Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is (γ =(Cp/Cv))

Question

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is (&gamma; =(Cp/Cv)) (A) (3&gamma; +5 /6) (B) (&gamma; +1/2) (C) (3 &gamma;-5/6) (D) (&gamma;-1/2)

Tardigrade Admin · Accepted Answer

mean free path λ=(1/√2 π d 2 n) n =( text no. of molecules / text volume ) v text avg. propto √ T T .V &gamma;-1= C t =(λ/ V text avg. ) propto ( V /√ T ) v arrow is volume ( V /√( C ) v r -1) propto V (&gamma;+1/2) vq propto v(&gamma;+1/2) q =(&gamma;+1/2)

Q. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $V^{q}$ , where $V$ is the volume of the gas. The value of $q$ is $(γ = \frac{C _{p}}{C _{v}})$

$\frac{3 γ + 5}{6}$

$\frac{γ + 1}{2}$

$\frac{3 γ - 5}{6}$

$\frac{γ - 1}{2}$

Q. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is (γ=Cv​Cp​​)

63γ+5​

2γ+1​

63γ−5​

2γ−1​

mean free path λ=2​πd2n1​ n= volume no. of molecules ​ vavg. ​∝T​T.Vγ−1=C t=Vavg. ​λ​∝T​V​v→ is volume vr−1C​​V​∝V2γ+1​ vq∝v2γ+1​ q=2γ+1​

Q. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $V^{q}$ , where $V$ is the volume of the gas. The value of $q$ is $(γ = \frac{C _{p}}{C _{v}})$

$\frac{3 γ + 5}{6}$

$\frac{γ + 1}{2}$

$\frac{3 γ - 5}{6}$

$\frac{γ - 1}{2}$