Consider a perfect gas contained in a cylinder and separate by a frictionless ad
ID: 493652 • Letter: C
Question
Consider a perfect gas contained in a cylinder and separate by a frictionless adiabatic piston into two sections A and B. All changes in B are isothermal; that is, a thermostat surrounds B to keep its temperature constant. There is 2.00 mol of the gas molecules in each section. Initially T_A = T_B = 300 K, V_A = V_B = 2.00 dm^3. Energy is supplied as heat to Section A and the piston moves to the right reversibly until the final volume of Section B is 1.00 dm^3. Calculate Delta S_A and Delta S_B, Delta A_A and Delta A_B, Delta G_A and Delta G_B, Delta S of the total system and its surroundings. If numerical values cannot be obtained, indicate whether the values should be positive, negative, or zero or are indeterminate from the information given. (Assume C_V, m = 20 JK^-1 mol^-1.)Explanation / Answer
Section A Initial V,T, p
Ti = 300 K
V i = 2 L
p i = 24.6 atm from ideal gas equationof state pV = nRT
Section B initial V,T,p
T i = 300 K
V i = 2 L total V = 4 L
p i = 24.6 atm from ideal gas equation of state pV = nRT
Section A Final V,T, p
V f = 3 L by difference
p f = same as B
Therefore, can find T f = 900K from ideal gas equation of state
Section B Final V,T, p
Tf = 300 K (isothermal)
V f = 1 L (given)
p f = 49.2 atm from ideal gas equation of state
SA and SB
SA
dS = (1/T)Cp dT - (V/T)p dp
SA = n Cp dT/T - n Rdp/p
= 2*28.3144* ln(900/300) - 2*8.3144 * ln(49.2/24.6)
= +50.7 J/K
SB
dS = (1/T)Cp dT - (V/T)p dp
as dT=0
S = - n Rdp/p
= - 2*8.3144 * ln(49.2/24.6)
= - 11.53 J/K
AA and AB
AA = U – TS by definition
A = U - (TS) need to calculate S at initial and final states, but we don’t have enough information here to get Si and Sf separately.
AB
AB = U - (TS) = U -TS
as T = constant
= 0 – 300*(-11.53) J
= 3.46 kJ
GA and GB
GA
G = H - (TS)
We need to calculate S at initial and final states, but we don’t have enough information here to
get Si and Sf separately.
GB
GB = H - (TS)
= H -TS
As T = constant
= -TS
= - 300 * (-11.53) J
= 3.46 kJ
Ssystem = SA + SB
= +50.7 -11.53
= + 39.2 J/K
Ssurround = 39.2 J/K since process was carried out reversibly.
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