Chapter 15, Problem 12P

(a)

To determine

To find: Q for path abc.

Answer to Problem 12P

Q for path abc is $Q_{abc}=-76\text{ J}$

Explanation of Solution

Given:

Work done by the gas, $W=-35\text{ J}$

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

Work done along path abc, $W=-48\text{ J}$

Calculation:

For curved path ac, the potential energy can be given as:

  $\begin{array}{l}\Delta U=Q_{ac}-W_{ac}\\ U_c-U_a=\left(-63\text{ J}\right)-\left(-35\text{ J}\right)\\ U_c-U_a=-28\text{ J}\end{array}$

For path abc, Q can be given as:

  $\begin{array}{l}{U}_c-U_a=Q_{abc}-W_{abc}\\ -28\text{ J}=Q_{abc}-\left(-48\text{ J}\right)\\ Q_{abc}=-76\text{ J}\end{array}$

Conclusion:

So, Q for path abc is $Q_{abc}=-76\text{ J}$

(b)

To determine

To find: W for path cda.

Answer to Problem 12P

W for path cda is $W_{cda}=24\text{ J}$

Explanation of Solution

Given:

  $P_c=\frac{1}{2}{P}_b$

Calculation:

Work done for path cda can be given by:

  $\begin{array}{l}{W}_{cda}=P_c\left(V_d-V_c\right)\\ W_{cda}=\left(\frac{1}{2}{P}_b\right)\left(V_a-V_c\right)\\ W_{cda}=\frac{1}{2}{W}_{ba}\\ W_{cda}=-\frac{1}{2}{W}_{ab}\\ W_{cda}=-\frac{1}{2}\left(-48\text{ J}\right)\\ W_{cda}=24\text{ J}\end{array}$

Conclusion:

So, W for path cda is $W_{cda}=24\text{ J}$

(c)

To determine

To find: Q for path cda.

Answer to Problem 12P

Q for path cda is $Q_{cda}=52\text{ J}$

Explanation of Solution

Given:

Work done by the gas, $W=-35\text{ J}$

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

Work done along path abc, $W=-48\text{ J}$

  $P_c=\frac{1}{2}{P}_b$

Calculation:

Using the first law of thermodynamics, we get:

  $\begin{array}{l}{U}_a-U_c=Q_{cda}-W_{cda}\\ -\left(U_c-U_a\right)=Q_{cda}-W_{cda}\\ -\left(-28\text{ J}\right)=Q_{cda}-24\text{ J}\\ Q_{cda}=52\text{ J}\end{array}$

Conclusion:

Hence, Q for path cda is $Q_{cda}=52\text{ J}$

(d)

To determine

To find: $U_a-U_c$

Answer to Problem 12P

  $U_a-U_c=28\text{ J}$

Explanation of Solution

Given:

Work done by the gas, $W=-35\text{ J}$

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

Work done along path abc, $W=-48\text{ J}$

  $P_c=\frac{1}{2}{P}_b$

Calculation:

Now, as calculated in the previous parts,

  $\begin{array}{l}{U}_a-U_c=-\left(U_c-U_a\right)\\ U_a-U_c=-\left(-28\text{ J}\right)\\ U_a-U_c=28\text{ J}\end{array}$

Conclusion:

Hence, $U_a-U_c=28\text{ J}$

(e)

To determine

To find: Q for path da.

Answer to Problem 12P

Q for path da is $Q_{da}=23\text{ J}$

Explanation of Solution

Given:

Work done by the gas, $W=-35\text{ J}$

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

Work done along path abc, $W=-48\text{ J}$

  $P_c=\frac{1}{2}{P}_b$

  $U_d-U_c=5\text{ J}$

Calculation:

There is no work done for path da, so:

  $\begin{array}{l}{U}_a-U_d=\left(U_a-U_c\right)+\left(U_c-U_d\right)\\ Q_{da}-W_{da}=\left(U_a-U_c\right)-\left(U_d-U_c\right)\\ Q_{da}-0=28\text{ J}-5\text{ J}\\ Q_{da}=23\text{ J}\end{array}$

Conclusion:

Hence, Q for path da is $Q_{da}=23\text{ J}$