#### Answer

a) 202 J
b) 500 J transferred out of gas

#### Work Step by Step

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.