Question

A monoatomic gas of molar mass \( m \) is kept in an insulated container. The container is moving with velocity \( v \). If the container is suddenly stopped, then the change in the temperature of the gas is:

• A. \( \frac{mv^2}{4R} \)
• B. \( \frac{mv^2}{2R} \)
• C. \( \frac{mv^2}{R} \)
• D. \( \frac{mv^2}{3R} \)

Hint:

For a monoatomic gas, the change in temperature due to the kinetic energy change when the container stops is directly proportional to the square of the velocity of the container, and inversely proportional to the gas constant \( R \).

Answer 1

The Correct Option is D

Step 1: When the container is moving, the gas molecules inside have kinetic energy due to their motion. For a monoatomic ideal gas, the total kinetic energy is given by the formula: \[ E_k = \frac{3}{2} n R T \] where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature of the gas. 
Step 2: When the container is suddenly stopped, the gas molecules stop moving in the direction of the container's motion, and the kinetic energy associated with the motion of the container is converted into internal energy, which causes the temperature to increase. 
Step 3: The total kinetic energy of the gas molecules is related to the motion of the container. The kinetic energy of the container is \( \frac{1}{2} m v^2 \), where \( m \) is the mass of the gas and \( v \) is the velocity of the container. 
Step 4: Since the system is insulated, the change in internal energy equals the change in kinetic energy. For an ideal monoatomic gas, the change in temperature is related to the change in internal energy. The equation for the change in temperature \( \Delta T \) is: \[ \Delta T = \frac{E_k}{n C_V} \] where \( C_V = \frac{3}{2} R \) is the molar specific heat at constant volume. 
Step 5: Substituting \( E_k = \frac{1}{2} m v^2 \) and solving for \( \Delta T \): \[ \Delta T = \frac{\frac{1}{2} m v^2}{n \cdot \frac{3}{2} R} = \frac{m v^2}{3 n R} \] Since \( n = \frac{m}{M} \) (where \( M \) is the molar mass), we get: \[ \Delta T = \frac{m v^2}{3 R} \] Thus, the change in the temperature of the gas is \( \boxed{\frac{mv^2}{3R}} \).