## Essential University Physics: Volume 1 (4th Edition)

a) Going from A to C is an isothermal process, so we find: $W = (250)(1)ln(\frac{5}{1})=402 \ J$ Going from C to D involves no work, for there is no change in volume. Going from D to A is isobaric, so it follows: $W = -P(V_f-V_0)=-(50)(5-1)=-200 \ J$ Thus, the net work is $+202J$. b) This process is a constant-volume process, so we know: $Q=nC_v\Delta T$ There are 5 degrees of freedom, so this becomes: $Q = \frac{5}{2}nRT$ Substituting in the ideal gas law, it follows: $Q = 2.5\Delta PV = 2.5(-200)(1)=-500 \ J$ Since this is negative, the energy is transferred out of the gas.