Question

Obtain an expression for the workdone by a gas in an isothermal process.

Answer 1

Work done by a gas in an isothermal process:

Consider the moles of a gas contained within a cylinder with a moveable, light, and frictionless piston. Let P_i, V_i, and T represent the gas's initial pressure, volume, and absolute temperature, respectively.

Consider an isothermal expansion (or compression) of a gas, where P_f, v_f, and T are the final pressure, volume, and absolute temperature of the gas, respectively.

For an isothermal change,

P_iV_i = P_fV_f = constant

If the gas behaves like an ideal gas, its equation of state is

PV = nRT = constant     ...(i)   ....(as T = constant, R is universal gas constant)

The work done in a minuscule isothermal expansion is given by

dW = PdV     ...(ii)

The total work done in completing the expansion from initial volume v_i to final volume V_f is denoted by

`W = int_(v_i)^(v_f) PdV`

∴ `W = nRT int_(v_i)^(v_f) (dV)/V`        ...[from(i)]

∴ W = nRT [In V_f - InV_i]

∴ W = nRT In `V_f/V_i`

∴ W = 2.303 nRT `log_10  V_f/V_i`

Answer 2

Consider the isothermal expansion of an ideal gas. During this process, small work is done, and it is given by

dW = PdV

We get the total work done by integrating the above equation with the limit V_i to V_f.

`W = int_(V_i)^(V_f)Pdv`    ...(i)

But we know that for an ideal gas, PV = nRT.

∴ Equation (i) becomes,

`W = int_(V_i)^(V_f)(nRT)/V dV`

`W = nRT int_(V_i)^(V_f)(dV)/V`

`W = nRT [log V]_(V_i)^(V_f)`

`W = nRT log_e (V_f/V_i)`

or `W = 2.303 nRT log_10 (V_f/V_i)`