When the pressure of the gas contained in a closed vessel is increased by 2.5 % , the temperature of the gas increases by 4 K . The initial temperature of the gas is

Step-by-step solution

The vessel is closed, maintaining constant volume. Applying Gay-Lussac's Law for ideal gases at constant volume: P₁ T₁ = P₂ T₂ The pressure increases by 2.5 % giving P₂ = 1.025P₁ , and the temperature increases by 4 K giving T₂ = T₁ + 4 . Substituting: P₁ T₁ = 1.025P₁ T₁ + 4 Canceling P₁ and solving: T₁ + 4 = 1.025T₁ 4 = 0.025T₁ T₁ = 4 0.025 = 160 K Initial temperature: 160 K