How is it possible to calculate the work done by a gas when pressure is not explicitly stated to be constant? The question states:

Calculate $\Delta H$, $Q$, and $W$ when $1\ \mathrm{mol}$ of He expands from $V = 5\ \mathrm L$ at $T = 298.15\ \mathrm K$ to $V = 10\ \mathrm L$ at $T = 373.15\ K$.

I am not told that the process is isobaric, adiabatic, or even to treat He as an ideal gas (although I know it roughly approximates one).

If I treat it as an ideal gas, I can find $\Delta H$ by using the equations $$H = U + pV$$ $$p = nRT/V$$ and $$U = (3/2)RT$$ but I have no ideas about finding $Q$ or $W$ since I cannot make any other assumptions. Is there an implicit assumption I should make about He gas (or ideal gases in general) expanding, or is there simply not enough information to solve this problem?

  • 1
    $\begingroup$ As you’ve said, the problem statement, which does not specify a path, is insufficient to calculate the path functions $q$ and $w$. $\endgroup$ Commented Sep 16, 2018 at 12:17
  • $\begingroup$ To add to what @a-cyclohexane-molecule said, it is possible to determine Q-W for any unspecified path between the two end states, but not Q and W individually (which are both non-unique). $\endgroup$ Commented Sep 16, 2018 at 12:30

1 Answer 1


You can calculate the work done and Heat exchange if you assume that the process is quassi-statically performed and it is reversible and most importantly done in a single step.

Assume that the process through which the $\ce{He}$ is expanded obeys the equation, $PV^x = K$(constant). This equation can also be alternatively stated as, $TV^{x-1} = K$. Now, suppose the gas is transformed to the state $(T_2,V_2)$ from the state $(T_1,V_1)$. So, we can thus write, $$T_2V_2^{x-1} = T_1V_1^{x-1}$$ so, putting the given conditions, we can find $x$ as, $$x= 1+ \frac{\log\left(\frac{T_2}{T_1}\right)}{\log\left(\frac{V_1}{V_2}\right)} \approx 0.676$$ Thus, after knowing $x$ ,you can calculate the work done by the gas as, $$W= \int_{V_1}^{V_2} P\mathrm{d}V = \int_{V_1}^{V_2} \frac{K}{V^x}\mathrm{d}V = \frac{P_2V_2 - P_1V_1}{1-x} = \frac{R(T_2-T_1)}{1-x}$$ We know all the value of $T_1, T_2 $ & $x$, and work done can be easily calculated. and we already know the change in Internal energy and thus heat exchange can be easily calculated and also $\Delta H $ can be calculated ($Q = (C_V + \frac{R}{1-x}) \Delta T$ and $\Delta H = C_P \Delta T $) and thus all the quantitites can be known.

  • $\begingroup$ While $PV^x = C$ is perhaps the simplest general assumption you can make, it is certainly not implied by the question statement. I know you mention that this is an assumption in your answer, but I’d like to emphasize it again. $\endgroup$ Commented Sep 16, 2018 at 12:25
  • $\begingroup$ Another path would be to assume that temperature is varied linearly with volume along the process path. This would lead to a different amount of work and heat. $\endgroup$ Commented Sep 16, 2018 at 12:28
  • $\begingroup$ @SoumikDas. So T = 223.15+15V doesn't satisfy the two end conditions? $\endgroup$ Commented Sep 16, 2018 at 16:04
  • $\begingroup$ @Chester Miller, Oh Sorry, I had some misunderstandings. Yes it is possible. Sorry for that... $\endgroup$
    – Soumik Das
    Commented Sep 16, 2018 at 17:57

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.