Question

Consider the slow adiabatic compression of a closed volume of gas for which C_p = 3.5 R. If the initial gas temperature is 320 K and the ratio of the final pressure to the initial pressure is 4, what is the change in enthalpy of the gas, the change in internal energy, the heat transferred Q, and the work W? Assume a basis of 1 mole of gas.

Answer 1

In an adiabatic process the relation between the temperature and pressure is given as

(T1/T2)^ϒ = (P2/P1)^1-ϒ

ϒ = ratio specific heats = (Cp/Cv)

T1 and P1 =initial temperature and pressure respectivelly

T2 and P2 = final temperature and final pressure repectivelly

Cp - Cv = R

Cv = Cp-R

Cv = 3.5R - R = 2.5 R

ϒ = 3.5R/2.5R = 1.4

(T1/T2)^ϒ = (P2/P1)^1-ϒ

(T1/T2) = (P2/P1)^1-ϒ/ϒ

320 / T2 = (4)^-0.285

320 / T2 = 0.674

T2 = 320 / 0.674 = 474.7K

Change in enthalpy (ΔH) = CpΔT

ΔT = T2 - T1 = 474.7 - 320 = 154.7K

ΔH = 3.5 X 8.314 X 154.7 = 4501.6J

Change in enternal energy (ΔU) = CvΔT

ΔU = 2.5 X 8.314 X 154.7 = 3215.4J

Heat change = 0 J As the process is adiabatic.

Work done (W)

ΔU = Q + W

Q = 0

ΔU = W = 3215.4J