#### Answer

The proofs are below.

#### Work Step by Step

a) In problem number 56, we found:
We know that the work is equal to the sum of all of the works done over each of the processes. Thus, we find:
1. The first step is adiabatic, so we find:
$P_1V_1^{\gamma}=P_2V_2^{\gamma}$
Using substitution for $\frac{V_1}{V_2}$, it follows:
$P_2=\frac{P_1}{r}$
This means:
$W=\frac{\frac{P_1}{r}V_2-P_1V_1}{\gamma-1}$
2. Pressure is constant, so this is isobaric. Thus, we find:
$W = -\frac{P_1}{r}(V_3-V_2)$
3. This step is adiabatic, so we find:
$P_1V_1^{\gamma}=P_2V_2^{\gamma}$
$P_f=\frac{P_1V_3}{V_1r}$
Thus, it follows:
$W=\frac{\frac{P_1V_3}{V_1r}V_1-V_3\frac{P_1}{r}}{\gamma-1}$
4. This step is constant volume, so no work is done. Thus, adding all of the individual works together and plugging in $\alpha=\frac{V_3}{V_2}$, we obtain:
$W = \frac{P_1V_1(r^{1-\gamma}(\alpha-1)\gamma-\alpha^{\gamma}+1)}{(\gamma-1)}$
Using this, we find that the net heat flow in is:
$Q_{in} = \frac{P_1V_1r^{1-\gamma}(\alpha-1)\gamma}{(\gamma-1)}$
b) An engine's efficiency is given by:
$e=\frac{W}{Q_h}$
Thus, using substitution, it follows:
$e_{diesel}=1-\frac{r^{1-\gamma}(\alpha^{\gamma}-1)}{(\alpha-1)\gamma}$