The temperature of a gas having 2.0 × 10²⁵ molecules per cubic meter at 1.38 atm (Given, k=1.38 × 10⁻²³ JK⁻¹) is :

Solution

Physics sir · 4 steps

Use the per-molecule form of the gas law, P = nkT, and change atm to pascal before you divide.

1. Gas law per molecule. Here you are given the number of molecules per cubic metre, n, and Boltzmann's constant k, not moles and R. So use the molecular form of the ideal gas equation.

   P = nkT ⇒ T = P/nk

2. Convert the pressure. With 1 atm ≈ 10⁵ Pa, the pressure is P = 1.38 × 10⁵ Pa. The denominator is nk = (2.0 × 10²⁵) × (1.38 × 10⁻²³) = 276 J m⁻³ K⁻¹.
3. Put in numbers. T = (1.38 × 10⁵)/276 = 500 K. If you use 1 atm = 1.013 × 10⁵ Pa instead, you get about 506 K, which is still nearest to 500 K.
4. So. The temperature of the gas is 500 K, option (a).

The idea to keep. When a question gives molecules per volume and k, use P = nkT. It is just PV = nRT written for single molecules. Always put pressure in pascal first.

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