The volume of a diatomic gas $(γ = 7 / 5)$ is increased two times in a polytropic process with molar heat capacity C = R. How many times will the rate of collision of molecules against the wall of the vessel be reduced as a result of this process ? The answer is $(x)^{4 / 3} t i m e s$ then x is

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Solution

Since $P V = n R T$ ,

$P d V + V d P = n R d T$

Rate of collision is N

$\frac{N Δ p}{A Δ t} = P$

$\frac{N 2 m v_{r m s}}{A Δ t} = P$

$N^{'} = \frac{N}{A Δ t} α \frac{P}{v_{r m s}} α \frac{P}{\sqrt{T}}$

As per given information and first law of thermodynamics,

$n R d T = 2.5 n R d T + P d V$

Hence,

$1.5 (P d V + V d P) = - P d V$

$5 P d V = - 3 V d P$

$P V^{5 / 3} = C$

Since $P V = n R T$

$T V^{2 / 3} = K$

On doubling volume,

$P^{'} = P 2^{- 5 / 3}$

$T^{'} = T 2^{- 2 / 3}$

Thus, ${N^{'}}_{n e w} = \frac{2^{- 5 / 3}}{2^{- 1 / 3}} N = 2^{- 4 / 3} N$

Thus, number of collisions per unit time per unit are are reduced by $2^{4 / 3}$ times.

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The volume of a diatomic gas (γ=7/5) is increased two times in a polytropic process with molar heat capacity C = R. How many times will the rate of collision of molecules against the wall of the vessel be reduced as a result of this process ? The answer is (x)4/3times then x is

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