Answer: [tex]49.7\ cm^3[/tex]
Explanation:
Given
At temperature [tex]T_1=289\ K[/tex]
Volume is [tex]V_1=42\ cm^3[/tex]
and the temperature changes to [tex]T_2=342\ K[/tex]
At this temperature volume is [tex]V_2[/tex]
Consider Pressure remains the constant
[tex]\therefore \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\\Rightarrow V_2=V_1\times \dfrac{T_2}{T_1}\\\\\text{Insert values}\\\\\Rightarrow V_2=42\times \dfrac{342}{289}\\\\\Rightarrow V_2=49.7\ cm^3[/tex]
The volume at this temperature is [tex]49.7\ cm^3[/tex]
