Law of Gay-Lussac

Tutorial 2

        

$ 10.0\;m^3$ of helium are at $12.0 \; ^oC $. What will be the temperature at which this gas occupies a volume of $ 10.8\; m^3 $, the pressure remaining constant? Solution (Click): Conversion to $K$: ..........$12.0\;^oC=285.15\;K$ Law of Gay-Lussac: ..........$\frac{V_1 }{T_1}=\frac{V_2 }{T_2}$ Calculation: ..........$T_1=\frac{V_1\cdot T_2}{V_2}$ = $\frac{10.8\cdot 285.15}{10.0} = 307.96\;K$ Conversion to $^oC$: ..........$307.96\;K=34.81\;^oC$