Question

Suppose that 132 moles of a monatomic ideal gas is initially contained in a piston with a volume of 0.94 m³at a temperature of 348 K. The piston is connected to a hot reservoir with a temperature of 1064 K and a cold reservoir with a temperature of 348 K. The gas undergoes a quasi-static Stirling cycle with the following steps:

1. The temperature of the gas is increased to 1064 K while maintaining a constant volume.
2. The volume of the gas is increased to 3.38 m³ while maintaining a constant temperature.
3. The temperature of the gas is decreased to 348 K while maintaining a constant volume.
4. The volume of the gas is decreased to 0.94 m³ while maintaining a constant temperature.

It may help you to recall that CV = 12.47 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022

Answer 1

Using ideal gas law;PV=nRT; P=nRT/V;Therefore P=406312.3253Pa;

energy transferred into the gas from the hot reservoir=nRT_H*ln(3.38/0.94)=1494430.595J;

energy transferred out of the gas into the cold reservoir=nRTc*ln(3.38/0.94)=488779.9316J;

Work done=Heat supplied-Heat rejected=1494430.595-488779.9316=1005650.663J;

Efficiency=work done/heat supplied = 1005650.663/1494430.595=67.29%;

Carnot efficiency = 1- Tc/T_H=1-(348/1064)=67.29%;

Q2;

Using PV=nRT;

T1=36.85K; T2=120.62K; T3=770.09K; T4=235.30K;

Using first law of Thermodynamics:Energy supplied(Q)=Change in internal Energy(U)+Work done by the system(W);

That is:Q=U+W;

U=nCv*(change in temperature);

W=P*(change in volume);

U1=584982.664J; W1=0; Q1=584982.664J;

U2=4535378.904J; W2=3024000J; Q2=7559378.904J;

U3=-3734545.53J; W3=0; Q3=-3734545.53J

U4=-1385816.04J W4=-924000J; Q4=-2309816.04J;

energy transferred into the gas from the hot reservoir=sum of +ve Q's=8144361.568J;

energy transferred out of the gas into the cold reservoir=sum of -ve Q's= 6044361.57J;

Work done = W1+W2+W3+W4=2100000J;

Efficiency = Work done/Energy supplied=2100000/8144361.568=25.78%