Question

Consider a monoatomic ideal gas of N moles in a gas cylinder eqilibrated at temperature T1 and pressure P1 by a mass placed on the piston. Upon removal of the mass , the gas reaches a new eqilibrium pressure P2 (<P1). Calculate the amount of work done by the gas on the surroundings for the following processes.

( You must express your answer in terms of the given variables.)

1. a nonquasistatic isothermal process (sudden removal of the mass)

2. a quasistatic isothermal process (gradual removal of the mass)

3. a nonquasistatic adiabatic process

4. a quasistatic adiabatic process

5. Show that for both isothermal and adiabatic processes, the quasistatic work is larger than the

nonquasistatic work.

Answer 1

consider a monoatomic gas of N moles

in a cylinder , temperature T1, pressure P1

on removing a mass

P2 < P1

a. for non quasi static isothermal process

the pressure shall suddenly drop to P2 and then the work is done

W = P2*(V2 - V1) = nRT1(1 - P1/P2)

b. for quasistatic isothermal process

initially

P1V1 = nRT1

V1= nRT1/P1

finally

P2V2 = nRT1

V2 = nRT1/P2

work done by gas on surroundings = integral(PdV)

P = nRT1/V

W = nRT1*ln(V2/V1) = nRT1*ln(P1/P2)

c. for non quasistatic adiabatic process

W = P2(Vf - Vi) = nRT1(1 - P1/P2)

d. for quasistatic adiabatic process

W = P2V2^gamma(V2^(1 - gamma) - V1^(1 - gamma))/(gamma - 1)

P2V2^gamma = P1V1^gamma

P1V1 = nRT1

P2V2 = nRT2

W = P1V1^gamma((V1(P1/P2)^(1/gamma))^(1 - gamma) - V1^(1 - gamma))/(gamma - 1)

W = nRT1((P1/P2)^((1 - gamma)/gamma)) - 1)/(gamma - 1)

as we can clearly see

comparingh case 1 and 2

W2 - W1 = nRT1*ln(P1/P2) - nRT1(1 - P1/P2)

dW = nRT1[ln(P1/P2) - 1 + P1/P2]

P1 > P2 ( given)

ln(P1/P2) > 0

P1/P2 > 1

P1/P2 - 1 > 0

P1/P2 - 1 + ln(P1/P2) > 0

hence

dW > 0

more work is done in quasi static process

for 3 and 4

dW = nRT1{[(P1/P2)^(1/gamma - 1) - 1]/(gamma - 1) - 1 + P1/P2}

P1/P2 > 1

gamma > 1

0 < 1/gammma < 1

-1 < 1/gamma - 1 < 0

hence

(P1/P2)^-1 > (P1/P2)^(1/gamma - 1)

hence

1 > P2/P1 > [(P1/P2)^(1/gamma - 1)]

hence

dW > 0

hence work donein quasistatic process 4 is more than the irreversible non quasistatic process 3