#### Question

Derive an expression for maximum work in isothermal reversible expansion of two moles of an ideal gas.

#### Solution

**a.** Consider ‘n’ moles of an ideal gas placed in a cylinder fitted with weightless, frictionless piston at a constant temperature ‘T’. If the gas expands isothermally and

reversibly from initial volume V_{1} to final volume V_{2} and if the expansion is by a small volume ‘dV’, then opposing pressure ‘P’ decreases by a small amount ‘dP’ producing small work done ‘dW’. Now, the pressure (pex) will be ‘P – dP’.**b.** This work done is given as:

Work done (dW) = pressure x change in volume

= -(P - dP) - dV

= -(PdV - dPdV)

both dP and dV are small, dP.dV is also very small and is neglected.

The above equation becomes dW =- PdV ….(1) **c.** When the expansion of a gas is carried out reversibly, there will be series of such ‘PdV’

terms. As the pressure decreases continuously, volume gradually increases from initial

volume ‘V_{1}’ to final volume ‘V_{2}’.

Thus, in order to obtain maximum work (W_{max)}, integrate equation (1) between the limits V_{1} to V_{2} as

`int_0^(wmax)dw=-int_(v_1)^(v_2)Pdv ..........(2)`

**d.** The general ideal gas equation for n mole is, PV = nRT

` or P=(nRT)/V`

Substituting (3) in equation (2),`

`int_0^(wmax)dw=-int_(v_1)^(v_2)(nRT)/Vdv`

Since the number of moles is constant, R is a gas constant and the process is isothermal,

therefore, T is constant. Hence, nRT = constant.

`int_0^(wmax)dw=-nRTint_(v_1)^(v_2)1/Vdv`

`[w]_0^(wmax)=-nRTlog[V]_(v_1)^(v_2)`

`[W_(max)-0]=-nRT[log(V_2)-log(V_1)]`

`W_(max)=-nRTlog(V_2/V_1)`

**e. **`W_(max)=-2.303nRTlog(V_2/V_1)`

**f**. Since, number of moles of ideal gas (n) = 2

`W_(max)=-4.606xxRT log_10(V_2/V_1)`