Q3 / 150 · MHT CET 2020 (14 Oct Shift 1) · Chemistry

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?

  • A
    $540.15^{\circ} \mathrm{C}$
  • B
    $400.15^{\circ} \mathrm{C}$
  • C
    $546 \cdot 15^{\circ}$ C
  • Correct answer
    D
    $273.15^{\circ}$ 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, $\mathrm{T} \mathrm{k}=\mathrm{t}^{\circ} \mathrm{C}+273.15$ $\therefore \mathrm{t}^{\circ} \mathrm{C}=\mathrm{T} \mathrm{k}-273.15$ $\therefore \mathrm{t}^{\circ} \mathrm{C}=546.3-273.15=273.15^{\circ} \mathrm{C}$

Hence, the correct answer is option: 'D'.

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