1 kg of air in a piston-cylinder apparatus can exchange heat only with a reservoir maintained at 300 K.

When 10 kJ of work is done on the air, its state is asserted to change from 1 bar 300 K to 2.5 bar, 310 K.

(a) What is the entropy change of the air?

(b) What is the heat transfer from the air?

for part a,

s = cpln(T2/T1) - Rln(p2/p1)

so the entropy change of the air is s = 1.004ln(310/300) - 0.287ln(2.5/1)

change in entropy for air = -0.23 kJ/K

For part b,

I am using the equation ΔS = ΔQ/T(average)

ΔQ = -0.23 x 305 = -70.15 kJ of heat transferred from the air.

My textbook says the correct answer for part b is -2.82 kJ/K

I am struggling to understand how to arrive at this answer and I would appreciate it if someone could explain. Thanks.


3 Answers 3


You can work out how much the internal energy of the air has changed ($mc_V\mathrm{\Delta}\!\!T$). You know how much work has been done on the air ($10\,\mathsf{kJ}$). The difference between the two is the heat transferred out of the air.

As to why the method you tried didn't produce the right answer: I think @RC_23 is right about "irreversible work". What that means is that the work-doing movement of the piston happens so fast that the air is no longer a single thermodynamic system at a uniform temperature (e.g. the piston moves supersonically and a shock front forms), and there are internal heat transfers from higher-temperature subsystems of the air to lower-temperature subsystems of the air. Those internal heat transfers increase entropy, but don't contribute anything to the heat transfer out of the air.

  • $\begingroup$ No, the are no shock fronts anywhere. Piston work is typically assumed to be quasi-static. What I meant is, in general the entropy would be a function of dQ/T, plus any irreversibility in the process which in this case is assumed to be zero. I was essentially pointing out the difference between isentropic and adiabatic, though in this case they can be considered the same. I think you hit on the better approach to the OP's problem, but wanted to clarify that point $\endgroup$
    – RC_23
    Commented Jan 11, 2022 at 15:44

It's been awhile since I did things like this, but I believe for part (b) you'd find the temperature of the air assuming that $\Delta S = 0$, then find the heat exchange required to go from that state to your actual state in part (a).

Entropy change is $\Delta S = \frac{\delta Q}{T}+(Irreversible$ $work)/T$ And in this case because it's within a piston assembly the final term is zero. It would be irreversible if it were free expansion (not against a piston) or some type of mixing.


I wouldn't bother trying to solve this problem, because the problem doesn't make sense as written. There are two issues:


If the system is in contact with a heat bath at 300 K, then its final temperature should be 300 K, not 310 K.

I suppose the idea could be that the compression is fast enough that the rate at which thermal energy is generated is greater than the rate at which it can flow out to the bath, and that the specified final state is a temporary non-equlibrium state, and that one is calculating the approximate entropy change between an equilibrium and non-equilibrium state. [I say approximate because the non-equilibrium state doesn't have a well-defined entropy.] But, if so, that should have been mentioned explicitly.


The 10 kJ specified for work can't be right. As you may know, the minimum possible compression work is that done with a reversible process. Using that, let's leave aside the details of 300 K vs. 310 K, and try to get a ballpark constraint on the work.

1 kg of air, at an average molecular weight of 28.96 g/mol, is 33.4 moles.

The reversible work to compress 33.4 moles of air at 300 K from 1 bar to 2.5 bar is:

$$w = - n R T \ln \frac{V_f}{V_i} = - n R T \ln \frac{p_i}{p_f}=- n R T \ln \frac{1}{2.5} = 76.3 kJ$$

[With constant T and n, $\frac{V_f}{V_i} = \frac{p_i}{p_f}.$]

This value, which is the work for a reversible process, gives us an approximate lower bound for what the compression work could be (playing with the difference between 300 K and 310 K isn't going to change things much). Thus the 10 kJ value specified in the problem, which is nearly an order of magnitude lower, clearly makes no sense.

Sure, you could get the work down to 10 kJ (indeed, down to 0) by cooling the air to a low temperature, letting it contract, stopping the piston, and heating the air back up. But that's not what the problem specifies.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.