The pressure and density of a diatomic gas - - 7-5 change adiabatically from P-d to P-d- if d-d - 32 then P-P should be

Volume of the gas V = md where , m is the mass and d is the density of the gas. nbsp; nbsp;and using PV^γ = constant nbsp; we get P'P = (VV' )^γ = (d'd )^ ...

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Standard XII

Chemistry

Relation between P, V, T, Gamma in Adiabatic Proceses

The pressure ...

Question

The pressure and density of a diatomic gas $(γ = 7 / 5)$ change adiabatically from (P,d) to (P',d') if $\frac{d^{'}}{d} = 32$ then $\frac{P^{'}}{P}$ should be

None of the above

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128

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\frac{1}{128}

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Solution

The correct option is B 128
Volume of the gas $V = \frac{m}{d}$ where , m is the mass and d is the density of the gas. and using $P V^{γ}$ = constant we get
$\frac{P^{'}}{P} = {(\frac{V}{V^{'}})}^{γ} = {(\frac{d^{'}}{d})}^{γ} = (32)^{\frac{7}{5}} = 128$

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Similar questions

Q. The pressure and density of a diatomic gas

(γ = 7 / 5)

change adiabatically from (P,d) to (P',d') if

\frac{d^{'}}{d} = 32

then

\frac{P^{'}}{P}

should be

Q. The pressure

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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 :

Q. If the pressure on density gas

(γ = \frac{7}{5})

in adiabatic process changes from (P, e) to (P', e'), where

P = 32 P^{'}

, then

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is equal to:

Q. A diatomic gas

(γ = \frac{7}{5})

is compressed to

\frac{1}{32}

times of its initial volume adiabatically and reversibly. If its initial pressure is

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, its new pressure will be :

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The pressure and density of a diatomic gas (γ=7/5) change adiabatically from (P,d) to (P',d') if d′d=32 then P′P should be

The correct option is B 128Volume of the gas V=md where , m is the mass and d is the density of the gas. and using PVγ = constant we get P′P=(VV′)γ=(d′d)γ=(32)75=128

The pressure and density of a diatomic gas $(γ = 7 / 5)$ change adiabatically from (P,d) to (P',d') if $\frac{d^{'}}{d} = 32$ then $\frac{P^{'}}{P}$ should be

The correct option is B 128
Volume of the gas $V = \frac{m}{d}$ where , m is the mass and d is the density of the gas. and using $P V^{γ}$ = constant we get
$\frac{P^{'}}{P} = {(\frac{V}{V^{'}})}^{γ} = {(\frac{d^{'}}{d})}^{γ} = (32)^{\frac{7}{5}} = 128$