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A certain mass of a gas occupies a volume of $2 \mathrm{dm}^{3}$ at STP. At what temperature the volume of…
A certain mass of a gas occupies a volume of $2 \mathrm{dm}^{3}$ at STP. At what temperature the volume of gas becomes double, keeping the pressure constant?
$540.15^{\circ} \mathrm{C}$ $400.15^{\circ} \mathrm{C}$ $546 \cdot 15^{\circ}$ C $273.15^{\circ}$ C Solution
$\begin{array}{l}
\mathrm{V}_{1}=2 \mathrm{dm}^{3}, \mathrm{~T}_{1}=273.15 \mathrm{~K} \\
\mathrm{~V}_{2}=4 \mathrm{dm}^{3}, \mathrm{~T}_{2}=?
\end{array}$
According to Charle's law,
$\begin{array}{l}
\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \quad \therefore T_{2}=\frac{V_{2} \times T_{1}}{V_{1}} \\
\therefore T_{2}=\frac{4 \times 273.15}{2}=546.3 \mathrm{~K}
\end{array}$
Now, $\mathrm{T} \mathrm{k}=\mathrm{t}^{\circ} \mathrm{C}+273.15$
$\therefore \mathrm{t}^{\circ} \mathrm{C}=\mathrm{T} \mathrm{k}-273.15$
$\therefore \mathrm{t}^{\circ} \mathrm{C}=546.3-273.15=273.15^{\circ} \mathrm{C}$
Asked in: MHT CET 2020 (14 Oct Shift 1)
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