Question

Suppose 0.540 mol of an ideal gas is isothermally and reversibly expanded in the four situations given below. What is the change in the entropy of the gas for each situation?

Situation (a) (b) (c) (d)

Temperature (K) 250 350 400 450

Initial volume (cm3) 0.200 0.200 0.450 0.350

Final volume (cm3) 0.900 0.700 1.20 1.25

deltaS (J/K) _____ _____ _____ _____

Answer 1

In an isothermally reversibly expansion, the change in heat energy = The work done.

a) T = 250, V2 = 0.9, V1 = 0.2, the work done in an isothermal process = nRTln(V2/V1) = 0.54*8.314*250*ln(4.5) = 1688.1614J, So deltaS = deltaQ/T = Workdone/T = 6.75265 J/K

b) T = 350, V2 = 0.7, V1 = 0.2, the work done in an isothermal process = nRTln(V2/V1) = 0.54*8.314*350*ln(3.5), So deltaS = deltaQ/T = Workdone/T = 5.62435 J/K

c) T = 400, V2 = 1.2, V1 = 0.45, the workdone in an isothermal process = nRTln(V2/V1) = 0.54*8.314*400*ln(2.667), So deltaS = deltaQ/T = Workdone/T = 4.4035 J/K

d) T = 450, V2 = 1.25, V1 = 0.35, the work done in an isothermal process = nRTln(V2/V1) = 0.54*8.314*450*ln(3.57143), So deltaS = deltaQ/T = Workdone/T = 5.71506 J/K