Answer:
V₂ = 457.49 mL
Explanation:
Given that,
Initial volume, [tex]V_1=543\ mL[/tex]
Initial temperature, [tex]T_1=75^{\circ} C=75+273=348\ K[/tex]
Final temperature, [tex]T_2=20.2^{\circ} C=20.2+273=293.2\ K[/tex]
We need to find the final volume of the gas. The relation between the volume and the temperature of the gas is given by :
[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]
Put the respected values,
[tex]V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{543\times 293.2}{348}\\\\V_2=457.49\ mL[/tex]
So, the final volume of the gas is equal to 457.49 mL.
