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 quantityδ=ΔV/VΔT varies with temperature as

From ideal gas equationPV=RT               ........(i)PΔV=RΔT           ........(ii) Dividing Eq. (ii) bv Eq. (i), we getΔVV=ΔTT⇒ΔVVΔT=1T=δ    (given)∴ δ=1TSo the graph between δ and T will be a rectangular hyperbola.