One mole of a monatomic ideal gas has an initial pressure, volume, and temperatu

Question

One mole of a monatomic ideal gas has an initial pressure, volume, and temperature of Po, Vo, and 335 K, respectively. It undergoes an isothermal expansion that triples the volume of the gas. Then, the gas undergoes an isobaric compression back to its original volume. Finally, the gas undergoes an isochoric increase in pressure, so that the final pressure, volume, and temperature are Po, Vo, and 335 K, respectively. Find the total heat (including the algebraic sign) for this three-step process.

Explanation / Answer

1 mole of a mono atomic gas is expanded isothermally to triple times the volume Therefore V1= 3V0 implies P1=P0/3(since isothermal). change in internal energy in an isothermal process is 0 and work done in isothermal process = n R T ln(V1/V0) n=1(since 1 mole) Therefore heat change = R T ln(3)= RT0ln(3) Next isobaric process from P1,V1,T1 to P2,V2,T2 where P1=P2=P0/3 V2= V0 =>T2=3T1=3T0 Heat supplied = Cp(T2-T1) Therefore heat supplied= internal energy + work done or = Cp(3T0-T0) =5RT0 Isochoric process from P2,V2,T2 to P3,V3,T3 where P3= P0 and T3=T0 Next isochoric process heat supplied = Cv(T3-T2)= 1.5 R(T0-3T0) =-3RT0 Therefore net heat supplied = RT0ln(3)+ 5RT0-3RT0 =3.098RT0= 3.098* 0.287*335 =297.916 KJ

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