Ill-posed problem about themodynamic work? (comparing reversible and irreversible for different final states)

Question

I'm not concerned with the answer itself or the points for this assignment. My interest is in whether this problem is ill-posed, or can in fact be solved with the given information.

I am under the impression this problem is ill-posed and unsolvable. Here is my reasoning, please correct me if you see an error. Let's examine processes A and B (assuming ideal gas behavior). We can compare the final temperatures by rearranging the first law and heat capacity:

$\Delta U = Q + W = W$

$\Delta U = n \int c_V dT = nc_V \Delta T$

which for process "i" yields

$T_i = \frac{W_i}{nc_V} + T_1$.

We can compare the final temperatures $T_A$ and $T_B$ based on the work values in the corresponding processes ($W_A$ and $W_B$). So this problem appears to depend on comparing work performed in an irreversible process (A) to work performed in a reversible process (B). In expansion, the work performed reversibly is greater than the work performed reversibly. However, I believe such a comparison of paths can only be made if the initial and final states are the same. I believe that is not the case here.

The two processes (A and B) have the same initial state, so they have the same number of moles. They also have the same final pressure ($P_{atm}$). Since they presumably have different final temperatures (the point of the problem), then they must have different final volumes:

$V_A = \frac{nRT_A}{P_{atm}} \neq V_B = \frac{nRT_B}{P_{atm}}$ when $T_A \neq T_B$.

Since the final volumes are different, then I do not believe we can compare the reversible and irreversible work. The irreversible work ($W_A$) could be larger than the reversible work ($W_B$) if the final volume of the irreversible process ($V_A$) was some key amount greater than the final volume of the irreversible process ($V_B$). Pardon the crude diagram, but it illustrates the point better than words can:

I cannot figure out a way to solve this problem without making the unjustified "reversible work is greater than irreversible work" comparison. Is this problem ill-posed, or have I missed something?

Answer 1

In cases A and B, assuming a massless frictionless piston, the initial pressure is $$P_i=\frac{Mg}{A_p}+P_{atm}$$where $A_p$ is the cross sectional area of the piston; and the final pressure is $$P_f=P_{atm}$$Case A is an irreversible expansion taking place at constant external pressure equal to $P_f$, and, from the first law of thermodynamics, we get: $$\Delta U=nC_v(T_f-T_i)=-P_f(V_f-V_i)=-P_f\left(\frac{nRT_f}{P_f}-\frac{nRT_i}{P_i}\right)$$This equation can be solved for the ratio $T_f/T_i$ as a function of $P_f/P_i$.

Case B is a reversible expansion taking place at gradually varying external pressure, and, from the first law of thermodynamics is governed by: $$nC_vdT=-PdV=-\frac{nRT}{V}dV$$Combined with the ideal gas law, this equation can also be solved for the ratio $T_f/T_i$ as a function of $P_f/P_i$.

The two different results for the two cases A and B can then be compared for their predictions of the final temperature. Similar analyses can be applied to cases C and D.

Answer 2

This answer is to draw a conclusion from the foundation laid by Chet Miller's answer, which I will accept as the correct answer.

For the irreversible expansion (process A), continuing from the equation Chet Miller wrote,

$$nc_v(T_f - T_i)=-P_f\left(\frac{nRT_f}{P_f}-\frac{nRT_i}{P_i}\right)$$

we can take $c_v = \frac{1}{\gamma-1}R := \alpha R$ (ideal gas) and perform some arithmetic to obtain the final temperature (with respect to the fixed initial temperature) in terms of the final pressure (with respect to the fixed initial pressure):

$$\frac{T_f}{T_i} = \frac{P_f/P_i+\alpha}{1+\alpha} \;\;\;\;\;\; \text{(irreversible expansion)}$$

where $\alpha$ is a constant (equal to half the molecular degrees of freedom of the gas, for instance $\alpha = 5/2$ for a diatomic gas).

For the reversible expansion (process B),

$$nC_vdT=-PdV=-\frac{nRT}{V}dV$$

separation of variables (and employing $c_v = \alpha R$) yields $\frac{dT}{T}=-\frac{1}{\alpha}\frac{dV}{V}$ which integrates to $\ln(\frac{T_f}{T_i})=-\frac{1}{\alpha}\ln(\frac{V_f}{V_i}) \implies \frac{T_f}{T_i}=(\frac{V_f}{V_i})^{\frac{-1}{\alpha}}=(\frac{nRT_f/P_f}{nRT_i/P_i})^{\frac{-1}{\alpha}}=(\frac{T_f}{T_i})^{\frac{-1}{\alpha}}(\frac{P_i}{P_f})^{\frac{-1}{\alpha}}$ which ultimately yields the desired equation for final temperature (with respect to the fixed initial temperature) in terms of the final pressure (with respect to the fixed initial pressure):

$$\frac{T_f}{T_i}=(\frac{P_f}{P_i})^{\frac{1}{\alpha+1}} \;\;\;\;\;\; \text{(reversible expansion)}$$

To gain some intuition, I graphed the (normalized) final temperature $T_f/T_i$ vs. the (normalized) final pressure $P_f/P_i$ for each process (taking $\alpha = 5/2$) and the result surprised me.

It seems that regardless of how different the final volumes might be, an ideal gas expanding irreversibly always has a higher final temperature than if the expansion were performed reversibly (assuming both processes are carried out to the same final pressure and the process is adiabatic). This is also true for adiabatic compression: the final temperature for the irreversible process is greater than that for the reversible process, regardless of the final volume (and assuming identical final pressures).

Thanks again to Chet Miller for helping me see that an analytical solution is indeed possible, and this problem was not "ill-posed".