Question

One mole of an ideal gas at 1.00 atm and 298K with C_{p,m=3.5R is put through the}

following cycle: a) (reversible) constant volume heating to twice its initial

temperature; b) reversible, adiabatic expansion back to its initial temperature;

c) reversible isothermal compression back to 1.00 atm. Calculate ΔS for each step and

overall. d) What is the change of entropy of the surroundings?

Answer 1

Initial volume:

PV = nRT

VA = nRT / P = 1 mol * 8.314 J/mol K * 298 K / 101325 Pa = 0.02445 m3

1. SB - SA = n Cv ln (TB/TA) + n R ln(VB/VA)

As VB = VA , ln(1) = 0, n = 1 mol

SB - SA = Cv ln (TB/TA)

For di-atomic gases Cp = 3.5 R and Cv = 2.5 R

Also, TB = 2TA

SB - SA = 2.5 mol R ln (2TA/TA) = 2.5 R ln (2) = 2.5 mol * 8.314 J/mol K * ln(2)

SB - SA = 14.4 J/K

2) Sc-SB = n Cv ln ( TC / TB) + n R ln ( VC / VB) = 0 (adiabatic process)

2TC = TB ; VB = VA ; n= 1; Cv = 2.5 R; VA = 0.02445 m3

2.5 R ln ( TC / 2TC) - R ln ( VC / VA) = 0

2.5 ln (½) + ln (VC/0.02445) = 0

ln VC = ln (0.02445) - 2.5 ln (½)

Vc = 0.13831 m3

3) SA - SC = n Cv ln ( TA / TC) + n R ln ( VA / VC)

TA / TC = 1; n= 1

SA - SC = R ln ( VA / VC) = 8.314 ln (0.02445/0.13831) = -14.4 J/K

Overall dS = 14.4 + 0 - 14.4 = 0

dSsurroundings = Qsurroundings / T = -W/T = - n R ln (V2/V1),

where 1 and 2 are the first and the last step. As V2 = V1 in our case:

dSsurroundings = 0