Question

During an adiabatic change the density becomes $\frac{1}{16}th$ of the initial value, then $\frac{P_1}{P_2}$ is : $(\gamma =1.5)$

• A. 16
• B. 4
• C. 32
• D. 64

Answer 1

The correct option is D 64

For ideal gas, $PM=dRT$

For adiabatic process, $T^{\gamma }{P}^{1-\gamma }=constant$

Thus, $T_1^{1.5}{P}_1^{-.5}=T_2^{1.5}{P}_2^{-.5}$

Also $\frac{P_1}{d_1{T}_1}=\frac{P_2}{d_2{T}_2}$

Given, $d_2=\frac{d_1}{16}$

$\frac{P_1}{P_2}=16\frac{T_1}{T_2}$....(i)

$\frac{P_2}{P_1}=\frac{T_2^3}{T_1^3}$...(ii)

On multiplying (i) and (ii), we get:

$1=16\frac{T_2^2}{T_1^2}$

$\frac{T_2}{T_1}=\frac{1}{4}$

$\Rightarrow \frac{T_1}{T_2}=4$

Substituting $\frac{T_1}{T_2}$ in equation (i)
From above equations,
$\frac{P_1}{P_2}=16\times 4=64$