A gas-tight frictionless piston of small thermal conductivity slides in a therma

Question

A gas-tight frictionless piston of small thermal conductivity slides in a thermally insulated cylinder, dividing it into two compartments, A and B, each containing nA=n and nB=1.95n moles of ideal monatomic gas (n moles). Initially the temperature of the gas is To in compartment A and 6To in B. Assume that the system is in mechanical equilibrium at all times and that the mass of the piston and the effect of gravity are negligible. Consider the process by which the system reaches thermal equilibrium:

(a) What is the final temperature, Tf?
To
(b) What is the ratio of the volume of A to that of B initially, ri?

(c) What is the ratio of the volume of A to that of B after thermal equilibrium is reached, rf?

(d) Calculate the work done on the gas in compartment A.
nRT
(e) Calculate the work done on the gas in compartment B. Does this answer make sense? Why or Why not?
nRT
(f) Calculate the change in entropy for A.
nR
(g) Calculate the change in entropy for B.
nR
(h) Calculate the change in entropy for the entire system. Does this answer make sense? Why or Why not?
nR

Explanation / Answer

First of all we try to understand the process, as this is mentioned that the system is thermally insulated then there will be no heat transfer from surrounding to the system. But heat transfer can take place between the two compartment. Since we have neglected the gravity and mass of piston and no pressure is applied on the piston other than atmospheric therefore this complete process will takes place under constant pressure.
So enthalpy in compartment A is mCpdt where m is mass Cp is specific heat at constant pressure and dt is the change in temp. similarly for compartment B.
Change in enthalpy in both the comaprtment is equal so we make the things equal like this
(mCp(Ti - Tf) )A = (mCp (Ti - Tf))B we can replace m by no. of moles directly because of equating the both will not effect the dimensions. Tf will be same for both compartment and we know the initial temperature of each compartment and Cp will get cancelled. therefore we have only one variable in the above eqaution therefore we can calculate it .
(b) From the ideal gas equation i.e. PV= n RT since pressure is constant throughout the process therefore we will take ratio for both compartments.
(VA/VB) = (nATiA) /(nBTiB) , we know all the values therefore we can calculate the ratio.
(c) when thermal equilibrium is reached then the temperature become equal for both compartment theredfore the volume ratio will be equal to their mole ratio.
(d) Work done will be equal to change in enthalpy therefore we can calculate mCpdt for compartment A .
(e) This answer does not make sense because work is done by the gas compartment B due to high temperature and mole value therefore the final temperature would be less than the initial temperature which shows that the work is done by compartment B.
(f) we have to apply the formula for change in entropy at constant pressure therefore for compartment A.
  it will be like this mCp ln(Tf/Ti)
(g) Similarly we calculate it for compartment B.
(h) this answer make sense because the enropy change for both the compartment will be different therefore the net change in entropy will have some value.

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