Question

A thermally insulated vessel contains an ideal gas of molecular weight M and specific heat ratio $\gamma$. If it is moving with speed V and suddenly brought to rest, the temperature increase is:

• A. $\left[\frac{(\gamma+1)MV^2}{2(\gamma+2)R}\right] K$
• B. $\left[\frac{(\gamma-1)MV^2}{2\gamma R}\right] K$
• C. $\left[\frac{\gamma MV^2}{2R}\right] K$
• D. $\left[\frac{(\gamma-1)MV^2}{2R}\right] K$

Hint:

Conversion of bulk kinetic energy to random molecular motion increases internal energy.

Answer 1

The Correct Option is D

Step 1: Picture the experiment.
An insulated vessel holds one mole of an ideal gas with molar mass $M$ and specific heat ratio $\gamma$. The whole vessel moves at speed $V$ and is suddenly stopped. We find the rise in temperature.

Step 2: Use energy conservation.
Since the vessel is insulated, no heat escapes. The ordered kinetic energy of the moving gas turns into extra internal (thermal) energy, which raises the temperature.

Step 3: Write the kinetic energy lost.
The bulk kinetic energy of one mole of mass $M$ moving at speed $V$ is $\tfrac{1}{2} M V^{2}$.

Step 4: Write the internal energy gained.
The internal energy rise for one mole is $C_v \Delta T$, where $C_v$ is the molar heat capacity at constant volume.

Step 5: Bring in the value of Cv.
For an ideal gas, $C_v = \dfrac{R}{\gamma - 1}$. Setting energy lost equal to energy gained: $\tfrac{1}{2} M V^{2} = \dfrac{R}{\gamma - 1}\,\Delta T$.

Step 6: Solve for the temperature rise.
Rearranging gives $\Delta T = \dfrac{(\gamma - 1) M V^{2}}{2 R}$. \[ \boxed{\dfrac{(\gamma - 1) M V^{2}}{2R}} \]