Answer
$ \epsilon = 13.1 \%$
Work Step by Step
To find the efficiency, we need to find the net work done and net input heat first
$W_{net} = nRT_H ln(\frac{V_{max}}{V_{min}}) + nR(T_L - T_H )$
$W_{net} = (3.4 mol)(8.31 J/mol.K) (500K) ln(\frac{5}{2}) + (3.4) (8.31 J/mol.K) (200K - 500K)$
$W_{net} = 12944 J+ (-8476 J)$
$W_{net} = 4468 J $
Now find the input heat, $Q_{in}$
$Q_{in} =nRT_H ln(\frac{V_{max}}{V_{min}}) + nC_V(T_H - T_L ) $
$Q_{in} = (3.4 mol)(8.31 J/mol.K ) (500K) ln(\frac{5}{2}) + (3.4 mol)(5/2 \times 8.31 J/mol.K) (500K - 200K)$
$Q_{in} = 12944 J + 21191 J $
$Q_{in} = 34135 J $
Now we can find the effieciency
$ \epsilon = \frac{W_{net}}{Q_{in}} \times 100 $
$ \epsilon = \frac{4468 J}{34135 J} \times 100 $
$ \epsilon = 13.1 \%$
