Calculate w,q and Delta Uwhen 0.75 mol of an ideal gasexapnds isothermally and reversibly at 27^(@) froma volume of 15 L to 25 L

Calculate w,q and Delta Uwhen 0.75 mol of an ideal gasexapnds isotherm

1.	Calculate w,q and Delta Uwhen 0.75 mol of an ideal gasexapnds isothermally and reversibly at 27^(@) froma volume of 15 L to 25 L
Answer» Solution :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`