Question

An insulated container contains a monoatomic gas of molar mass 'm'. The container is moving with velocity 'V'. If it is stopped suddenly, the change in temperature of the gas is (R-gas constant)

• A. \(\frac{mv^2}{R}\)
• B. \(\frac{mv^2}{2R}\)
• C. \(\frac{mv^2}{3R}\)
• D. \(\frac{mv^2}{5R}\)

Hint:

Bulk kinetic energy becomes internal energy of the gas.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The container is insulated, so no heat leaves. When it stops suddenly, the bulk kinetic energy of the gas turns into random thermal energy of molecules.

Step 2: Write the energy balance:
Let the mass of the gas be \(M\) and its molar mass be \(m\), so the number of moles is \(n=\dfrac Mm\). Kinetic energy lost \(=\dfrac12MV^2\).

Step 3: Internal energy gained:
For a monatomic gas, \(\Delta U=n\cdot\dfrac32R\,\Delta T=\dfrac Mm\cdot\dfrac32R\Delta T\).

Step 4: Solve:
\(\dfrac12MV^2=\dfrac{3MR}{2m}\Delta T\), so \(\Delta T=\dfrac{mV^2}{3R}\). Option C.

Step 5: Why the other options are wrong.
Options A and B ignore the \(\dfrac32R\) per mole for a monatomic gas (they use \(C_v=R\) or \(\dfrac R2\) instead). Option D uses \(\dfrac52R\), the value for a diatomic gas.

Final Answer:
The temperature rise is m V^2 / (3R). \[ \boxed{\text{(C) }\dfrac{mV^2}{3R}} \]