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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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A closed vessel moving with a velocity $v$ (with zero acceleration) is stopped suddenly. If the vessel contains a gas of molecular mass $M$, what will the rise in temperature of the gas be?

$\begin{array}{1 1} \large\frac{Mv^2(\gamma-1)}{2R} \\ \large\frac{Mv^2(\gamma+1)}{2R} \\ \large\frac{2R}{Mv^2(\gamma-1)} \\ \large\frac{2R}{Mv^2(\gamma+1)}\end{array}$

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  • Since internal energy depends only on the temperature, as the temperature changes, the change in internal energy is $\Delta U$= $\large\frac{n\;f\;R\;\Delta T}{2}$
Since internal energy depends only on the temperature, as the temperature changes, the change in internal energy is = $\large\frac{n\;f\;R\;\Delta T}{2}$
Vessel contains a gas of Mass M, change in kinetic energy = $\large\frac{nMv^2}{2}$
$\large\frac{n\;f\;R\;\Delta T}{2} =$ $\large\frac{nMv^2}{2}$
Since, $f = \large\frac{2}{\gamma-1}$ where $\gamma = C_p/C_v$, substituting,we get, $\Delta T = \large\frac{Mv^2(\gamma-1)}{2R}$
answered Jan 22, 2014 by balaji.thirumalai
 

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