Question

An ideal diatomic gas undergoes a process AB, for which PV indicator diagram is shown. Temperature at state A is $T_0$. When process changes from endothermic to exothermic,
temperature of the gas is $\frac{175\alpha }{864}{T}_0$. Find the value of $\alpha$.

Answer 1

Equation of straight line, AB P = $-\frac{2P_0}{V_0}V+5P_0$
Slope of the straight line, AB = $-\frac{2P_0}{V_0}$
When process changes from endothermic to exothermic, slope should be
equal to adiabatic P-V curve slope =$\frac{\gamma P}{V}=\frac{7}{5}\frac{P}{V}$
$\frac{2P_0}{V_0}=\frac{7P}{5V},P=\frac{10}{7}\frac{P_0}{V_0}V$
$P=\frac{10}{7}\frac{P_0}{V_0}V$
$=-\frac{2P_{0}V}{V_0}+5P_0$
$P=\frac{10}{7}\frac{P_0}{V_0}\frac{35V_0}{24}$
$3P_0{V}_0=nR{T}_0$,
Also,
$\left(\frac{25}{12}{P}_0\right)\left(\frac{35}{24}{V}_0\right)$
$=\frac{875}{288}{P}_0{V}_0=nRT$
$\frac{T}{T_0}=\frac{875}{288\left(3\right)}=\frac{875}{864}$
$T=\frac{875}{864}{T}_0$