Based on the question, it is evident that this question rests on the premise of Charle's Law, which essentially states that temperature is proportional to volume once pressure and mass remain constant.
Thus by Charle's Equation: [tex]\frac{ V_{1} }{ T_{1} } = \frac{ V_{2} }{ T_{2} } [/tex]
Since the initial volume = 2.2 L; the initial temperature = 20 C; and the final volume = 2.6 L,
then V₁ = 2.2 L; T₁ = 20 C; V₂ = 2.6 L and what we would be
finding is T₂ (the final/new temperature)
Now, [tex]T_{2} = \frac{(V_{2}) (T_{1}) }{V_{1}} [/tex]
⇒ [tex]T_{2} = \frac{(2.6 L) (20 ^{o} C) }{2.2 L} [/tex]
⇒ [tex]T_{2} = 23.64 ^{o}C[/tex]
∴ the new temperature is ≈ 23.6 °C
