A vessel contains 14 g of nitrogen gas at a temperature of 27° C. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :
Take R=8.32 J mol⁻¹ k⁻¹.

Solution

Physics sir · 3 steps

Doubling vᵣₘₛ means four times the temperature. Then it is just nC_VΔ T in a closed vessel.

1. Speed to temperature. Since vᵣₘₛ = √(3RT/M), the speed goes as √T. To double the speed, T must become 4 times larger: from 300 K to 1200 K, so Δ T = 900 K.
2. Which heat capacity. The gas sits in a closed vessel, so its volume stays fixed and all the heat goes into internal energy. Nitrogen is diatomic, so C_V = 5/2R. The amount is n = 14/28 = 0.5 mol.

   Q = nC_VΔ T = 0.5 × 5/2 × 8.32 × 900

3. So. Q = 10.4 × 900 = 9360 J, which is option (c).

The idea to keep. Speed goes as √T, so multiplying speed by k multiplies the temperature by k². A closed vessel means constant volume: use C_V, and 5/2R for a diatomic gas.

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