A) \[\frac{M{{v}^{2}}(\gamma -1)}{2R(\gamma +1)}\]
B) \[\frac{M{{v}^{2}}(\gamma -1)}{2R}\]
C) \[\frac{M{{v}^{2}}}{2R(\gamma +1)}\]
D) \[\frac{M{{v}^{2}}}{2R(\gamma -1)}\]
Correct Answer: B
Solution :
[b] If m is the total mass of the gas, then its kinetic energy\[=1/2m{{v}^{2}}\]when the vessel is suddenly stopped, total kinetic energy will increase the temperature of the gas (because process will be adiabatic), i.e., \[\frac{1}{2}m{{v}^{2}}=\mu {{C}_{v}}\Delta T=\frac{m}{M}{{C}_{v}}\Delta T\] \[\Rightarrow \frac{m}{M}\frac{R}{\gamma -1}\Delta T=\frac{1}{2}m{{v}^{2}}\] \[\left( As\,{{C}_{V}}=\frac{R}{^{\gamma -1}} \right)\] \[\Rightarrow \Delta T=\frac{M{{v}^{2}}(\gamma -1)}{2R}\]
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