1. For a van der Waals gas πT=a/Vm2. Calculate ΔUm for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm3 to 24.8 dm3at 298K. What are values for q and w? 2. Relevant equations I looked up a for N2 gas to be 1.352 atm*dm6/mol2 dU(V,T)=(∂U/∂V)TdV+(∂U/∂T)VdT (∂U/∂V)T=πT (∂U/∂T)V=Cv 3. The attempt at a solution I start with (∂U/∂V)T=πT and substitute in πT and Cv which gives dU(V,T)=πTdV+CvdT Since this is isothermal dt=0 so dU(V,T)=πTdV I integrate both sides ∫dU=∫πTdV. πT= a/Vm2. a is a constant so it comes out of the integral leaving ΔU=a∫dV/Vm2.= a*n2 ∫ dV/V2 Evaluating the integral ΔU= -((a*n2)/2)*V-3. After plugging in values for a, Vi, and Vf I get: ΔU=-((1.352 *n2)/2)*[(1/24.83)-(1/13)] I get 0.676 *n 2 J/mol2. I just don't know what to do with the moles. I know how to get work which is just w=-nrt∫dV/V and the solution is -7.95 kJ of work but without ΔU I cannot calculate q.