JEE Main 2024 Physics question
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
Answer: (C)
Solution
P = nkT, so 1.38 × 101325 = 2 × 10²⁵ × 1.38 × 10⁻²³ × T, giving T = 101325/200 ≈ 500 K.