Thermal Physics, work done / internal energy help

Hey! I do CAIE exam board and I'm struggling in No. 2 of this paper https://papers.gceguide.xyz/A%20Levels/Physics%20(9702)/9702_w18_qp_41.pdf

Can someone help me with answering the table? I really don't know how to do it Thank you!

Hi, I am not sure why the helper deleted the post. If you still need help on the question, let me know. I would provide some explanation.

Hey! I do CAIE exam board and I'm struggling in No. 2 of this paper https://papers.gceguide.xyz/A%20Levels/Physics%20(9702)/9702_w18_qp_41.pdf

Can someone help me with answering the table? I really don't know how to do it Thank you!

hello! yes please! here's what i think i know, please tell me if i'm correct.

for P->Q, work done on gas is 0 because volume doesnt change. since delta U = W + Q, therefore increase in internal energy is +97.0 + 0 = +97.0 J

for Q->R, U=W + Q so U = -42.5 + 0 = -42.5 J

i don't know how to find for R->P though, i found that solving it like hess' law from chemistry gives the right answer, but is there another way to do it?

Hi, thank for the reply. Your answer and explanation for the change P→Q and Q→R is correct.



As for the change R→P, you need to make use your answer for 2(b)(i), the total change in internal energy of the gas during the complete cycle PQRP is zero.

Using this info, you can find the increase in internal energy of gas for the change R→P.

Then apply the first law of thermodynamics

for the change R→P to find thermal energy transferred to gas.

so since internal energy of P->Q is 97.0, and Q->R is -47.5, we have 97.0 + (-42.5) = 54.5 J, but since total internal energy is 0, then internal energy of R->P would have to be -54.5 J yes?

Correct.