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:

Question

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: (A) 326 (B) (1/32) (C) 32 (D) 128

Tardigrade Admin · Accepted Answer

In adiabatic proccss PV &gamma;= constant P((m/ρ))&gamma;= constant as mass is constant P propto ρ&gamma; (Pf/Pi)=((ρf/ρi))&gamma; =(32)7 / 5=27=128

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:

326

$\frac{1}{32}$

32

128

In adiabatic proccss
$P Vγ =$ constant
$P (\frac{m}{ρ})^{γ} =$ constant
as mass is constant
$P \propto ρ^{γ}$
$\frac{P _{f}}{P _{i}} = (\frac{ρ _{f}}{ρ _{i}})^{γ}$
$= (32)^{7/5} = 2^{7} = 128$

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:

326

321​

32

128

In adiabatic proccss PVγ= constant P(ρm​)γ= constant as mass is constant P∝ργ Pi​Pf​​=(ρi​ρf​​)γ =(32)7/5=27=128

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:

$\frac{1}{32}$

In adiabatic proccss
$P Vγ =$ constant
$P (\frac{m}{ρ})^{γ} =$ constant
as mass is constant
$P \propto ρ^{γ}$
$\frac{P _{f}}{P _{i}} = (\frac{ρ _{f}}{ρ _{i}})^{γ}$
$= (32)^{7/5} = 2^{7} = 128$