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A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats \gamma . It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by

Option: 1

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


Option: 2

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


Option: 3

\frac{\gamma Mv^{2} }{2R}K


Option: 4

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


Answers (1)

best_answer

Basically here kinetic energy tends to zero.

if heat exchange is zero so, only internal energy will increase means Temperature will increase.

so, internal energy = kinetic energy.

\textup{internal energy}=\frac{1}{2}mv^{2}  

This will convert into a change in internal energy.

\Delta U=nC_{v}\Delta T=\frac{nR}{\gamma-1}\Delta T

\therefore\ \; \frac{1}{2}mv^{2}=\frac{nR}{\gamma-1}\Delta T

 

Posted by

Pankaj Sanodiya

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