Question

A sample of diatomic gas at pressure p, volume V and temperature T is compressed to V/2 isothermally. Its pressure is found to be $p_i$. If it is compressed to V/2 adiabaticaly, its pressure comes out to be $p_a$. What is the relation between $p_i$ and $p_a$?

• A. $p_i$ = $p_a$
• B. $2p_i=$ $p_a$
• C. $p_i$ = $2\cdot 6p_a$
• D. $2\cdot 6p_i$ = $2p_a$

Answer 1

The correct option is D $2\cdot 6p_i$ = $2p_a$

Given:

a diatomic gas,

Pressure = p; Volume =V; Temperature =T

Compressed isothermally

$v_i=\frac{V}{2}$ and pressure = $p_i$

Compressed adiabatically,

$v_a=\frac{V}{2}$ and pressure = $p_a$

To find:

Relation between $p_i$ and $p_a$

We know,

For isothermal,

$\frac{P_1}{P_2}=\frac{V_2}{V_1}⟹\frac{p_i}{p}=2⟹p=\frac{p_i}{2}......\left(i\right)$

For adiabatic,

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }⟹p_a{\left(\frac{V}{2}\right)}^{\gamma }=p(V{)}^{\gamma }⟹p_a=p{(V\times \frac{2}{V})}^{1.4}⟹p_a=p(2{)}^{1.4}⟹p_a=p\left(2.6\right)⟹p=\frac{p_a}{2.6}.....\left(ii\right)$

Equating equation (i) and (ii), we get,

$\frac{p_i}{2}=\frac{p_a}{2.6}⟹2.6p_i=2p_a$

is the correct answer.