Chemistry - Thermodynamics - I Question with Solution | TestHub

For a reversible adiabatic process:

$T{V}^{\gamma -1}=constant$ and $P{V}^{\gamma }=constant$

Given $\gamma =4/3⟹\gamma -1=1/3$

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

$800\cdot V_1^{1/3}=200\cdot V_2^{1/3}$

$\frac{V_2^{1/3}}{V_1^{1/3}}=\frac{800}{200}=4$

$\frac{V_2}{V_1}=4^3=64$

Volume ratio:-64 → (A) correct

pressure ratio:

$P{V}^{\gamma }=constant⟹P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

$\frac{P_2}{P_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }={\left(\frac{1}{64}\right)}^{4/3}=(64{)}^{-4/3}$

$64^{4/3}=(4^3{)}^{4/3}=4^4=256$

$\frac{P_2}{P_1}=\frac{1}{256}$

Pressure ratio:- 1/256 → (B) correct