Answer:
[tex]T_2=261.46\ K[/tex]
Explanation:
It is given that,
Original temperature, [tex]T_1=323^{\circ}C=596.15\ K[/tex]
Original volume, [tex]V_1=2.85\ L[/tex]
We need to find the temperature if the volume of the balloon to be shrink to 1.25 L.
According to Charles law, at constant pressure, [tex]V\propto T[/tex]
It would means, [tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}[/tex]
T₂ = ?
[tex]T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{1.25\times 596.15}{2.85}\\\\T_2=261.46\ K[/tex]
So, the new temperature is 261.46 K.
