Question

10.0 L of an ideal gas at 0°C and 10.0 bar are expanded to a final pressure of 1.00 bar C_V = 3/2 R. Calculate deltaU, deltaH, q, w, and deltaS if the process is:

a) reversible and isothermal

b) irreversible and adiabatic

Answer 1

Initial condition: Volume, V1 = 10.0 L

temperature, T1 = 0 C = 273 K

pressure, P1 = 10.0 bar = 9.87 atm

moles of ideal gas, n = PV / RT = (9.87 atm x 10.0 L) / (0.0821 L.atm.mol-1K-1 x 273K)

= 4.4036 mol

Given Cv = 3/2 R

=> Cp = Cv + R = 3/2R + R = 5/2R

Final condition: Pressure,P2 = 1.00 bar = 0.987 atm

(a): Reversible and isothermal:

For isothermal process, dT = 0

=>deltaU = n*Cv*dT = 0 (answer)

deltaH = n*Cp*dT = 0 (answer)

w = - nRT*ln(Pi/Pf) = - (4.4036 mol * 8.314 J.K-1mol-1*273 K)*ln(10.0 bar / 1.00 bar)

=> w = - 23014 J or 23.01 KJ (answer)

From 1st law of thermodynamics, deltaU = q + w

=> q = deltaU - w = 0 - (- 23014 J)

=> q = + 23014 J (answer)

deltaS = q(rev) / T = + 23014 J / 273 K = 84.3 J.K^-1 (answer)

(b):  Irreversible and adiabatic process:

Heat supplied, q = 0 (answer)

Irreversible and adiabetic expansion is also called adiabetic free expansion of the gas. Hence in irreversible and adiabetic expansion the gas undergoes expansion freely in vacuum without any external pressure. Hence hence work done on the system and by the system is 0 i.e

w = 0 (answer)

From 1st Law of thermodynamics,

deltaU = q + w = 0+0 = 0 (answer)

also deltaH = 0

deltaS = nR*ln(Pi/Pf) = (4.4036 mol * 8.314 J.K-1mol-1)*ln(10.0 bar / 1.00 bar)

=> deltaS = 84.3 J.K^-1 (answer)