Question

Question: (III) When a gas is taken from a to c along the curved path in Fig. 15–24, the work done by the gas is \(W = - 35\;{\rm{J}}\) and the heat added to the gas is \(Q = - 175\;{\rm{J}}\). Along path abc, the work done by the gas is \(W = - 56\;{\rm{J}}\). (That is, 56 J of work is done on the gas.) (a) What is Q for path abc? (b) If \({P_{\rm{c}}} = \frac{1}{2}{P_{\rm{b}}}\), what is W for path cda? (c) What is Q for path cda? (d) What is \({U_{\rm{a}}} - {U_{\rm{c}}}\) ? (e) If \({U_{\rm{d}}} - {U_{\rm{c}}} = 42\;{\rm{J}}\), what is Q for path da?

Short answer

(a)The value of Q for path abc is \( - 196\;{\rm{J}}\).

(b) The value of W for path cda is \(28\;{\rm{J}}\).

(c) The value of Q for path cda is \(168\;{\rm{J}}\).

(d) The value of \({U_{\rm{a}}} - {U_{\rm{c}}}\) is \(140\;{\rm{J}}\).

(e) The value of Q for path da is \(98\;{\rm{J}}\).

Step-by-step solution

First law of thermodynamics

According to first law of thermodynamics, the relationship between the heat transfer, work done and change in the internal energy of any closed system is as follows:

\(\Delta U = Q - W\)

Here,\(\Delta U\)is the change in the internal energy,\(Q\)is the heat transfer, and\(W\)is the work done.

Given information

Given data:

The work done in process a to c is \({W_{{\rm{ac}}}} = - 35\;{\rm{J}}\).

The heat transfer in process a to c is \({Q_{{\rm{ac}}}} = - 175\;{\rm{J}}\).

The work done in process abc is \({W_{{\rm{abc}}}} = - 56\;{\rm{J}}\).

The change in internal energy from d to c is \({U_{\rm{d}}} - {U_{\rm{c}}} = 42\;{\rm{J}}\).

Evaluation of the value of Q for path abc

(a)

The internal energy is a state function that means it is independent of the path followed by the system. Hence, the internal energy will be the same for each path.

The change in internal energy along the path ac can be calculated as:

\(\begin{aligned}{c}\Delta {U_{{\rm{ac}}}} &= {Q_{{\rm{ac}}}} - {W_{{\rm{ac}}}}\\\Delta {U_{{\rm{ac}}}} &= \left( { - 175\;{\rm{J}}} \right) - \left( { - 35\;{\rm{J}}} \right)\\\Delta {U_{{\rm{ac}}}} &= - 140\;{\rm{J}}\end{aligned}\)

Since, the internal energy change is same for each path. Therefore, the internal of the path abc will be:

\(\begin{aligned}{c}\Delta {U_{{\rm{abc}}}} &= \Delta {U_{{\rm{ac}}}}\\\Delta {U_{{\rm{abc}}}} &= - 140\;{\rm{J}}\end{aligned}\)

The value of Q for path abc can be calculated as:

\(\begin{aligned}{c}\Delta {U_{{\rm{abc}}}} &= {Q_{{\rm{abc}}}} - {W_{{\rm{abc}}}}\\{Q_{{\rm{abc}}}} &= \Delta {U_{{\rm{abc}}}} + {W_{{\rm{abc}}}}\\{Q_{{\rm{abc}}}} &= \left( { - 140\;{\rm{J}}} \right) + \left( { - 56\;{\rm{J}}} \right)\\{Q_{{\rm{abc}}}} &= - 196\;{\rm{J}}\end{aligned}\)

Here, a negative sign indicates that heat is released from the system.

Thus, the value of Q for path abc is \( - 196\;{\rm{J}}\).

Evaluation of the value of W for path cda

(b)

Work done along the path abc is equal to the sum of work done along the path ab and work done along the path bc. But work done along the bc is zero, since there is no volume change along the path. That is:

\(\begin{aligned}{c}{W_{{\rm{abc}}}} &= {W_{{\rm{ab}}}} + {W_{{\rm{bc}}}}\\{W_{{\rm{abc}}}} &= {W_{{\rm{ab}}}} + 0\\{W_{{\rm{abc}}}} &= {W_{{\rm{ab}}}}\end{aligned}\)

Similarly, work done along the path cda is equal to the sum of work done along the path cd and work done along the path da. But work done along the da is zero, as there is no volume change along this path. That is:

\(\begin{aligned}{c}{W_{{\rm{cda}}}} &= {W_{{\rm{cd}}}} + {W_{{\rm{da}}}}\\{W_{{\rm{cda}}}} &= {W_{{\rm{cd}}}} + 0\\{W_{{\rm{cda}}}} &= {W_{{\rm{cd}}}}\end{aligned}\)

The change in volume along the path ab and cd are same henceforth the difference in the work done along these paths only because of the difference in pressure in each path. That is:

\(\frac{{{W_{{\rm{cd}}}}}}{{{W_{{\rm{ab}}}}}} = \frac{{{P_{\rm{d}}}}}{{{P_{\rm{b}}}}}\)

Since, \({P_{\rm{d}}} = {P_{\rm{c}}}\) and \({P_{\rm{c}}} = \frac{1}{2}{P_{\rm{b}}}\). Therefore, \({P_{\rm{d}}} = \frac{1}{2}{P_{\rm{b}}}\).

Hence, the work done along the path cda can be calculated as:

\(\begin{aligned}{c}\frac{{{W_{{\rm{cd}}}}}}{{{W_{{\rm{ab}}}}}} &= - \frac{{\left( {\frac{1}{2}{P_{\rm{b}}}} \right)}}{{{P_{\rm{b}}}}}\\{W_{{\rm{cd}}}} &= - \frac{1}{2}{W_{{\rm{ab}}}}\\{W_{{\rm{cda}}}} &= - \frac{1}{2}{W_{{\rm{abc}}}}\\{W_{{\rm{cda}}}} &= - \frac{1}{2}\left( { - 56\;{\rm{J}}} \right)\\{W_{{\rm{cda}}}} &= 28\;{\rm{J}}\end{aligned}\)

Thus, the value of W for path cda is \(28\;{\rm{J}}\).

Evaluation of the value of Q for path cda

(c)

The value of the Q for path cda can be calculated as:

\(\begin{aligned}{c}{Q_{{\rm{cda}}}} &= \Delta {U_{{\rm{cda}}}} + {W_{{\rm{cda}}}}\\{Q_{{\rm{cda}}}} &= \left( {140\;{\rm{J}}} \right) + \left( {28\;{\rm{J}}} \right)\\{Q_{{\rm{cda}}}} &= 168\;{\rm{J}}\end{aligned}\)

Thus, the value of Q for path cda is \(168\;{\rm{J}}\).

Evaluation of the value of \({U_{\rm{a}}} - {U_{\rm{c}}}\)

(d)

The value of \({U_{\rm{a}}} - {U_{\rm{c}}}\) can be calculated as:

\(\begin{aligned}{c}\Delta {U_{{\rm{ac}}}} &= {Q_{{\rm{ac}}}} - {W_{{\rm{ac}}}}\\{U_{\rm{a}}} - {U_{\rm{c}}} &= \left( { - 175\;{\rm{J}}} \right) - \left( { - 35\;{\rm{J}}} \right)\\{U_{\rm{a}}} - {U_{\rm{c}}} &= 140\;{\rm{J}}\end{aligned}\)

Thus, the value of \({U_{\rm{a}}} - {U_{\rm{c}}}\) is \(140\;{\rm{J}}\).

Evaluation of the value of Q for path da

(e)

The value of Q for path da can be calculated as:

\(\begin{aligned}{c}{Q_{{\rm{da}}}} &= \Delta {U_{{\rm{da}}}}\\{Q_{{\rm{da}}}} &= \Delta {U_{{\rm{ac}}}} - \Delta {U_{{\rm{cd}}}}\\{Q_{{\rm{da}}}} &= \left( {{U_{\rm{a}}} - {U_{\rm{c}}}} \right) - \left( {{U_{\rm{d}}} - {U_{\rm{c}}}} \right)\\{Q_{{\rm{da}}}} &= \left( {140\;{\rm{J}}} \right) - \left( {42\;{\rm{J}}} \right)\\{Q_{{\rm{da}}}} &= 98\;{\rm{J}}\end{aligned}\)

Thus, the value of Q for path da is \(98\;{\rm{J}}\).