Question

A 0.520-mol sample of an ideal diatomic gas at 432 kPa and 324 K expands quasi-statically until the pressure decreases to 144 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is the following.

a) isothermal and adiabatic final temperature volume of the gas wrok done by the gas heat absorbed?

K=?, L=?, work done?, heat absorb?

Answer 1

PV = nRT

V = 0.52*8.314*324/(432*10^3) = 3.242*10^-3 m^3 = 3.242 L

A. Isothermal Ti = Tf = 324 K

Vf = Vi*(Pi/Pf) = 3.242*(432/144) = 9.726 L

Wby gas = nRT*ln(Vf/Vi) = 0.52*8.314*324*ln(9.726/3.242) = 1538.87 J

dU = n*Cv*dT = 0

as dT = 0

Q = dU - Won =

Q = 0 - (-1538.87) = 1538.87 J

B. Adiabatic

in this case y = 5/3

Vf = Vi*(Pi/Pf)^(1/y)

Vf = 3.242*(432/144)^(3/5) = 6.267 L

Tf = Pf*Vf/nR = 144*10^3*6.267*10^-3/(0.52*8.314) = 208.74 K

In adiabatic Qin = 0

W = dU - Q = dU

dU = nCv*dT = 1.5*n*R*dT

W = dU = 1.5*0.52*8.314*(208.74 - 324) = -747.45 J

Won = -747.45

Wby gas = 747.45 J