An ideal gas is initially at a temperature T and volume V. Its volume is increased by ΔV due to an increase in temperature ΔT, pressure remaining constant. The quantity δ=ΔV/(VΔT) varies with temperature as
A
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B
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C
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D
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Solution
The correct option is C From ideal gas equation PV=nRT(1)PΔV=nRΔT(2)
Dividing equation (2) by equation (1), we get
ΔVV=ΔTT⇒ΔVVΔT=1T=δ
δ=1T
So the graph between δ and T will be a rectangular hyperbola.