Question

A certain amount of a monoatomic ideal gas undergoes a process $\rho U^{\eta }=C$, where$\rho$ is the density of the gas and $U$ is the internal energy. It was found that the ratio $r=\frac{\Delta W}{\Delta Q}$ for the process was $r=2/3$. What is the value of $\eta$?

Answer 1

Given, $\rho U^{\eta }=constant$
Whenever the type of process is not specified we try to convert the information in the form of $P{V}^{\alpha }=constant$
Using ideal gas equation,
$PV=nRT$
$P=\frac{\rho RT}{M}\Rightarrow \rho =\frac{PM}{RT}$
The internal energy $U=n{C}_{v}T$
$\rho U^{\eta }=\frac{PM}{RT}(n{C}_v\Delta T{)}^{\eta }=$ constant
$P^1{T}^{\eta -1}\frac{M{C}_v^{\eta }{n}^{\eta }}{R}=constant$
$\Rightarrow P{T}^{\eta -1}=constant$
$\because$ all other values are constants
Substituting $T=\frac{PV}{nR}$ in the above equation,
$\Rightarrow P(\frac{PV}{nR}{)}^{\eta -1}=constant$
$\Rightarrow P^{\eta }{V}^{\eta -1}=constant$
Taking ${\eta }^{th}$ root on both sides,
$\Rightarrow P{V}^{\frac{\eta -1}{\eta }}=k=constant$
now, $\alpha =1-\frac{1}{\eta }$...(i)

Given ratio, $\frac{2}{3}=\frac{\Delta W}{\Delta Q}=\frac{\Delta W}{\Delta U+\Delta W}$
$2\Delta U=\Delta W$...(ii)

In general for any polytropic process, $\Delta W=\int PdV$
WKT, $P{V}^{\alpha }=k\Rightarrow P=\frac{k}{V^{\alpha }}$
$\Delta W=\int PdV={\int }_{V_i}^{V_f}\frac{k}{V^{\alpha }}$
$\Delta W={\left[\frac{V^{-\alpha +1}}{-\alpha +1}k\right]}_{V_i}^{V_f}$
here, $k=P{V}^{\alpha }$
$\Delta W={\left[\frac{PV}{1-\alpha }\right]}_i^f={\left[\frac{nRT}{1-\alpha }\right]}_{T_i}^{T_f}$
$=\frac{nR\Delta T}{1-\alpha }$...(iii)

Equation (ii) $\Rightarrow 2n{C}_v\Delta T=\frac{nR\Delta T}{1-\alpha }$
$\Rightarrow 2n\frac{3}{2}R\Delta T=\frac{nR\Delta T}{1-\alpha }$
$\alpha =\frac{2}{3}$
Equation (i) $\Rightarrow \alpha =1-\frac{1}{\eta }$
$\Rightarrow \eta =3$