Question

Derive the expression for maximum work.

Answer 1

Here work done is calculated with the help of  

Work = Pressure * change in volume

Work = -(-P – dP) – dV

Work = -(-PdV – dPdV)

For the expansion of the gas, the expansion will be taken into consideration, so integrate the over equation,  

Integrate to o to wmax dw = integrate to v1 to v2 PdV

Ideal gas, P = nRT/V

Integrate o to wmax dw = integrate v1 to v2 nRT/V dt

On calculation, Wmax = -4.606 RT Log10 (V2/V1)

Answer 2

an expresion for maximum work in isothermal reversible expansion of two mole of ideal gas : We can also relate work to internal pressure of the system under reversible condition if volume change from V1 to V2

Wrev=−∫V2V1Pindv

we know,

Wrev=−∫V2V1Pexdv=−∫V2V1(Pin±dp)dV

=−∫V2V1(PindV±dp.dV)

dp and dV are infinite small so that dp.dV is neglected

Wrev=−∫V2V1PindV

Now pressure of gas

Pin=P

can be expressed in terms of its volume through gas equation

PV=nRT

P=nVRT

Therefore, at constant temperature

Wrev=−∫V2V1nRT dV÷V