An ideal gas is initially at temper...
Question
An ideal gas is initially at temperature T and volume V. Its volume is increased by ΔV due to an increase in temperature △ T, pressure remaining constant. The quanti
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An ideal gas is initially at temperature ( T ) and volume ( V ). Its volume is increased by ( Delta V ) due to an increase in temperature ( triangle ) T, pressure remaining constant. The quantity ( delta=frac{Delta V}{(V Delta T)} ) varies with temperature as

JEE/Engineering Exams
Physics
Solution
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An ideal gas is initially at temperature T and volume V. Its volume is increased by ΔV due to an increase in temperature △ T, pressure remaining constant. The quantity δ=ΔV/(VΔT) varies with temperature as
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Correct Option (c)
Explanation:
From ideal gas equation ( mathrm{PV}=mathrm{RT} )
or ( P Delta V=R Delta T )
Dividing equation (ii) by (i) we get ( Delta V / V ) ( =Delta mathrm{T} / mathrm{T}=Delta mathrm{V} / mathrm{V} Delta mathrm{T}=1 / T=delta )
( therefore quad delta=1 / mathrm{t} . ) So the graph between ( delta ) and ( T )
will be rectangular hyperbola.

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