The pressure P 1 and density d 1 of a diatomic gas - - 7 5 change to P 2 and d 2 during an adiabatic operation- If d 2 d 1 - 32- then p 2 p 1 is -

The correct option is C 128Given, γ=75 d_2d_1=32 P_2P_1=? For adiabatic process, PV^γ=constant P(md)^γ=constant m is a constant te ...

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Standard XII

Chemistry

First Law of Thermodynamics

The pressure ...

Question

The pressure $P_{1}$ and density $d_{1}$ of a diatomic gas ( $γ = \frac{7}{5}$ ) change to $P_{2}$ and $d_{2}$ during an adiabatic operation. If $\frac{d_{2}}{d_{1}}$ = 32, then $\frac{p_{2}}{p_{1}}$ is :

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128

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168

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298

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Solution

The correct option is C 128
Given,
$γ = \frac{7}{5}$
$\frac{d_{2}}{d_{1}} = 32$
$\frac{P_{2}}{P_{1}} = ?$
For adiabatic process,
$P V^{γ} = c o n s t a n t$

$P {(\frac{m}{d})}^{γ} = c o n s t a n t$
$m$ is a constant term

$P d^{- γ} = c o n s t a n t$

$P_{1} d_{1}^{- γ} = P_{2} d_{2}^{- γ}$

$\frac{P_{2}}{P_{1}} = (\frac{d_{1}}{d_{2}})^{- γ}$

$\frac{P_{2}}{P_{1}} = (\frac{1}{32})^{- 7 / 5}$

$\frac{P_{2}}{P_{1}} = 128$
The correct option is B.

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The pressure P1 and density d1 of a diatomic gas (γ=75) change to P2 and d2 during an adiabatic operation. If d2d1 = 32, then p2p1 is :

The correct option is C 128Given,γ=75d2d1=32P2P1=?For adiabatic process,PVγ=constantP(md)γ=constantm is a constant termPd−γ=constantP1d−γ1=P2d−γ2P2P1=(d1d2)−γP2P1=(132)−7/5P2P1=128The correct option is B.

The pressure $P_{1}$ and density $d_{1}$ of a diatomic gas ( $γ = \frac{7}{5}$ ) change to $P_{2}$ and $d_{2}$ during an adiabatic operation. If $\frac{d_{2}}{d_{1}}$ = 32, then $\frac{p_{2}}{p_{1}}$ is :