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#Mechanics

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Comments

The average time between two collisions is given by , $\tau=\frac{1}{\sqrt{2} \pi n v_{rms}d^{2}}$

$n=\frac{N}{V}$ So, from this $\tau \propto \frac{V}{\sqrt{T}}$ For an adiabatic process, $TV^{\gamma-1}=k$ By substituting, $\tau \propto \frac{V}{\sqrt{\frac{1}{V^{\gamma-1}}}}$ Hence, $\tau \propto V^{\frac{\gamma+1}{2}}$

Also, try my problem on profile.

Sahil Silare - 3 years, 2 months ago

Thank you very much for your response! Btw, I'm not using my facebook lately so you can use this discussion page to tell me anything important.

Tapas Mazumdar - 3 years, 2 months ago

Uh, check out my problem boi, Thermodynamics.

Sahil Silare - 3 years, 2 months ago

I think the answer should be $\frac{\gamma-1}{2}$ . We can use the expression for collision frequency. The average collision time can be said to be the reciprocal of the same. We find that

$t_{av} \alpha v$

where v is the average speed of the gas molecules. This means that

$t_{av} \alpha \sqrt{t}$

Now we can use the condition for an adiabatic process, namely

$TV^{\gamma-1}=const$

This can give you the result. Hope it helps!

Kanad Pardeshi - 3 years, 2 months ago

Thank you very much for your response. However the answer $\dfrac{\gamma + 1}{2}$ as explained above seems correct to me. I don't have the answer key unfortunately.

Tapas Mazumdar - 3 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$