Q.

In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is :

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a

32

b

132

c

326

d

128

answer is D.

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Detailed Solution

If density of certain amount of gas becoming 32 times  ⇒V2=V1/32 and it is given that diatomic gas undergoes adiabatic processPVγ=constantPmργ=constantPργ=constant∵ρ=mV, m is constant P∝ργP2P1=ρ2ρ1γnP1P1=32ρ1ρ175n=3275=2575=27=128
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