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One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature T. The face of the piston opens into the atmosphere. What is the heat supplied in the process to expand the gas from volume $V_1$ to $V_2$ isothermally ? Friction of the piston against the cylindrical wall is negligible is small

Now thermodynamics is a common topic to both physics and chemistry and as per my understanding of thermodynamics I am getting different results for the heat that will need to be supplied.

Chemistry says $$∆U =∆Q +∆W $$ and defines work as the work done on the gas and this will give $$W = -P_{atm}(V_2-V_1) $$ and as for isothermal process$$ ∆U =0 , ∆Q =P_{atm}(V_2-V_1)$$

According to physics on the other hand $$∆Q =∆U+∆W$$ and here the work done is the work done by the gas. That is

$$W = \int_{V_1}^{V_2} PdV=\int_{V_1}^{V_2} \frac{RT}{V}dV = RT \ln V_2-RT \ln V_1$$ and as for isothermal process $$∆U =0 ,\, ∆Q =RT \ln \left(\frac{V_2}{V_1}\right)$$

I am also uncertain that maybe neither of these is the correct expression and the network ok is actually in that work from both these forces and we have to take the sum of these to yield $$∆Q =RT\ln \left(\frac{V_2}{V_1}\right)-P_{atm}(V_2-V_1)$$

Now the value of Q should be unique because in the real world when we perform an experiment, only one value of Q would be supplied.

So which one of these is correct and where am I lacking in my understanding of thermodynamics from a chemistry standpoint ??

Now as this question links both physics and chemistry, I want to post it on both sites (I hope it's fine) and the link to it is here

Buck Thorn
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Mr.HiggsBoson
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    It's not particularly fine - crossposting is frowned upon. – Mithoron May 23 '20 at 16:26
  • Sorry, but actually it doesn't... I'm not asking exactly why they are different , but just that due to different conventions, what I understand is we take different pressure (ie internal P v/s external P) and thus get different values of Q, whereas there should only be one value(As far as I understand) – Mr.HiggsBoson May 23 '20 at 16:44
  • Hmm. I would suggest you to read the definitions of first law of thermodynamics again as you are making a common misconception! BOTH equations are equivalent. The ∆W in physics is work done by the gas, whereas ∆W used in chemistry is work done on the gas. As these are always equal in magnitude and opposite in sign you always have the exact same equation!. You may use whichever definition you prefer in whichever subject you wish! Just ensure you don't mix up the sign anywhere/ calculate one quantity instead of the other. – user600016 May 23 '20 at 17:16
  • At last, some questions they may ask just to calculate "∆W" without mentioning work done on/by the gas. Only in this(annoying) case do you have to answer as per the "subject" this question is from. – user600016 May 23 '20 at 17:17
  • I understand they only differ in sign , but I want to know , which one is correct ? – Mr.HiggsBoson May 23 '20 at 18:26
  • @RandomAspirant - RE: "... which one is correct?" – Are you a chemist or a physicist? – MaxW May 23 '20 at 19:37

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The problem is that both the problem statement you presented and the chemistry solution you showed are fallacious. The fallacy is that, for an ideal gas $V=nRT/P$, so that, if the pressure and temperature are both constant, the volume can't change. If the volume does change, then either the pressure must change or the temperature must change.

In the "chemistry solution" you presented, the pressure is taken as constant (atmospheric). So the temperature must change. But if the temperature changes, then $\Delta U$ is not zero (as they assume). So the solution is incorrect. The correct solution, based on the 1st law, is $$\Delta U=nC_v\Delta T=Q-P_{atm}(V_2-V_1)$$or equivalently, $$\Delta H=\Delta U+P_{atm}(V_2-V_1)=nC_p(T_2-T_1)=Q$$Also, since $$n=\frac{P_{atm}V_1}{RT_1}$$we have $$\frac{T_2}{T_1}=1+\frac{QR}{C_pP_{atm}V_1}$$

In the "physics solution" you presented, they make the opposite assumption that, rather than the pressure remaining constant, the temperature remains constant. For the temperature remaining constant, the solution you presented is correct. But the final pressure is: $$P_2=P_{atm}\frac{V_1}{V_2}$$

Chet Miller
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