1
$\begingroup$

I've been thinking to this for the last two hours and haven't been able to come with a solution.

Problem. A mole of gas initially at pressure $P_A = 2 \text { atm}$ and occupying a volume $V_A=20\,\text {L}$, passes through an irreversible transformation in which it absorbs $Q=1200J$ as heat and ends with pressure $P_B=3/2 P_A$. Then it expands reversibly and in adiabatic conditions, until its pressure is again $P_A$ ($P_C=P_A$). Finally, it has an isobaric transformation that brings the volume back to $V_A$.

Given that the absolute values $|W_{CA}|=300J$ and $|\Delta U _{CA}|=450J$, determine the thermodynamic coordinates $P,V,T$ in the three states $A,B,C$.

I think that the only way to find the coordinates is to work in reverse: finding $V_C$ from $$W_{CA}=p_A(V_A-V_C),$$ and then $V_B$ using the adiabatic's $P|V$ law.

Now, the information on the absolute values tells us that $$Q_{\text {tot}}=Q_{AB}+Q_{CA}>0,$$ since $Q_{CA}\geq-750J$ and so $W_{\text{tot}}>0$. Having noted this, I don't see how does it put a constraint on the sign of $W_{CA}$. Since the transformation $A\to B$ is totally unspecified, I think that the volume $V_B$ could be everything without any contradiction, and this affects the sign of the work in last transformation.

Have any idea?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.