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An ideal gas is initially at temperature T and volume V. Its volume is increased by `DeltaV` due to an increase in temperature `DeltaT,` pressure remaining constant. The quantity `delta=(DeltaV)/(VDeltaT)` varies with temperature as
A. image
B. image
C. image
D. image

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Correct Answer - C
c. From ideal gas equation
`PV = RT`
`P Delta T = R Delta T`
Dividing Eq. (ii) by Eq. (i), we get
`(Delta T)/(V) = (Delta T)/(T) implies (Delta V)/(V Delta T) = (1)/(T) = delta` (given)
`:. Delta = (1)/(T)`
So the graph between `delta` and `T` will be a rectangular hyperbola.

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