The temperature of a gas is -78° C and the average translational kinetic energy of its molecules is K. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2 K is :

Solution

Physics sir · 3 steps

Translational KE goes as absolute temperature, so convert to kelvin first and then double it.

1. KE depends on T. The average translational kinetic energy of a molecule is 3/2k_B T, with T in kelvin. It depends only on temperature, not on the type of gas.

   K₂/K₁ = T₂/T₁ ⇒ T₂ = 2T₁

2. Convert to kelvin. T₁ = -78 + 273 = 195 K. Doubling the energy doubles the absolute temperature, so T₂ = 2 × 195 = 390 K.
3. So. T₂ = 390 - 273 = 117°C, which is option (d).

The idea to keep. Any ratio law from kinetic theory, such as K ∝ T or vᵣₘₛ ∝ √T, works only in kelvin. Convert first, use the ratio, then convert back.

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