A gas expands with temperature according to the relation V = KT^(2//3) where K is a constant. The work done by the gas when the temperature changes by 60 K is

A gas expands with temperature according to the relation V = KT^(2//3)

1.	A gas expands with temperature according to the relation V = KT^(2//3) where K is a constant. The work done by the gas when the temperature changes by 60 K is
Answer» 10 R 30 R 40 R 20 R Solution :`dW = PdV = (RT)/(V)DV` …(i) As `V = KT^(2//3) :. dV = K(2)/(3)T^(-1//3) dT` `:. (dV)/(V) = (K(2)/(3)T^(-1//3)dT)/(KT^(2//3)) = (2)/(3) (dT)/(T)` ….(II) From (i), `W = underset(T_(1))overset(T_(2))int RT(dV)/(V) = underset(T_(1))overset(T_(2))int RT (2)/(3)(dT)/(T) = (2)/(3)R[T]_(T_(1))^(T_(2))` (Using (ii)) `= (2)/(3)R(T_(2) - T_(1)) = (2)/(3) R xx 60 = 40R`