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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?

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    $\begingroup$ As you’ve said, the problem statement, which does not specify a path, is insufficient to calculate the path functions $q$ and $w$. $\endgroup$
    – a-cyclohexane-molecule
    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). $\endgroup$
    – Chet Miller
    Sep 16, 2018 at 12:30

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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.

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  • $\begingroup$ 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. $\endgroup$
    – a-cyclohexane-molecule
    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. $\endgroup$
    – Chet Miller
    Sep 16, 2018 at 12:28
  • $\begingroup$ @SoumikDas. So T = 223.15+15V doesn't satisfy the two end conditions? $\endgroup$
    – Chet Miller
    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
    Sep 16, 2018 at 17:57

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