Question

A piston contains 410 moles of an ideal monatomic gas that initally has a pressure of 2.89

Answer 1

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=nC_v*(CHANGE IN TEMPERATURE);W=P*(CHANGE IN VOLUME);

FROM 1----->2;

U(1---->2)=nC_v(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)=nC_v(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)=nC_v(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)=nC_v(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%