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:
$(γ = \frac{C_{P}}{C_{v}})$

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 $V^{q}$ , where $V$ is the volume of the gas. The value of $q$ is:

$(γ = \frac{C_{P}}{C_{v}})$

TestHub · Accepted Answer

For an adiabatic process T V γ − 1 = constant. We know that average time of collision between molecules τ = 1 n π 2 v rms d 2 where, n = number of molecules per unit volume V rms = rms velocity of molecules As n ∝ 1 V and v rms ∝ T τ ∝ V T Thus, we can write n = K 1 V - 1 and V rms = K 2 T 1 2 . where, K 1 and K 2 are constants. For adiabatic process, T V γ - 1 = constant. Thus, we can write τ ∝ V T - 1 2 ∝ V V 1 - γ - 1 2 or τ ∝ V γ + 1 2

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