Consider a monoatomic ideal gas of N moles in a gas cylinder eqilibrated at temp
ID: 1884490 • Letter: C
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.
Explanation / Answer
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
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.