TWo containers each containing 2 m oles of th same ideal gas at the same pressure and tempreature are kept isolated from one another. Two procedures are performed:300 J of heat enery is added to the gas in container 1 and container two is allowed to expand, the gas inside doing 300 j of work work on the container walls. What is the difference between internal energies of the two samples of gas after these procedures?


This question is from Kaplan Q bank and the answer is 600J of work. I dont understand kaplan's explanation for this question.
Regards.

container 1 is not allowed to expand and will not feel any effects when any other energies are added to it. But, container 2 is allowed to expand. since there are 2 moles of the gas, adding 300 J/mol will give you 600 J of work

the wording is a bit confusing, but I think the energy added is 300 J and the other container work is being done so the W=PV volume is getting bigger so the work is positive done by the system. So there is a 600 J difference.

W=q-w
W=q-PV
W=0-300 (Beaker 1) -> -300J
W=300-0 (Beaker 2) -> +300J

Difference between the two, 300-(-300)=600J difference in energy.

hope this helps