Question

Writing the first- and second-law relations and simplifying, obtain the reversible work relation for a closed system that exchanges heat with the surrounding medium at \(T_{0}\) in the amount of \(Q_{0}\) as well as a heat reservoir at \(T_{R}\) in the amount of \(Q_{R^{*}}\) (Hint: Eliminate \(Q_{0}\) between the two equations.)

Short answer

In this exercise, we derived the reversible work relation for a closed system that exchanges heat with the surrounding medium at temperature \(T_{0}\) and a heat reservoir at temperature \(T_{R}\). Our final result is the expression: \[ W_{rev} = W - T_{0} \left( dS - \frac{dU}{T_{0}} - \frac{Q_{R^{*}}}{T_{R}} \right) + Q_{R^{*}}. \] The reversible work relation is important because it provides a quantitative expression for the maximum work that can be extracted from a closed system while accounting for the heat exchange with its surroundings.

Step-by-step solution

Write the first-law equation for the system

According to the first law of thermodynamics, the change in internal energy of a closed system is equal to the heat added to the system, minus the work done by the system. Since there are two heat exchanges taking place (\(Q_{0}\) and \(Q_{R^*}\)), we can write the first-law equation as: \[ dU = Q_{0} + Q_{R^{*}} - W. \]

Write the second-law equation for the system

According to the second law of thermodynamics, the change in entropy of a closed system is equal to the entropy transfer with the two heat reservoirs minus the reversible work component: \[ dS = \frac{Q_{0}}{T_{0}} + \frac{Q_{R^{*}}}{T_{R}} - \frac{W_{rev}}{T_{0}}. \]

Solve for \(Q_{0}\) in the first-law equation

In order to eliminate \(Q_{0}\), we need to solve for it in the first-law equation. Rearranging the equation gives us: \[ Q_{0} = dU - Q_{R^{*}} + W. \]

Substitute for \(Q_{0}\) in the second-law equation

Now, let's substitute the expression for \(Q_{0}\) we found in step 3 into the second-law equation: \[ dS = \frac{dU - Q_{R^{*}} + W}{T_{0}} + \frac{Q_{R^{*}}}{T_{R}} - \frac{W_{rev}}{T_{0}}. \]

Rearrange the equation to get the reversible work relation

We want to find the reversible work relation, which means we need to have \(W_{rev}\) alone on one side of the equation. Let's rearrange the equation to obtain the desired relation: \[ W_{rev} = W - T_{0} \left( dS - \frac{dU}{T_{0}} - \frac{Q_{R^{*}}}{T_{R}} \right) + Q_{R^{*}}. \] This is the reversible work relation for the closed system that exchanges heat with the surrounding medium at \(T_{0}\) and a heat reservoir at \(T_{R}\).