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Since this is a cyclic process:- Change in internal energy=0. In the process C-A the work done W is zero, thus the heat Q is zero since it is an isochoric process.

Next in the process A-B, I got the heat Q to be equal to 0.5 L-atm and Work done W equal -0.5 L-atm . (Area under the graph and expansion of gas gives negative work done here)

For the process B-C its given Work done W is 1 L-atm (Positive since compression) and since change in internal energy =0, the heat must be negative of work done thus Q = -1 L-atm

Therefore the total heat exchange corresponding to each process should be -1+0.5+0=-0.5. The absolute value being 0.5 L-atm.

But the answer gives 1.5 L-atm as the total heat exchange. How? Please explain


closed as off-topic by Mithoron, Mathew Mahindaratne, airhuff, Todd Minehardt, Nuclear Chemist Sep 8 at 17:26

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  • $\begingroup$ The change in internal energy for each step of the process is not zero; just the overall process. In CA, the change in internal energy is negative, as is Q. Also, apparently in this problem, the criterion that work done on the system is considered positive is being used. $\endgroup$ – Chet Miller Sep 6 at 17:51

Since the change in internal energy for the cycle is zero, if you are using the form of the 1st law written as $\Delta U=Q+W$ (where work done on the system is regarded as positive), the heat for the entire cycle must equal minus the work for the entire cycle. The work done on the system in AB must be -(1)(1.5-1)=-0.5 L-atm. Therefore, the work done on the system for the entire cycle is 1-0.5=0.5 L-atm. So, the heat transferred to the system for the entire cycle must be -0.5 L-atm.

  • $\begingroup$ Exactly, thats what I coclude as well! But the given answer is +1.5 L-atm $\endgroup$ – Asad Ahmad Sep 7 at 5:55
  • $\begingroup$ There are multiple issues with this problem statement. Our answer is correct and the "given answer" is incorrect. $\endgroup$ – Chet Miller Sep 7 at 13:06

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