Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from τ1 to τ2. If (Cp/Cv)=γ for this gas then a good estimate for (τ2/τ1) is given by :

Question

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from τ1 to τ2. If (Cp/Cv)=&gamma; for this gas then a good estimate for (τ2/τ1) is given by : (A) ((1/2))(&gamma;+1/2) (B) (1/2) (C) ((1/2))&gamma; (D) 2

Tardigrade Admin · Accepted Answer

t propto (V/√T) ....(1) TV&gamma;-1 = constant ....(2) ∴ t propto V(&gamma;+1/2)

Q. Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $τ_{1}$ to $τ_{2} .$ If $\frac{C _{p}}{C _{v}} = γ$
for this gas then a good estimate for $\frac{τ _{2}}{τ _{1}}$ is given by :

$(\frac{1}{2})^{\frac{γ + 1}{2}}$

$\frac{1}{2}$

$(\frac{1}{2})^{γ}$

$2$

$t \propto \frac{V}{T} .... (1)$
$T V^{γ - 1} =$ constant $.... (2)$
$∴ t \propto V^{\frac{γ + 1}{2}}$

Q. Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from τ1​ to τ2​. If Cv​Cp​​=γ for this gas then a good estimate for τ1​τ2​​ is given by :

(21​)2γ+1​

21​

(21​)γ

2

t∝T​V​....(1) TVγ−1= constant ....(2) ∴t∝V2γ+1​

Q. Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $τ_{1}$ to $τ_{2} .$ If $\frac{C _{p}}{C _{v}} = γ$
for this gas then a good estimate for $\frac{τ _{2}}{τ _{1}}$ is given by :

$(\frac{1}{2})^{\frac{γ + 1}{2}}$

$\frac{1}{2}$

$(\frac{1}{2})^{γ}$

$2$

$t \propto \frac{V}{T} .... (1)$
$T V^{γ - 1} =$ constant $.... (2)$
$∴ t \propto V^{\frac{γ + 1}{2}}$