Question

Two samples of gases 1 and 2 are initially kept in the same state. Sample 1 is expanded through an isothermal process whereas sample 2 through an adiabatic process up to the same final volume. Let $P_1$ and $P_2$ be the final pressure of the samples 1 and 2 respectively then

• A. $P_1<P_2$
• B. $P_1=P_2$
• C. $P_1>P_2$
• D. the relation between $P_1$ and $P_2$ cannot be deduced

Answer 1

The correct option is C $P_1>P_2$

For isothermal process temperature is constant,

$\therefore P_O{V}_O=P_1{V}_1$

$P_1=\frac{P_0{V}_0}{V_1}⟶\left(1\right)$

For adiabatic process,

$P_0{V}_0^4=P_2{V}_2^Y$

$P_2=P_0{\left[\frac{V_0}{V_2}\right]}^Y$

Given, $V_2=V_1$

$P_2=P_0{\left[\frac{V_0}{V_2}\right]}^Y⟶\left(2\right)$

dividing (1) and (2)

$\frac{P_1}{P_2}=\frac{\frac{P_0{V}_0}{V_1}}{P_0{\left[\frac{V_0}{V_1}\right]}^Y}$

$\frac{P_1}{P_2}={\left[\frac{V_0}{V_1}\right]}^{1-Y}$

As the gas in expanded, $V_0<V_1$

$\frac{V_0}{V_1}<1$

As, $Y>1$

$1-Y<0$

$\therefore {\left[\frac{V_0}{V_1}\right]}^{1-Y}>1$

$\frac{P_1}{P_2}>1$

$\therefore P_1>P_2\left(C\right)$

OPTION : C