Question

Calculate w,q and `Delta U ` when 0.75 mol of an ideal gas exapnds isothermally and reversibly at `27^(@)` from a volume of 15 L to 25 L

Answer 1

For isothermal reversible expansion of an ideal gas,
`w= -nRT ln (V_(2))/(V_(1))= -2.303 nRT log. (V_(2))/(V_(1))`
Putting `n= 0.75` mol, `V_(1) =15 L, V_(2) = 25 L, T= 27 + 273 = 300 K` and `R = 8.314 J K^(-1) mol^(-1)` , we get `w = -2.303 xx 0.75 xx 8.314 xx 300 log. (25)/(15) = -955.5 J` `( -` ve sign represents work of expansion)
For isothermal expansion of an ideal gas, `Delta U = 0`
`:. Delta U = q+w` gives ` q= -w =+ 955.5 J`