy which must be added to the system by heat along the red path ABC.

$W_{ABC}$ is the work done on the gas along the red path ABC.

Write the equation of conservation of energy.

${\left(\Delta E_{\mathrm{int}}\right)}_{ABC}={\left(\Delta E_{\mathrm{int}}\right)}_{AC}$

Here,

${\left(\Delta E_{\mathrm{int}}\right)}_{ABC}$ is the change in internal energy along the red path ABC.

${\left(\Delta E_{\mathrm{int}}\right)}_{AC}$ is the change in internal energy along the blue path AC.

Substitute $+800\text{J}$ for ${\left(\Delta E_{\mathrm{int}}\right)}_{AC}$ in the above equation to get the change in internal energy along ABC.

${\left(\Delta E_{\mathrm{int}}\right)}_{ABC}=+800\text{J}$

Substitute $+800\text{J}$ for ${\left(\Delta E_{\mathrm{int}}\right)}_{ABC}$ and $-500\text{J}$ for $W_{ABC}$ in the equation (1) to get the energy $Q_{ABC}$ which must be added to the system by heat along ABC.

$\begin{array}{c}{Q}_{ABC}=\left(+800\text{J}\right)-\left(-500\text{J}\right)\\ =800\text{J}+500\text{J}\\ \text{=1300}\text{J}\end{array}$

Conclusion:

Therefore, the energy which must be added to the system by heat which goes from A through B to C is $1300\text{J}$

To determine

The work done on the system from C to D if pressure at A is five times that of point C.

Answer to Problem 22P

The work done on the system from C to D if pressure at A is five times that of point C is $\text{100}\text{J}$.

Explanation of Solution

Given info: The change in internal energy of a gas which is taken along the blue path from A to C is $+800\text{J}$, the work done on the gas along the red path ABC is $-500\text{J}$, The pressure at point A is five times that of point C.

Write the equation to calculate the work done on the system from C to D.

$W_{CD}=-P_C\Delta V_{CD}$ (2)

Here,

$W_{CD}$ is the work done on the system from C to D.

$P_C$ is the pressure at point C.

$\Delta V_{CD}$ is the change in volume of the gas.

Here,

The pressure at point A is five times that of point C.

$\begin{array}{c}{P}_A=5P_C\\ P_C=\frac{P_A}{5}\end{array}$

And the volume is,

$\begin{array}{l}\Delta V_{AB}=-\Delta V_{CD}\\ \Delta V_{CD}=-\Delta V_{AB}\end{array}$

Substitute $\left(\frac{P_A}{5}\right)$ for $P_C$ and $-\Delta V_{AB}$ for $\Delta V_{CD}$ in equation (2) to find the work done on the system from C to D.

$\begin{array}{c}{W}_{CD}=\left(\frac{P_A}{5}\right)\left(-\Delta V_{AB}\right)\\ =-\frac{1}{5}\left(P_A\Delta V_{AB}\right)\\ =-\frac{1}{5}{W}_{AB}\end{array}$

Substitute $-500\text{J}$ for $W_{AB}$ in the above equation to find the work done on the system from C to D.

$\begin{array}{c}{W}_{CD}=-\frac{1}{5}\left(-500\text{J}\right)\\ =\frac{500\text{J}}{5}\\ =100\text{J}\end{array}$

Conclusion:

Therefore, the work done on the system from C to D if pressure at A is five times that of point C is $\text{100}\text{J}$.

To determine

The energy exchanged with the surroundings by heat along the green path from C to A.

Answer to Problem 22P

The energy exchanged with the surroundings by heat along the green path from C to A is $-900\text{J}$.

Explanation of Solution

Given info: The change in internal energy of a gas which is taken along the blue path from A to C is $+800\text{J}$, the work done on the gas along the red path ABC is $-500\text{J}$.

Write the equation of first law of thermodynamics.

$\begin{array}{c}{\left(\Delta E_{\mathrm{int}}\right)}_{CA}=Q_{CA}+W_{CDA}\\ Q_{CA}={\left(\Delta E_{\mathrm{int}}\right)}_{CA}-W_{CDA}\end{array}$ (3)

Here,

${\left(\Delta E_{\mathrm{int}}\right)}_{CA}$ is the change in the internal energy along the green path from C to A.

$Q_{CA}$ is the energy exchanged with the surroundings by heat along the green path from C to A.

$W_{CDA}$ is the work done on the gas along the green path from C to A.

Write the equation to calculate work done along the green path from C to A.

$\begin{array}{c}{W}_{CDA}=W_{CD}+W_{DA}\\ =W_{CD}+P\Delta V_{CDA}\\ =W_{CD}+P\left(V_D-V_A\right)\end{array}$

Here, the volume $V_D$ and $V_A$ are equal. Thus, the difference $\Delta V_{CDA}$ is zero.

Substitute 0 for $\Delta V_{CDA}$ in the above equation.

$\begin{array}{l}{W}_{CDA}=W_{CD}+P\left(0\right)\\ W_{CDA}=W_{CD}\end{array}$

Substitute $W_{CD}$ for $W_{CDA}$ in equation (3).

$Q_{CA}={\left(\Delta E_{\mathrm{int}}\right)}_{CA}-W_{CD}$

Substitute $-800\text{J}$ for ${\left(\Delta E_{\mathrm{int}}\right)}_{CA}$ and $100\text{J}$ for $W_{CD}$ in the above equation to find the energy exchanged with the surroundings by heat along the green path from C to A.

$\begin{array}{c}{Q}_{CA}=\left(-800\text{J}\right)-\left(100\text{J}\right)\\ =-900\text{J}\end{array}$

Conclusion:

Therefore, the energy exchanged with the surroundings by heat along the green path from C to A is $-900\text{J}$.

To determine

The energy which must be added to the system by heat which goes from C to D.

Answer to Problem 22P

The energy which must be added to the system by heat which goes from C to D

is $-1400\text{J}$.

Explanation of Solution

Given info: The change in internal energy of a gas which is taken along the blue path from A to C is $+800\text{J}$, the work done on the gas along the red path ABC is $-500\text{J}$.

Write the equation of first law of thermodynamics.

$\begin{array}{c}{\left(\Delta E_{\mathrm{int}}\right)}_{CD}=Q_{CD}+W_{CD}\\ Q_{CD}={\left(\Delta E_{\mathrm{int}}\right)}_{CD}-W_{CD}\\ Q_{CD}=\left[{\left(\Delta E_{\mathrm{int}}\right)}_{CDA}-{\left(\Delta E_{\mathrm{int}}\right)}_{DA}\right]-W_{CD}\end{array}$

Here,

${\left(\Delta E_{\mathrm{int}}\right)}_{CD}$ is the change in internal energy along CD.

$Q_{CD}$ is the energy which must be added to the system by heat which goes from C to D.

$W_{CD}$ is the work done on the gas along CD.

${\left(\Delta E_{\mathrm{int}}\right)}_{CDA}$ is the change in internal energy along CDA.

${\left(\Delta E_{\mathrm{int}}\right)}_{DA}$ is the change in internal energy along DA.

Substitute $-800\text{J}$ for ${\left(\Delta E_{\mathrm{int}}\right)}_{CDA}$, $500\text{J}$ for ${\left(\Delta E_{\mathrm{int}}\right)}_{DA}$ and $W_{ABC}$ and $100\text{J}$ for $W_{CD}$ in the above equation to get the energy which must be added to the system by heat which goes from C to D.

$\begin{array}{c}{\left(\Delta E_{\mathrm{int}}\right)}_{CD}=\left[\left(-800\text{J}\right)-\left(500\text{J}\right)\right]-\left(100\right)\\ =-1300\text{J}-100\text{J}\\ \text{=}-\text{1400}\text{J}\end{array}$

Conclusion:

Therefore, the energy which must be added to the system by heat which goes from C to D is $-1400\text{J}$.