Question

m grams of a gas of molecular weight M is flowing in an isolated tube velocity v. If the gas flow is suddenly stopped the rise in the temperature is: (y = ratio of specific heats, R = universal gas constant, J = mechanical equivalent of heat).

• A. $ \frac{M{{v}^{2}}(\gamma -1)}{2RJ} $
• B. $ \frac{m}{M}\frac{{{v}^{2}}(\gamma -1)}{2RJ} $
• C. $ \frac{m{{v}^{2}}\gamma }{2RJ} $
• D. $ \frac{M{{v}^{2}}\gamma }{2RJ} $

Answer 1

Answer: (A)

$ \Delta U=\frac{\mu RJ\Delta \Tau }{(\gamma -1)} $ $ \Rightarrow $ $ \Delta T=\frac{(\gamma -1)\Delta U}{\mu RJ} $ $ \Rightarrow $ $ \Delta T=\frac{(\gamma -1)\times \frac{1}{2}m{{v}^{2}}}{(m/M)RJ} $ $ =\frac{M(\gamma -1){{v}^{2}}}{2RJ}=\frac{M{{v}^{2}}(\gamma -1)}{2RJ} $