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

Question

The pressure and density of a diatomic gasleft - gamma -dfrac-7-5- right - change adiabatically from -P-1-rho-1- to -P-2-rho-2- If dfrac-rho -2-rho -1-32 - then dfrac-P -2-P -1- should be126128124146

Toppr Admin · Accepted Answer

For an adiabatic process-PV-x3B3-constantP-x3C1-x2212-x3B3-constantP2P1-x3C1-1-x3C1-2-x2212-x3B3-132-x2212-7-5-32-7-5-128Option 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
126
128
124
146

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.

The pressure and density of a diatomic gas(γ=75) change adiabatically from (P1,ρ1) to (P2,ρ2). If ρ2ρ1=32, then P2P1 should be126128124146

For 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
126
128
124
146

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.