Chapter 20, Problem 40P

(a)

To determine

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

Answer to Problem 40P

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

Write the equation of first law of thermodynamics.

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

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.

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

Conclusion:

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

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

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

(b)

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 40P

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

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

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

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 (II) to find the work done on the system from C to D.

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

Conclusion:

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

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

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

(c)

To determine

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

Answer to Problem 40P

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

Explanation of Solution

Write the equation of first law of thermodynamics.

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

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 (III).

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

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 $-900\text{J}$.

(d)

To determine

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

Answer to Problem 40P

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

is $-1400\text{J}$.

Explanation of Solution

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.

Conclusion:

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.

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

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