When temperature of a gas, contained in a closed vessel is increased by 5`C, it`s pressure increases by 1%. The original temperature of gas was approximately:

Question

Arun · Answer

As vessel is closed , so volume of the container is constant.

P proportional to T
where P = pressure
and T= temperature

so,
P1/P2 = T1/T2

T2/T1 = P2/P1

T2/T1 -1 = P2/P1 -1
T2 -T1/T1 = P2 -P1/P1

(5/T1)*100 = 1

T1 = 500°c

Regards

Arun (askIITians forum expert)

Piyush Upadhyay · Answer

There is a little correction the answer is 500K or 227`C. You forgot that we have assumed T1 and T2 to be absolute temperatures 😄😦😳😱

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When temperature of a gas, contained in a closed vessel is increased by 5`C, it`s pressure increases by 1%. The original temperature of gas was approximately:

When temperature of a gas, contained in a closed vessel is increased by 5`C, it`s pressure increases by 1%. The original temperature of gas was approximately:

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When temperature of a gas, contained in a closed vessel is increased by 5`C, it`s pressure increases by 1%. The original temperature of gas was approximately:

When temperature of a gas, contained in a closed vessel is increased by 5`C, it`s pressure increases by 1%. The original temperature of gas was approximately:

2 Answers

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Other Related Questions on Thermal Physics

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