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

We know the following equation: $P_B=P_A(\frac{V_A}{V_B})^{\gamma}$ Plugging in the known values, we obtain: $P_B=696 \ kPa$ Now that we know this, we can break the work up into two paths. The work done overall is given by: $W=\frac{P_BV_B-P_AV_A}{\gamma-1}-nrTln(\frac{V_A}{V_C})$ Plugging in the known values gives: $\fbox{W=186 Joules}$