Answer:
T₂ = 128.19 K
Explanation:
Given that,
The initial volume, V₁ = 12.5 ft³
Initial temperature, T₁ = -173 °C = 100.15 K
Final volume, V₂ = 16 ft³
We need to find the new temperature. The relation between temperature and volume is given by :
[tex]V\propto T\\\\\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}[/tex]
Put all the values,
[tex]T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{100.15\times 16}{12.5}\\\\T_2=128.19\ K[/tex]
So, the new temperature is 128.19 K.
