Question: 1 mol of a monoatomic gas at 298 Kelvin acquires a volume of 3 litres. It is expanded adiabatically and reversibly to a pressure of of 1 atm. It is then compressed isothermally and reversibly until its volume becomes 3 litres. Calculate the change in entropy.

What I tried: As both the processes are reversible, and since entropy is a state function rather than a path function, we can ignore the middle step and only consider the initial and final state.

Considering $S$ to be the change in entropy:

Thus: $$S = nC_vln({T_2/T_1})$$

where $C_v$ represents the heat capacity at constant volume, which for a monoatomic ideal gas is $1.5R$. However, I am unable to understand how to get the final temperature. I tried applying several concepts but still couldn't obtain the final temperature of the gas. Please help.


closed as off-topic by John Rennie, ACuriousMind, JamalS, Danu, Kyle Kanos Nov 23 '14 at 11:56

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  • $\begingroup$ I definitely did provide some info on how I attempted the question. Don't think the close is exactly fair. $\endgroup$ – Gummy bears Nov 23 '14 at 12:44

The gas is expanded adiabatically and then isothermally. Thus the temperature it has at the end of adiabatic expansion stays the same even after the isothermal process.

Ideal Gas equation after adiabatic expansion: $p_aV_a=nRT_a$, where index "a" shows after.

You do not have $V_a, T_a$ in this equation.

However, another equation you can write down is the adiabatic process equation: $TV^{γ-1}=const$, where $γ=5/3$, because the gas is monoatomic and ideal.

Now you have a system of two equations: $$T_bV_b^{γ-1}=T_aV_a^{γ-1}$$ and $$p_aV_a=nRT_a$$, where index a means "after", index b means "before" adiabatic expansion. You have $V_b, T_b, n, R, p_a$, so this is solvable.

Solve for $T_a$ and you should get the correct value.

  • $\begingroup$ I see. The second equation, involving gamma, didn't strike me. Is there any way to do this question without the use of that equation though? $\endgroup$ – Gummy bears Nov 23 '14 at 12:43
  • $\begingroup$ Well you could try using the first law of thermodynamics maybe, since for adiabatic process $dQ=0$, thus $dU=-dW$; Since the gas is ideal and monoatomic you can say $dU=/frac32 * nR dT$. I'm not sure how would you get the $dW$ however, and after getting there you would need to integrate. I think using the adiabatic process equation is by far the easiest approach. $\endgroup$ – Henrikas Nov 23 '14 at 15:13

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