If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1oC, the initial temperature must be :
Answer & explanation
Correct answer: option 1
\(P_1 = P, T_1 = T, P_2 = P + (0.4% \text{ of } P)\)
\(\Rightarrow P_2 = P + \frac{4}{1000} P\)
\(\Rightarrow P_2 = P + \frac{P}{250}\)
and \(T_2 = T + 1\)
From Gay-Lussac's Law : \(\frac{P_1}{P_2} = \frac{T_1}{T_2}\)
\(\frac{P}{P + \frac{P}{250}} = \frac{T}{T+1}\) ... [as V = constant for closed vessel]
By solving, we get : T = 250 K