Essential University Physics: Volume 1 (4th Edition)

a) We know that pressure times volume is constant in this case. Thus, it follows: $(60)(5) = P_f (1) \\ P_f=300 \ kPa$ b) We know that the work done from A to B is the same as the sum of the works to go from A to C and from C to B. Going from C to B involves no volume change, so no work is done. Thus, we find the work involved in going from A to C to obtain: $W_{AC}=(60)(5-1)=\fbox{240 J}$ *Note, we have to go against the arrow in the diagram, for the work when going from C to A is the negative work when going from A to B.