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 528 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 528 K, respectively. Find the total heat (including the algebraic sign) for this three-step process.

Thanks in advance.

Explanation / Answer

Ok, for an ideal gas, let's remember both things :

Po*Vo = n*R*T

n = 1 >>> Po*Vo = R*528

R = 0.082

Po*Vo = 43.296

Heat = Work + internal energy

For an ideal gas : internal energy = n*cv*dT

For the isothermal expansion : Vo to 3Vo

Work = Po*(3Vo - Vo) = 2*Vo*Po

Isothermal process, then : Internal energy = 0

Heat = 2*Po*Vo

For the isobaric process, the pressure is the same, then :

Work = Po*(3Vo -Vo) = -2*Vo*Po

In the isochoric process, the work = 0 Joules, because there is no change of volume.

Total heat = 2*Po*Vo - 2*Po*Vo + 0 = 0 Joules

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