According to Charle's law,
V∝T at constant pressure
Hence increase in temperature, increases the volume. Hence to maintain constant pressure, some volume of the gas is expelled.
Initial volume of the gas V1=3L
Initial temperature of the gas T1=27∘C=300K
Final volume V2=initial volume+volume expelled
=3+ΔV L
Final temperature T2=?
Determination of ΔV at T2:
Given ΔV L at 17∘C or 290 K = 1.45 L
Accoridng to Charle's law At constant pressure V1T2=V2T1
ΔV L at temperature T2
ΔV = 1.45×T2290
Hence final volume V2 = 3 + 1.45×T2290
Again applying Charle's law,
∴T2=V2T1V1
=(3+1.45×T2290)×3003=600 K
∴T2=600 K = 327∘C