A container contains 1 text{ mole} of a gas w | vedclass

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(B) The kinetic energy of the container is $K = \frac{1}{2} M v^2$ (since it contains $1 	ext{ mole}$,mass $m = M$).When the container stops,this kin

Vedclass · Accepted Answer

$\frac{Mv^2(\gamma - 1)}{2R}$

Vedclass · Answer

(B) The kinetic energy of the container is $K = \frac{1}{2} M v^2$ (since it contains $1 	ext{ mole}$,mass $m = M$).When the container stops,this kinetic energy is converted into the internal energy of the gas,$\Delta U = \mu C_V \Delta T$.Since $\mu = 1 	ext{ mole}$,we have $\Delta U = C_V \Delta T$.We know that $C_V = \frac{R}{\gamma - 1}$.Equating the kinetic energy to the change in internal energy:$\frac{1}{2} M v^2 = C_V \Delta T$$\frac{1}{2} M v^2 = \frac{R}{\gamma - 1} \Delta T$Solving for $\Delta T$:$\Delta T = \frac{M v^2 (\gamma - 1)}{2R}$.

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