Given,
The initial volume of the air, V₁=30 cm³=30×10⁻6 m³
The initial temperature of the air, T₁=29 °C=302.15 K
The final temperature of the air, T₂=4 °C=277.15 K
From Charles' law, we have,
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]Where V₂ is the final volume of the air.
On substituting the known values,
[tex]\begin{gathered} \frac{30\times10^{-6}}{302.15}=\frac{V_2_{}}{277.15} \\ \Rightarrow V_2=\frac{30\times10^{-6}\times277.15}{302.15} \\ =27.52\times10^{-6}m^3 \\ =27.52cm^3 \end{gathered}[/tex]Thus the volume of the ai after reducing the temperature is 27.52 cm³
