Answer
a) $W_t=\frac{1}{1-\gamma}(\frac{R(T_3-T_2)}{5^{\gamma-1}}-R(T_3-T_2)$
b) $ \frac{1}{5^{(\gamma-1)}}T_{min}$
c) 0
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:
$W_{12}=\frac{1}{1-\gamma}(\frac{RT_3}{5^{\gamma-1}}-RT_3)$
$W_{34}=-\frac{1}{1-\gamma}(\frac{RT_2}{5^{\gamma-1}}-RT_2)$
Using factoring, these two equations combine to:
$W_t=\frac{1}{1-\gamma}(\frac{R(T_3-T_2)}{5^{\gamma-1}}-R(T_3-T_2))$
We can simplify the above equation to be:
$\frac{R(T_3-T_2)}{1-\gamma}(\frac{1}{5^{(\gamma-1)}}-1)$
We know that the value of Q is the negative coefficient of this equation, so this can be simplified to:
$e=\frac{W}{Q}=(-\frac{1}{5^{(\gamma-1)}}+1)$
b) We find:
$T_{max}=((-\frac{1}{5^{(\gamma-1)}}+1)(-1)+1)T_{min}=\frac{1}{5^{(\gamma-1)}}T_{min}$
c) The efficiency of a Carnot engine is determined by how many times greater the maximum temperature is than the minimum temperature. Thus, if both temperatures were the same, the efficiency would be 0.
