Calculate the work done in isothermal reversible change.

Explanation

We know that `dw = pdV`
Also we know that `pV = nRT`
So `p = (n)/(V)RT`
Substitutinh in eq. (i) we get `dw = (nRT)/(V)dV`
Since work is of expansion,
`therefore` it should be with negative sign. `dw = (nRT)/(V)dV`
Integrating both sides we get
`int dw = - int (nRT)/(V)dV`
`or W = -nRT int_(v_(l))^(v_(f)) (dV)/(V)`
`or W=-nRT[lnV]_(v_(l))^(v_(f))`
`or W =- nRT(lnV_(f)-lnV_(l)]`
or `W = - nRT ln ""(V_(f))/(V_(l))`
It is isothermal reversible change.