A van der Waal's gas obeys the equation of state ( P + n ² a V ² )( V - nb )= nRT . Its internal energy is given by U = CT - n ² a V . The equation of a quasistatic ad iabat for this gas is given by-

Step-by-step solution

For adiabatic process dQ =0 and - dU = dW - nC _ V T = P V or - nC _ V dT = PdV when change is very small now given U = CT - n ² a V d U = CdT + n ² a V ² ~d ~V put this value of dU in - dU = d W - ( CdT + n ² a V ² ~d ~V )= PdV also P = ( nRT V - nb )- n ² a V ² replace it in (1) - (C d T+ n² a V² d V )= ( ( n R T V-n b )- n² a V² ) d V array l - CdT = ( nRT V - nb ) dV - C nR dT T = dV V - nb array Integrating we get - n T ^ C n R = n ( V - nb )+ k ( k constant of integration) ( T ^ C / nR )( V - nb )=- k or ( T