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A certain sample of gas has a volume of $0.2 \mathrm{~L}$ at one atmosphere pressure and 273.15 K. What is the volume of gas at $273.15^{\circ} \mathrm{C}$ at same pressure?
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The correct answer is:
$0.4 \mathrm{~L}$
$\mathrm{V}_{1}=0.2 \mathrm{~L} \quad, \quad \mathrm{~T}_{1}=237.15 \mathrm{~K}$
$\mathrm{~V}_{2}=? \quad, \quad \mathrm{~T}_{2}=273.15+273.15=546.30 \mathrm{~K}$
$\therefore \quad$ According to Charle's law,
$\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}} \quad$ (at constant $\mathrm{P}$ and $\mathrm{n}$ )
$\therefore \quad \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \times \mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.2 \times 546.30}{273.15}=0.4 \mathrm{~L}$
$\mathrm{~V}_{2}=? \quad, \quad \mathrm{~T}_{2}=273.15+273.15=546.30 \mathrm{~K}$
$\therefore \quad$ According to Charle's law,
$\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}} \quad$ (at constant $\mathrm{P}$ and $\mathrm{n}$ )
$\therefore \quad \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \times \mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.2 \times 546.30}{273.15}=0.4 \mathrm{~L}$
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