Question

A certain volume of the gas was found to be at a pressure of 1200 mm of mercury. When the pressure was decreased by 500 mm the gas occupied a volume of 2400 cm3. Calculate the initial volume occupied by the gas if the change was done at a constant temperature?

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Answer 1

$\huge{ \underline{ \bf{Given \: : }}}$

Let,

• $\tt{v_1 = ? cm {}^{3} }$
• $\tt \: p_1 = 1200mm$
• $\tt \: v_2 = 2400cm {}^{3}$
• $\tt \: p_2 = 700mm$

By Boyles Law

$\color{purple} \leadsto \red{ \boxed{ \tt{ \green{p_1 \: v_1 = p_2 \: v_2}}}}$

Substituting the values

$\tt \to1200 \times x = 700 \times 2400$

$\tt \to \: x = \frac{700 \times \cancel{2400 }}{ \cancel{1200}}$

$\tt \to \: x = 700 \times 2$

$\tt \to \: x = 1400cm {}^{3}$

$\therefore \tt \: v_1 = 1400cm {}^{3}$

Therefore the initial volume occupied by the gas is 1400 cm3.

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Know more about

Boyles Law

Boyle's law, also referred to as the Boyle–Mariotte law, or Mariotte's law (especially in France), is an experimental gas law that describes how the pressure of a gas tends to increase as the volume of the container decreases.

Answer 2

$\huge{\fcolorbox{pink}{indigo}{\fcolorbox{lime}{purple}{\fcolorbox{cyan}{black}{\red{Good\: Morning}}}}}$

$\Large\bf\pink{We\: know\: that}$

$\longmapsto\:\:\tt{Pressure\:\propto\:\dfrac{1}{Volume}\:}$

$\longmapsto\:\:\bf{\dfrac{P_1}{P_2}\:=\:\dfrac{V_2}{V_1}\:}$

$\Large\bf{\underline{\red{Where,}}}$

• $\bf{P_1\:=\:1200\:mm\:Hg}$

• $\bf{P_2\:=\:700\:mm\:Hg}$

• $\bf{V_1\:=\:?}$

• $\bf{V_2\:=\:2400\:cm^3}$

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$:\implies\:\:\rm{\dfrac{1200}{700}\:=\:\dfrac{2400}{V_1}\:}$

$:\implies\:\:\rm{\dfrac{12}{7}\:=\:\dfrac{2400}{V_1}\:}$

$:\implies\:\:\rm{V_1\:=\:\dfrac{2400}{12}\times{7}\:}$

$:\implies\:\:\rm{V_1\:=\:200\times{7}\:}$

$:\implies\:\:\rm\green{V_1\:=\:1400\:cm^3\:}$

____________________

$\Large\bf{Know\: more\: about}$

$\Large\bf\red{Boyles \:Law}$

➝ Boyle's law, also referred to as the Boyle–Mariotte law, or Mariotte's law (especially in France), is an experimental gas law that describes how the pressure of a gas tends to increase as the volume of the container decreases.

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$\Large\mathbb\green{HOPE\: IT'S\: HELPFUL}$