If pressure of a gas contained in a closed vessel is increased by 0-4 - when heated by 1- C- the initial temperature must beA- 2500 KB- 250 KC- 250- CD- 25- C

The correct option is B 250 KP_1=P,T_1=T_1,P_2=P+(0.4%of P)=P+0.4/100P=P+P/250 T_2=T+1 From Gay Lussac's law P_1/P_2T_1/T_2⇒ P/P+p250T/T+1 [As V= constant for ...

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Standard XII

Chemistry

Gay Lussac's Law, Avagadro's Law

If pressure o...

Question

If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by $1^{\circ} C$ , the initial temperature must be

250 K

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2500 K

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Solution

The correct option is A 250 K
$P_{1} = P, T_{1} = T_{1}, P_{2} = P + (0.4 %$ of P)= $P + \frac{0.4}{100} P = P + \frac{P}{250} T_{2} = T + 1$
From Gay Lussac's law $\frac{P_{1}}{P_{2}} \frac{T_{1}}{T_{2}} \Rightarrow$ $\frac{P}{P + \frac{p}{250}} \frac{T}{T + 1}$ [As V= constant for closed vessel]
By solving we get T=250 K.

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If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1∘C, the initial temperature must be

The correct option is A 250 KP1=P,T1=T1,P2=P+(0.4%of P)=P+0.4100P=P+P250T2=T+1 From Gay Lussac's law P1P2T1T2⇒ PP+p250TT+1 [As V= constant for closed vessel] By solving we get T=250 K.

If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by $1^{\circ} C$ , the initial temperature must be