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This is a doubt on question 5.13 from the 5th Chapter of the book Heat and Thermodynamics by Zemansky and Dittman. The question says we have a cylinder with a movable piston, on either side of which we have a monoatomic ideal gas. It is given that heat is supplied slowly to the left side, due to which the nonconducting piston expands. My doubt is not about the calculations, I am just not able to understand what thermodynamic process takes place on the left side. On the right side, we see there is no addition or loss of heat, so it's an adiabatic process, but what about the left side? I am guessing it is an isothermal process due to slow heat addition, is it correct?

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  • $\begingroup$ Slow heat addition doesn't imply an isothermal process. Just because the temperature is uniform doesn't mean it's constant. $\endgroup$
    – Chemomechanics
    Feb 19 at 19:41

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I checked my copy of Zemansky. You left out a couple of important points. One is not just the piston is non conducting, but also the cylinder which, with the exception of one end cap where heat can be added, is thermally insulated. The second is the piston is frictionless which, together with the process being carried out slowly (quasi-statically) means both the expansion on the left and the compression on the right is reversible.

In any case the reversible expansion process on the left side can't be isothermal. A reversible isothermal expansion requires the external pressure be slowly reduced such that, on the left side, $PV=$constant. Here the external pressure is that on the right side of the piston, which increases from 1 Atm to 7.9 Atm during the compression, not decreases.

But to answer the questions you don't need to know the process on the left side since the process on the right side is a reversible adiabatic compression. There is an equation for that process and the work done by that process. By combining the ideal gas law and the equation for a reversible adiabatic process, together with the given initial and final conditions, you should be able to answer all the questions.

Hope this helps.

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  • $\begingroup$ Thanks for the reply. But we are also asked to find the final temperature on the left side right? To do that, should I assume that pressure on both sides is equal(in this case, 7.59 atm) due to equilibrium so that I can apply the ideal gas law? $\endgroup$
    – V Govind
    Feb 20 at 15:25
  • $\begingroup$ Yes. The pressure on both sides has to be equal at all times in order for the processes to be reversible $\endgroup$
    – Bob D
    Feb 20 at 15:40
  • $\begingroup$ Ok, now it's clear! $\endgroup$
    – V Govind
    Feb 20 at 15:56
  • $\begingroup$ So my answer is acceptable to you? $\endgroup$
    – Bob D
    Feb 20 at 18:44
  • $\begingroup$ Yes, it is acceptable, makes sense now. $\endgroup$
    – V Govind
    Feb 21 at 1:25
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There is no reason the process on the left should be isothermal. It is a quasi-static (if slow enough) heating process. The presence of an ideal gas implies that the temperature will increase (the ideal gas has no phase transition). Thermodynamic processes are not limited to isothermal and adiabatic transformations.

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