Chapter 15, Problem 11P

(a)

To determine

The value of Q for path abc.

Answer to Problem 11P

The value of Q for path abc is $-76\text{J}$

Explanation of Solution

Given:

The given path is shown below.

  

Also, the work done by the gas, from a to c is $\mathrm{W}=-35\text{J}$

Heat added to the gas is $\mathrm{Q}=-63\text{J}$

Work done along path abc is ${\mathrm{W}}_{\text{abc}}=-48\text{J}$

Formula used:

According to first law of thermodynamics,

  $\mathrm{U}=\mathrm{Q}-\mathrm{W}$

Where,

U is change in internal energy.

Q is heat added.

W is work done by the system.

Calculation:

Consider the given graph.

For path ac ,

  $\begin{array}{c}{\mathrm{U}}_{\text{ac}}={\mathrm{Q}}_{\text{ac}}-{\mathrm{W}}_{\text{ac}}\\ =\left(-63\text{J}\right)-\left(-35\text{J}\right)\\ =-63\text{J+35}\text{J}\\ =-28\text{J}\end{array}$

Also, change in internal energy for closed path is zero. So,

  $\begin{array}{c}{\mathrm{U}}_{\text{ac}}+{\mathrm{U}}_{\text{cba}}=0\\ {\mathrm{U}}_{\text{cba}}=-{\mathrm{U}}_{\text{ac}}\\ =28\text{J}\\ {\mathrm{U}}_{\text{abc}}=-{\mathrm{U}}_{\text{cba}}\\ =-28\text{J}\end{array}$

For path abc ,

  $\begin{array}{c}{\mathrm{Q}}_{\text{abc}}={\mathrm{W}}_{\text{abc}}+{\mathrm{U}}_{\text{abc}}\\ =-48\text{J+}\left(-28\text{J}\right)\\ =-76\text{J}\end{array}$

Conclusion:

Therefore, Q for path abc is $-76\text{J}$

(b)

To determine

The value of W for path cda.

Answer to Problem 11P

The value of W for path cda is 24 J.

Explanation of Solution

Given:

The given path is shown below.

  

Also, the work done by the gas, from a to c is $\mathrm{W}=-35\text{J}$

Heat added to the gas is $\mathrm{Q}=-63\text{J}$

Work done along path abc is ${\mathrm{W}}_{\text{abc}}=-48\text{J}$

  ${\mathrm{P}}_{\text{c}}=\frac{1}{2}{\mathrm{P}}_{\text{b}}$

Formula used:

The work done is equal to the product of pressure and the change in the volume.

It is given as,

  $\mathrm{W}=\mathrm{P}\times \mathrm{\Delta }\mathrm{V}$

Where,

W is work done

P is pressure

  $\mathrm{\Delta }\mathrm{V}$ is change in volume.

Calculation:

Work done along path cda is equal to work done along path cd. There will be no work done across path da, because volume is constant. So,

  $\begin{array}{c}{\mathrm{W}}_{\text{cda}}={\mathrm{W}}_{\text{cd}}+{\mathrm{W}}_{\text{da}}\\ ={\mathrm{P}}_{\text{c}}\left({\mathrm{V}}_{\text{d}}-{\mathrm{V}}_{\text{c}}\right)+0\\ ={\mathrm{P}}_{\text{c}}\left({\mathrm{V}}_{\text{d}}-{\mathrm{V}}_{\text{c}}\right)\end{array}$

Similarly, work done along path abc is equal to work done along path ab . There will be no work done across path bc , because volume is constant. So,

  $\begin{array}{c}{\mathrm{W}}_{\text{abc}}={\mathrm{W}}_{\text{ab}}+{\mathrm{W}}_{\text{bc}}\end{array}$

  $\begin{array}{c}={\mathrm{P}}_{\text{b}}\left({\mathrm{V}}_{\text{b}}-{\mathrm{V}}_{\text{c}}\right)+0\end{array}$

  $\begin{array}{c}={\mathrm{P}}_{\text{b}}\left({\mathrm{V}}_{\text{b}}-{\mathrm{V}}_{\text{c}}\right)\end{array}$

  $\begin{array}{c}=\left(2{\mathrm{P}}_{\text{c}}\right)\left(-\left({\mathrm{V}}_{\text{d}}-{\mathrm{V}}_{\text{c}}\right)\right)\because {\mathrm{V}}_{\text{b}}-{\mathrm{V}}_{\text{c}}=-\left({\mathrm{V}}_{\text{d}}-{\mathrm{V}}_{\text{c}}\right)\end{array}$

  $\begin{array}{c}=-2{\mathrm{W}}_{\text{cda}}\end{array}$

  $\begin{array}{c}{\mathrm{W}}_{\text{cda}}=-\frac{1}{2}{\mathrm{W}}_{\text{abc}}\end{array}$

  $\begin{array}{c}=-\frac{1}{2}\times \left(-48\right)\end{array}$

  $\begin{array}{c}=24\text{J}\end{array}$

Conclusion:

Therefore, the value of W for path cda is 24 J.

(c)

To determine

The value of Q for path cda.

Answer to Problem 11P

The value of Q for path cda is 52 J.

Explanation of Solution

Given:

The given path is shown below.

  

Also, the work done by the gas, from a to c is $\mathrm{W}=-35\text{J}$

Heat added to the gas is $\mathrm{Q}=-63\text{J}$

Work done along path abc is ${\mathrm{W}}_{\text{abc}}=-48\text{J}$

Formula used:

According to first law of thermodynamics,

  $\mathrm{U}=\mathrm{Q}-\mathrm{W}$

Where,

U is change in internal energy.

Q is heat added.

W is work done by the system.

Calculation:

From part (b),

The value of W for cda is 24 J.

The change in internal energy for closed path is 0. So,

  $\begin{array}{c}{\mathrm{U}}_{\text{cda}}+{\mathrm{U}}_{\text{ac}}=0\\ {\mathrm{U}}_{\text{cda}}=-{\mathrm{U}}_{\text{ac}}\\ =-\left(-28\text{J}\right)\\ =28\text{J}\end{array}$

So for path cda ,

  $\begin{array}{c}{\mathrm{Q}}_{\text{cda}}={\mathrm{U}}_{\text{cda}}+{\mathrm{W}}_{\text{cda}}\\ =28+24\\ =52\text{J}\end{array}$

Conclusion:

Therefore, the value of Q for path cda is 52 J.

(d)

To determine

The value of ${\mathrm{U}}_{\text{a}}-{\mathrm{U}}_{\text{c}}$ .

Answer to Problem 11P

The change in internal energy of the gas is 28 J.

Explanation of Solution

Given:

The given path is shown below.

  

Also, the work done by the gas, from a to c is $\mathrm{W}=-35\text{J}$

Heat added to the gas is $\mathrm{Q}=-63\text{J}$

Work done along path abc is ${\mathrm{W}}_{\text{abc}}=-48\text{J}$

  ${\mathrm{U}}_{\text{d}}-{\mathrm{U}}_{\text{c}}=5\text{J}$

Formula used:

According to first law of thermodynamics,

  $\mathrm{U}=\mathrm{Q}-\mathrm{W}$

Where,

U is change in internal energy.

Q is heat added.

W is work done by the system.

Calculation:

Consider the given graph.

For path ac ,

  $\begin{array}{c}{\mathrm{U}}_{\text{a}}-{\mathrm{U}}_{\text{c}}=-\left({\mathrm{U}}_{\text{c}}-{\mathrm{U}}_{\text{a}}\right)\\ =-{\mathrm{U}}_{\text{ac}}\\ =-\left({\mathrm{Q}}_{\text{ac}}-{\mathrm{W}}_{\text{ac}}\right)\\ =-\left(\left(-63\text{J}\right)-\left(-35\text{J}\right)\right)\\ =-\left(-63\text{J+35}\text{J}\right)\\ =-\left(-28\text{J}\right)\\ =28\text{J}\end{array}$

Conclusion:

Therefore, the change in internal energy of the gas is 28 J.

(e)

To determine

The value of Q for path da.

Answer to Problem 11P

The value of Q for path da is 23 J.

Explanation of Solution

Given:

The given path is shown below.

  

Also, the work done by the gas, from a to c is $\mathrm{W}=-35\text{J}$

Heat added to the gas is $\mathrm{Q}=-63\text{J}$

Work done along path abc is ${\mathrm{W}}_{\text{abc}}=-48\text{J}$

Formula used:

According to first law of thermodynamics,

  $\mathrm{U}=\mathrm{Q}-\mathrm{W}$

Where,

U is change in internal energy.

Q is heat added.

W is work done by the system.

Calculation:

Consider the given graph.

The change in internal energy for closed path is 0. So,

  $\begin{array}{c}{\mathrm{U}}_{\text{cd}}+{\mathrm{U}}_{\text{da}}+{\mathrm{U}}_{\text{ac}}=0\\ {\mathrm{U}}_{\text{da}}=-{\mathrm{U}}_{\text{ac}}-{\mathrm{U}}_{\text{cd}}\\ =-\left(-28\text{J}\right)-\left({\mathrm{U}}_{\text{d}}-{\mathrm{U}}_{\text{c}}\right)\\ =28\text{J}-\left(5\text{J}\right)\\ =23\text{J}\end{array}$

So for path da,

  ${\mathrm{Q}}_{\text{da}}={\mathrm{U}}_{\text{da}}+{\mathrm{W}}_{\text{da}}$

There will be no work done across path da, because volume is constant. So,

  $\begin{array}{c}{\mathrm{Q}}_{\text{da}}=23+0\\ =23\text{J}\end{array}$

Conclusion:

Therefore, the value of Q for path da is 23 J.