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I am having difficulty understanding this problem.

Given question :-

A mass of 0.25 kg of air in a closed system expands from 2 bar, 60°C to 1 bar, 40°C while receiving 1.005 kJ of heat from a reservoir at 100°C. The surrounding atmosphere is at 0.95 bar and 27°C.

  • Determine the maximum work.

  • How much of this work would be done on the atmosphere?

(Take R = 287 J/kg K and $c_v$ = 0.712 kJ/kg K)

Given solution :-

$$ Let :- \\ m = 0.25 kg, \\ p_1 = 2bar, T_1 = 60°C, \\ p_2 = 1 bar, T_2 = 40°C, \\ p = 0.95 bar, \\ Q = 1.005 kJ \\ \text{Initial volume of air } (v_1) = \frac{mRT_1}{p_1} = 0.119 m^3 \\ \text{Final volume of air } (v_2) = \frac{mRT_2}{p_2} = 0.224m^3 \\ \text{Work done on the atmosphere }(W_1) = p(v_2 - v_1) = \boxed{9.975 kJ} \\ \text{Change in internal energy }(dU) = mc_v(T_2 - T_1) = -3.56 kJ \\ \text{Net workdone }(W_2) = Q - dU = 4.565 kJ \\ \text{Maximum workdone}(W) = W_1 + W_2 = \boxed{14.54 kJ} $$

Why are two different works being done here, and what is the meaning of maximum work ?

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Name of the book : A Textbook of Thermal Engineering

Name of the authors : R.S. Khurmi, J.K. Gupta

Name of the publisher: S. Chand

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  • $\begingroup$ I have revised my answer. I now believe you are correct in questioning this problem. $\endgroup$ – Bob D Sep 6 '19 at 18:13
  • $\begingroup$ Where did the problem come from? A textbook? If so, what's the title and whose the author? $\endgroup$ – Bob D Sep 13 '19 at 17:44
  • $\begingroup$ I edited the post $\endgroup$ – arandomguy Sep 14 '19 at 10:39
  • $\begingroup$ Thanks for accepting my answer $\endgroup$ – Bob D Sep 18 '19 at 3:35
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Why are two different works being done here, and what is the meaning of maximum work?

As I previously pointed out in one of my earlier answers, the example problem does not make sense. First of all, the work being done by the gas on the atmosphere (9.975 kJ) is greater than the heat added (1.005 kJ) plus the reduction in internal energy (3.56 kJ), in violation of the first law (conservation of energy). Secondly, adding the net work done (4.565 kJ) to the work done on the atmosphere (9.975 kJ) to come up with something called "maximum work" makes no sense.

I ran this example by a thermo expert friend of mine and he said, "The question makes no sense to me, and the ‘given solution’ makes even less sense”.

That being said, in order to satisfy the first law, either more heat is needed or, as @user8736288 suggested, and unknown external force is doing work on the gas at the same time the gas is doing work on the atmosphere. We can rule out heat coming from the atmosphere since its temperature is less than the gas at all times. The diagram below shows the possibility of an unknown external force doing 5.41 kJ of work on the gas by way of a second piston.

Although the diagram would satisfy the first law, it should not be necessary to assume the existence of an unknown external force. If that is what the example wanted you to do, it is a very poor example.What's more, even if we do to satisfy the first law it doesn't address the "maximum work" calculated in the solution. There is no logic, that I can think of, for adding those two quantities and calling it "maximum work".

Hope this helps.

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My interpretation is that there is an unknown external force (pressure $P_{ext}$) to balance the pressure acting on the system, the system would otherwise never be at equilibrium. Since $p$ is a constant offset, the situation is equivalent to assuming that the internal pressure of the system is $P_{sys}-p$ and that the system is in equilibrium with $P_{ext}$. The "net" work is then: $ W_{2} = \int (P_{sys} - p) dV = \int P_{ext} dV$

So we don't need to know about $P_{ext}$ and the total work ( maximum?) work done by the system $\int P_{sys} dV$ is the derived quantity.

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  • $\begingroup$ Could you elaborate on the last sentence, and if is intended to explain why the solution adds the net work done and work done on the atmosphere and calls it "maximum work"? $\endgroup$ – Bob D Sep 14 '19 at 14:55
  • $\begingroup$ To be clearer I was just imagining $P_{ext}$ to be produced by the reaction of an elastic container, the mechanical work done on it involves the net pressure $P_{sys} - p$. So the intuition is that the work done by the system is effectively used to expand the container and "to push" the atmosphere. This is as far as I got to interpret the given solution and I'm not sure it really makes sense and why this work should be named "maximum"... $\endgroup$ – user8736288 Sep 14 '19 at 15:36
  • $\begingroup$ Got it. The whole example doesn't make sense, as I indicated in my answer. I envisioned a double piston arrangement. I'm not sure that makes sense either. We are just coming up with contrivances to somehow "fix" a bad example. Anyway, I'm up voting your answer because it also essentially says the example makes no sense as written. $\endgroup$ – Bob D Sep 14 '19 at 15:40
  • $\begingroup$ I agree with you and upvoted you answer. I would still love it if somebody came with a better answer! $\endgroup$ – user8736288 Sep 14 '19 at 15:59
  • $\begingroup$ Me too. But consider this, there simply may not be another better answer than yours or mine. The thermodynamics of the example are simply wrong, as far as I am concerned. $\endgroup$ – Bob D Sep 14 '19 at 16:33
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The gas has received energy (heat from reservoir), has exerted work (expanded) and has changed its internal energy. The work of expanding does not include the work exerted on the change of internal energy, and therefore needs to be calculated separately.

The change in internal energy is correct, but the formula for calculating the work assumes that this work was done under constant pressure (exterior), which is not correct since the gas pressure is larger than the exterior during the whole process. The correct calculation of the work should account for the pressure dependence on the volume:

$W = \int p(V) dV$

which can be deduced from the ideal gas law but requires knowing the rate at which the heat was transferred to the gas.

In any case, the work of expanding was mainly performed at the expense of the internal energy, and the heat received is a smaller amount.

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  • $\begingroup$ The work done always depends on the external pressure. Only when the system is in equilibrium with the surroundings during the entire process is the internal pressure throughout the gas is the same as the external pressure. You can't deduce the work done from the ideal gas law since it only gives you the equilibrium end points of the process and no details on the path. $\endgroup$ – Bob D Sep 5 '19 at 18:45
  • $\begingroup$ You can deduce it indirectly by calculating the change in internal energy (for an ideal gas which depends only on temperature) and coupling that with knowledge of the amount of heat transferred per the first law. The first law doesn't care about the rate of heat transfer. The only thing wrong with the problem description is that the energy transfers doe not balance per the first law. I think the error is in the amount of heat stated in the problem. $\endgroup$ – Bob D Sep 5 '19 at 18:47
  • $\begingroup$ The mechanical work of the gas is the integral in the answer, regardless of the process involved. The change in internal energy and heat are related to the work as you said when there is equilibrium. The system in the problem is clearly never in equilibrium at any point since the internal pressure and temperature differ from the external ones. $\endgroup$ – rmhleo Sep 5 '19 at 23:09
  • $\begingroup$ I agree with everything you just said. Work is always the integral. Where we part ways is that pressure in the integral is always the external pressure, and it is always the external pressure in the PV diagram. I learned this from Chet Miller, who I consider a thermodynamics guru on the SE. If the process is carried out reversibly, then it is also the equilibrium pressure of the gas. It the process is irreversible, as in this example, the gas pressure is not in equilibrium. Either way, pressure comes out of the integral if the external pressure is constant. $\endgroup$ – Bob D Sep 6 '19 at 13:48
  • $\begingroup$ The work done is always the area under the PV path. So here the work is clearly $P\Delta v$. The PV diagram alone does not tell you if the process is reversible or irreversible. In this example the graph is a vertical constant volume line from the original gas pressure to the atmospheric pressure, followed by a horizontal constant pressure line to the final volume. Here the external pressure drop is sudden, causing the process to be irreversible. $\endgroup$ – Bob D Sep 6 '19 at 13:48

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