A) \[\frac{M{{v}^{2}}}{2R(\gamma +1)}\]
B) \[\frac{M{{v}^{2}}}{2R(\gamma -1)}\]
C) \[\frac{M{{v}^{2}}(\gamma -1)}{2R}\]
D) \[\frac{M{{v}^{2}}(\gamma +1)}{2R}\]
Correct Answer: C
Solution :
If m is the total mass of the gas then its kinetic energy\[=\frac{1}{2}m{{v}^{2}}\] when the vessel is suddenly stopped, then the total kinetic energy will increase the temperature of the gas. Hence \[\frac{1}{2}m{{v}^{2}}=\mu {{C}_{V}}\Delta \Tau \] or \[\frac{1}{2}m{{v}^{2}}=\frac{m}{M}{{C}_{V}}\Delta \Tau \] But \[{{C}_{v}}=\frac{R}{\gamma -1}\] \[\therefore \] \[\frac{1}{2}m{{v}^{2}}=\frac{m}{M}\times \frac{R}{\gamma -1}\times \Delta T\] \[\Delta T=\frac{M{{v}^{2}}}{2R}(\gamma -1)\]
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