The kinetic energy of translation of the molecules in 50 g of CO₂ gas at 17°C is :

Solution

Physics sir · 4 steps

Translation always gets exactly 3 degrees of freedom, even for CO2, so use 3/2 nRT and nothing more.

1. Only translation counts. CO₂ is linear, so it also has 2 rotational degrees of freedom. The question asks only for translational kinetic energy, and every molecule has exactly 3 translational degrees of freedom. Each one carries 1/2RT per mole.

   Eₜᵣₐₙₛ = 3/2 nRT

2. Moles and temperature. The molar mass of CO₂ is 44 g mol⁻¹, so n = 50/44 = 1.136 mol. The temperature in kelvin is T = 17 + 273 = 290 K.
3. Put in numbers. Take R = 8.3 J K⁻¹ mol⁻¹: Eₜᵣₐₙₛ = 3/2 × 50/44 × 8.3 × 290 = 4102.8 J. With R = 8.314 you get about 4110 J, which is still closest to (d).
4. So. The translational kinetic energy is about 4102.8 J, option (d).

The idea to keep. Translational KE is always 3/2nRT, whatever the molecule. Add rotational or vibrational terms only when the question asks for total kinetic energy.

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