I really don\'t understand these two problems and wonder if someone can help me?

Question

I really don't understand these two problems and wonder if someone can help me? Thank you!

1) I need part B and C

2) The info is given for this problem. But the answer I put in is wrong...

Mazur 19:HT1 Two containers of ideal gas A and B have initial temperatures 250K and 300 K each with the same number of atoms -0.200 moles. The containers are put together and heat is allowed to flow between the two containers until they reach equilibrium. The containers A and B together are an isolated system (no energy or Q flows out of the total system A and B) Part A Containers A and B have pressure of 10*5 Pa. They are not allowed to change volume. What is the equilibrium temperature for box A? 275 K Submit My Answers Give Up Correct Part B Assuming the container volumes do not change, What is the entropy change to reach equilibrium? My Answers Give Ue Submit Incorrect; Try Again; 4 attempts remaining Part C Assume instead that the containers are pistons that are thermally isolated but still allowed the container to change volume. The pressure on the other side of the piston is always 1 atm=1.01 X 105 Pa. What is the change in entropy in reaching equilibrium? Submit My Answers Give Up

Explanation / Answer

Sorry we are allowed to solve one question at a time
so for
(b) It is given that the process is at constant volume
So entropy change will be
dS = mCVln(T2/T1)A + mCln(T2/T1)B  where Subscript A and B denotes the container A and B
since both have same no. of moles therefore mC willl be common
dS = mCV(ln(275/250) + ln(275/300)) = 0.0083mCV
Since the value of C is not provided in the data, so please plug the value.
(c) For the constant pressure
dS = mCP ln(T2/T1)A + mCP ln(T2/T1)B
In this case
dS = 0.0083 mCP

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