0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127°C. Th amount of heat required to double the speed of its molecules is ____________ k cal.
Take R = 2 cal mole⁻¹ K⁻¹)

Solution

Physics sir · 4 steps

Speed goes as the square root of T, so doubling the speed means four times the temperature.

1. Moles and temperature. Nitrogen is N₂ with molar mass 28 g mol⁻¹, so n = 56/28 = 2 mol. The starting temperature is 127 + 273 = 400 K.
2. New temperature. Molecular speed follows v = √(3RT/M), so v ∝ √T. To double v, T must become 4 × 400 = 1600 K, giving Δ T = 1200 K.
3. Heat at constant volume. The gas is in a closed vessel, so its volume is fixed and Q = nC_VΔ T. Nitrogen is diatomic with 5 degrees of freedom, so C_V = 5/2R = 5 cal mol⁻¹ K⁻¹.

   Q = 2 × 5 × 1200 = 12000 cal

4. So. The heat required is 12000 cal, that is 12 kcal.

The common slip: 4. Taking a doubled speed to mean a doubled temperature gives Δ T = 400 K and Q = 4000 cal, so 4 kcal instead of 12.

The idea to keep. Any molecular speed (rms, average or most probable) goes as √T, with T in kelvin. A gas in a closed vessel heats at constant volume, so use C_V, which is 5/2R for a diatomic gas.

Answer from the NTA final answer key. Solution by PrepWiser.