A piston contains 540 moles of an ideal monatomic gas that initally has a pressu

Question

A piston contains 540 moles of an ideal monatomic gas that initally has a pressure of 2.49 × 105 Pa and a volume of 2.9 m3. The piston is connected to a hot and cold reservoir and the gas goes through the following quasi-static cycle accepting energy from the hot reservoir and exhausting energy into the cold reservoir.

The pressure of the gas is increased to 5.49 × 105 Pa while maintaining a constant volume.

The volume of the gas is increased to 9.9 m3 while maintaining a constant pressure.

The pressure of the gas is decreased to 2.49 × 105 Pa while maintaining a constant volume.

The volume of the gas is decreased to 2.9 m3 while maintaining a constant pressure.

It may help you to recall that CVCV = 12.47 J/K/mole and CPCP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.

Explanation / Answer

number of moles=n=540
initial pressure=P1=2.49*10^5 Pa
initial volume=V1=2.9 m^3
Cv=12.47 J.K/mole

Cp=20.79 J/K/mole

then initial temperature=T1=P1*V1/(n*R)=160.84 K

process 1:

pressure is increased to P2=5.49*10^5 Pa at constant volume

then work done=W1=P*dV=0

new temperature=T2=P2*V1/(n*R)=354.62 K

for constant volume process, heat supplied to the system=Q1=n*Cv*(T2-T1)=1.3049*10^6 J

process 2:

volume is increased to V2=9.9 m^3 at constant pressure.

then work done by the gas=W2=P2*(V2-V1)=3.843*10^6 J

temperature at this stage=T3=P2*V2/(n*R)=1210.6 K

heat supplied to the system=Q2=n*Cp*(T3-T2)=9.6098*10^6 J

process 3:

pressure is decreased to P1 at constant volume V2.

work done=W3=0

temperature=T4=P1*V2/(n*R)=549.07 K
heat energy added to the system=Q3=n*Cv*(T4-T3)=-4.4546*10^6 J

-ve means heat is transferred out of the system

process 4:

volume is decreased to V1 at constant pressure of P1.

temperature at this stage will be T1.

then work done by the gas=W4=P1*(V1-V2)=-1.743*10^6 J

heat energy added to the system=Q4=n*Cp*(T1-T4)=-4.3585*10^6 J

part a:

energy transferred into the gas=Q1+Q2=1.3049*10^6 +9.6098*10^6=10914700 J

part b:

energy transferred out of the gas=-(Q3+Q4)=-(-4.4546*10^6 -4.3585*10^6)=8813100 J

part c:

work done by the gas=W1+W2+W3+W4=3.843*10^6-1.743*10^6=2.1*10^6 J

part d: efficiency=work done/total energy input=2.1*10^6/10914700=19.24%

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