Answer:
[tex]T_2=238.57^{\circ} C[/tex]
Explanation:
Given that,
Initial volume, V₁ = 43 L
Initial temperature, T₁ = 225℃ = 498 K
Final volume, V₂ = 20.6 L
We need to find the final temperature. We know that, the relation between volume and temperature is given by :
[tex]V\propto T\\\\\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}[/tex]
Put all the values,
[tex]T_2=\dfrac{498\times 20.6}{43}\\\\T_2=238.57^{\circ} C[/tex]
So, the final temperature is equal to [tex]238.57^{\circ} C[/tex].
