The pressure and density of a diatomic gas - -75 change adiabatically from P1-1 to P2-2- If - 2- 1-32 - then P 2P 1 should be

The correct option is B 128For an adiabatic process, PV^γ= constant Pρ^ γ= constant P_2P_1=(ρ _1ρ _2)^ γ=(132)^ 7/5=(32)^7/5=128 Option B. ...

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Standard XII

Chemistry

System, Surroundings, Intensive, Extensive Properties

The pressure ...

Question

The pressure and density of a diatomic gas $(γ = \frac{7}{5})$ change adiabatically from $(P_{1}, ρ_{1})$ to $(P_{2}, ρ_{2})$ . If $\frac{ρ_{2}}{ρ_{1}} = 32,$ then $\frac{P_{2}}{P_{1}}$ should be

126

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128

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146

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124

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Solution

The correct option is B 128
For an adiabatic process,
$P V^{γ} =$ constant
$P ρ^{- γ} =$ constant
$\frac{P_{2}}{P_{1}} = {(\frac{ρ_{1}}{ρ_{2}})}^{- γ} = {(\frac{1}{32})}^{- 7 / 5} = (32)^{7 / 5} = 128$
Option B.

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The pressure and density of a diatomic gas $(γ = \frac{7}{5})$ change adiabatically form $(p_{1}, d_{1})$ to $(p_{2}, d_{2})$ . If $\frac{d_{2}}{d_{1}} = 32, t h e n \frac{p_{2}}{p_{1}}$ is

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The pressure and density of a diatomic gas(γ=75) change adiabatically from (P1,ρ1) to (P2,ρ2). If ρ2ρ1=32, then P2P1 should be

The correct option is B 128For an adiabatic process,PVγ=constantPρ−γ=constantP2P1=(ρ1ρ2)−γ=(132)−7/5=(32)7/5=128Option B.

The pressure and density of a diatomic gas $(γ = \frac{7}{5})$ change adiabatically from $(P_{1}, ρ_{1})$ to $(P_{2}, ρ_{2})$ . If $\frac{ρ_{2}}{ρ_{1}} = 32,$ then $\frac{P_{2}}{P_{1}}$ should be

The correct option is B 128
For an adiabatic process,
$P V^{γ} =$ constant
$P ρ^{- γ} =$ constant
$\frac{P_{2}}{P_{1}} = {(\frac{ρ_{1}}{ρ_{2}})}^{- γ} = {(\frac{1}{32})}^{- 7 / 5} = (32)^{7 / 5} = 128$
Option B.