Suppose we have a fairly well insulated frictionless piston that divides a conta

Question

Suppose we have a fairly well insulated frictionless piston that divides a container into two volumes, A and B. Volumes A contains nA=n moles of an ideal monatomic gas and Volume B nB=1.2n moles of ideal diatomic gas (rigid rotator). Initially the temperature of the gas is To in volume A and 5To in volume B. Assume that the system is always in mechanical equilibrium. Furthermore assume the piston mass and gravitational effects are negligible. When the system reaches thermal equilibrium:

(a) Calculate the final temperature, Tf.
  To
(b) Calculate the ratio of the volume of A to that of B initially, ri.
  
(c) Calculate the ratio of the volume of A to that of B after thermal equilibrium is reached, rf.
  
(d) The only information about the pressures of the gases in this problem is that the system is always in equilibrium. A relationship that can be established between the pressures before and after the process is PAi/PAf=PBi/PBf, where the letter i in the subscript represents initial state and f represents final state. To calculate the total work done by any of the gases one must specify the path taken form the initial state to the final state, otherwise the work done by the gas cannot be calculated since it is NOT a state function. Suppose the process takes place isobarically. What is the work done by the gas in compartment A?
  nRT
(e) What is the work done by the gas in compartment B? Does this answer make sense? Does this answer say anything about whether or not this process can take place isobarically? 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

given the piston is well insulated

Divides the container into two volmes, volume A and volume B

Volume A -> n moles, temperature To, Volume = Va

Volume B -> 1.2n moles, temperature 5To, Volume = Vb

For mechanical equilibrium the pressure in both the compartments should be the same

so, using idfeal gas law PV = nRT

nR*To/Va = 1.2*5*n*RTo/Vb

Vb/VA = 6

A. NOW WHEN THE SYSTEM REACHES THERMAL equilibrium

heat given by the chamber with 5To temperature = heat recieved by the other chamber

for same Cp for same gases

n(T - To) = 1.2n(5To - T) where T is equilibrium temp[erature]

T - To = 6To - 1.2T

2.2T = 7To

T = 3.1818 To

Tf = 3.1818To

B. initial volume ratio Vb/Va = 6

Va/Vb = ri = 1/6

C. after thermal equilibrium is reached

nRTf/Va = 1.2nRTf/Vb

Va/Vb = 10/12 = 5/6 = rf

D. For isobaric case

Work done by gas in compartment A = Pi*(Vf - Vi)

now, PiVi = nRTo

PiVf = nR(3.1818To)

so work done = 2.1818nRTo

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