Practicing Success
If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1oC, the initial temperature must be : |
250 K 250oC 2500 K 25oC |
250 K |
\(P_1 = P, T_1 = T, P_2 = P + (0.4% \text{ of } P)\) \(\Rightarrow P_2 = P + \frac{4}{1000} P\) \(\Rightarrow P_2 = P + \frac{P}{250}\) 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 |