#### Answer

$1-r^{1-\gamma}$

#### Work Step by Step

a) Simplifying the equation for work done, we find that the total work done is equal to:
$W = W_{12}+W_{34}$
We now use what we know about the work done to find the two equations for work we need:
$W_{12}=\frac{1}{1-\gamma}(\frac{RT_3}{r^{\gamma-1}}-RT_3)$
$W_{34}=-\frac{1}{1-\gamma}(\frac{RT_2}{r^{\gamma-1}}-RT_2)$
Using factoring, these two equations combine to:
$W_t=\frac{1}{1-\gamma}(\frac{R(T_3-T_2)}{r^{\gamma-1}}-R(T_3-T_2))$
We can simplify the above equation to be:
$\frac{R(T_3-T_2)}{1-\gamma}(\frac{1}{r^{(\gamma-1)}}-1)$
We know that the value of Q is the negative coefficient of this equation, so this can be simplified to:
$e=\frac{W}{Q}=(-\frac{1}{r^{(\gamma-1)}}+1)$
$e=1-r^{1-\gamma}$