Obtain an expression for work done by an ideal gas in an isothermal expansion.

Explanation

The process during which the temperature of system remains constant is known as isothermal process.
Equation of state for an ideal gas in isothermal process is PV = constant
`therefore P prop 1/V`
mean pressure of a given amount of gas is inversely proportional to its volume. This is Boyle.s law.
Suppose, an ideal gas goes isothermally (at constant temperature T) from its initial state `[P_1, V_1]` to the final state [`P_2, V_2`].
Suppose, at any intermediate stage with pressure P and volume change from V to V `+DeltaV`
`therefore` Work done in this process,
`DeltaW = PDeltaV`...(1)
Taking `(DeltaV to 0)` and summing the quantity `DeltaW` over the entire process,
`W=int DeltaW`
` =int_(V_1)^(V_2) pdV`..(2) `[lim_(DeltaV to 0)` taking `DeltaV=dV`]
From `PV=muRT, P=(muRT)/V`
`therefore W=muRT int_(V_1)^(V_2) (dV)/V =muRT ln [V]_(V_1)^(V_2)`
`therefore W=muRT ln V_2/V_1`...(3)
From first law of thermodyanmics,
`DeltaQ=DeltaU+DeltaW`
Here, internal energy depends only on temperature and temperature remains constant and hence internal energy remains constant.
So, `DeltaU=0`
`therefore DeltaQ=DeltaW`
Indicate that work is done on the gas by the environment and heat is released.
`therefore` Q=W
From equation (3) that for `V_2 gt V_1`, (when gas expands ) `W gt 0` and gas absorbs heat but if `V_2 lt V_1` (gas compressed ) `W lt 0` and gas released heat.
`therefore` For compression ,
`therefore W=-muRT(V_2/V_1)`
`therefore W=muRT(V_2/V_1)^(-1)`
`therefore W=muRT (V_1/V_2)`