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Heat and Work

During an adi...

Question

During an adiabatic reversible change, the density becomes $\frac{1}{16}$ th of the initial value. Then find $\frac{P_{1}}{P_{2}}$ is : $(γ = 1.5)$

A

16

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B

4

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C

32

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D

64

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Solution

The correct option is D $64$
For ideal gas, $P M = d R T$
For adiabatic reversible process, $T^{γ} P^{1 - γ} = constant$
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, $d_{2} = \frac{d_{1}}{16}$
$\frac{P_{1}}{P_{2}} = 16 \frac{T_{1}}{T_{2}} . . . (i)$
$\frac{P_{2}}{P_{1}} = \frac{T_{2}^{3}}{T_{1}^{3}} . . . (i i)$
On multiplying (i) and (ii), we get
$1 = 16 \frac{T_{2}^{2}}{T_{1}^{2}}$
$\frac{T_{1}}{T_{2}} = \frac{1}{4}$
$\Rightarrow \frac{T_{1}}{T_{2}} = 4$
Substituting $\frac{T_{1}}{T_{2}}$ in equation (i)
From above equations.
$\frac{P_{1}}{P_{2}} = 16 \times 4 = 64$

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