A piston contains 150 moles of an ideal monatomic gas that initally has a pressu

Question

A piston contains 150 moles of an ideal monatomic gas that initally has a pressure of 1.61 × 105 Pa and a volume of 1.4 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 4.61 × 105 Pa while maintaining a constant volume.

The volume of the gas is increased to 8.4 m3 while maintaining a constant pressure.

The pressure of the gas is decreased to 1.61 × 105 Pa while maintaining a constant volume.

The volume of the gas is decreased to 1.4 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.

1)

How much energy is transferred into the gas from the hot reservoir?

2)

How much energy is transferred out of the gas into the cold reservoir?

3)

How much work is done by the gas?

4)

What is the efficiency of this cycle?

Explanation / Answer

I have solved a very similar problem here. You can follow the same procedure with your values. I hope that was helpful, please rate....

No. of moles, n=410

Given:P1=2.89*105 Pa;V1=3.1m3;using ideal gas equation PV=nRT;we get T1=262.809K=-10.35 C

P2=5.89*105 Pa;V2=3.1m3;using ideal gas equation PV=nRT;we get T2=535.622K=262.462 C

P3=5.89*105 Pa;V3=10.1m3;using ideal gas equation PV=nRT;we get T3=1745.09K=1471.93 C

P4=2.89*105 Pa;V4=10.1m3;using ideal gas equation PV=nRT;we get T4=856.25K=583.09 C

USING FIRST LAW OF THERMODYNAMICS:HEAT GIVEN(Q)=CHANGE IN INTERNAL ENERGY OF THE SYSTEM(U) + WORK DONE BY THE SYSTEM.(W)

THEREFORE Q=U+W

U=nCv*(CHANGE IN TEMPERATURE);W=P*(CHANGE IN VOLUME);

FROM 1----->2;

U(1---->2)=nCv(T2-T1)=(410)*(12.47)*(535..622-262.809)=1394.811kJ

W(1---->2)=0(SINCE ZERO CHANGE IN VOLUME)

Q1=1394.811kJ USING Q=U+W

FROM 2----->3;

U(2---->3)=nCv(T3-T2)=(410)*(12.47)*(1745.09-535..622)=6183.647kJ

W(2----->3)=P2(V3-V2)=(5.89*105)*(10.1-3.1)=4123kJ

Q2=10306.647kJ USING Q=U+W

FROM 3----->4;

U(3---->4)=nCv(T4-T3)=(410)*(12.47)*(856.25-1745.09)=-4544.372kJ(NEGATIVE SIGN MEANS HEAT GIVEN OUT)

W(3---->4)=0(SINCE ZERO CHANGE IN VOLUME)

Q3=-4544.372kJ USING Q=U+W

FROM 4----->1;

U(4---->1)=nCv(T1-T4)=(410)*(12.47)*(262.809-856.25)=-3034.086kJ

W(4----->1)=P1(V1-V4)=(2.89*105)*(3.1-10.1))=-2023kJ(NEGATIVE SIGN MEANS WORK DONE ON THE SYSTEM)

Q4=-5057.086kJ USING Q = U+ W;

NOW; HEAT TRANSFERRED TO THE SYSTEM = SUM OF POSITIVE Q's=Q1+Q2=1394.811+10306.647=11701.458kJ

HEAT TRANSFERRED OUT OF THE SYSTEM=SUM OF NEGATIVE Q's=Q3+Q4=5057.086+4544.372=9601.458kJ;

WORK DONE BY THE GAS = W(1----->2)+W(2----->3)+W(3----->4)+W(4----->1)=0+4123+0-2023=2100kJ

EFFICIENCY= OUTPUT/INPUT=WORK DONE BY GAS/HEAT INPUT TO THE GAS =2100kJ/11701.458kJ=0.17946=17.95%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.