Consider a well-insulated container of total volume V0. (a) Initially, the conta
ID: 2994548 • Letter: C
Question
Consider a well-insulated container of total volume V0.
(a) Initially, the container is divided into two equal parts, using a very thin partition. One side is filled with a mole of argon at temperature T0and pressure P0 whereas the other side is under vacuum. At t = 0, the partition is suddenly removed. After the system reaches equilibrium, calculate the final temperature, and pressure. Calculate the change in the entropy of the system and of the universe. How would your answer change if the system were in thermal contact with a bath at temperature T0?
(b) Suppose the partition is replaced with an adiabatic piston (see figure). If the gas is allowed to expand reversibly from initial volume V0/2 until it occupies the final volume V0, calculate the final temperature, pressure and the work done on the piston by the gas. Calculate the change in the entropy of the system and of the universe. How would your answer change if the system were in thermal contact with a bath at temperature T0?
Assume argon to be an ideal gas and its heat capacity to be independent of
temperature. R = 8.314 J/ (mol K)
Explanation / Answer
Part A and Part B are Similar Questions
a) Initially
P1 = Po
T1 = To
V1 = Vo/2
Finally
V2 = V0
P2 = ?
T2 = ?
The Process is Isentropic
P*V^1.667 = Constant
P1*V1^1.667 = P2*V2^1.667
P2 = 0.315 Po
From Ideal gas Equation
P1*V1 = n*R*T1
Po*Vo/2 = n*R*T0
n*R = Po*Vo/(2*To)------(1)
P2*V2 = n*R*T2
0.315*Po*Vo = Po*Vo/(2*To) *T2
T2 = 0.63 To
Since the Process is Isentropic (Heat Transfer,dQ = 0)
Entropy of the system(dS1) = dQ/T2 = 0
Entropy of the Surrounding(dS2) = -dQ/T(inf) = 0
Entropy of Universe,(dS) = dS1 + dS2 = 0
When temperature is kept constant
P1*V1 = P2*V2
Po*Vo/2 = P2*Vo
P2 = Po/2
T2 = To
Change in Internal energy,dU = 0 (since temperature is constant)
Work done in an Isothermal Process
W = n*R*T1 ln(V2/V1) = P1*V1*ln2 = 0.347*Po*Vo
From the first law
dQ = dU + W
dQ = W = 0.347*Po*Vo
Entropy of the System,dS1 = dQ/To = 0.347*(PoVo/To)
Entropy of the Surroundings,dS2 = -dQ/To = - 0.347*(PoVo/To)
Entropy of the Universe,dS = dS1 + dS2 = 0