Respuesta :
[tex]\bf \qquad \qquad \textit{inverse proportional variation}
\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------[/tex]
[tex]\bf \stackrel{\textit{the volume of a gas varies inversely with applied pressure}}{v=\cfrac{k}{p}} \\\\\\ \textit{we also know that } \begin{cases} v=60\\ p=1 \end{cases}\implies 60=\cfrac{k}{1}\implies 60=k \\\\\\ \boxed{v=\cfrac{60}{p}} \\\\\\ \stackrel{\textit{now if we apply 3 atmospheres, p=3, what is \underline{v}?}}{v=\cfrac{60}{3}}[/tex]
[tex]\bf \stackrel{\textit{the volume of a gas varies inversely with applied pressure}}{v=\cfrac{k}{p}} \\\\\\ \textit{we also know that } \begin{cases} v=60\\ p=1 \end{cases}\implies 60=\cfrac{k}{1}\implies 60=k \\\\\\ \boxed{v=\cfrac{60}{p}} \\\\\\ \stackrel{\textit{now if we apply 3 atmospheres, p=3, what is \underline{v}?}}{v=\cfrac{60}{3}}[/tex]
Answer:
The new volume that the gas occupy is:
20 cubic meters
Step-by-step explanation:
Boyle’s law states that the volume of a gas varies inversely with applied pressure.
This means that there exist a constant k such that:
[tex]v=\dfrac{k}{p}\\\\i.e.\\\\pv=k[/tex]
where v is the volume of the gas and is the pressure applied.
This means that if firstly the volume and pressure exerted are:
[tex]v_1\ and\ p_1[/tex] respectively.
and afterwards the pressure applied and the volume of the gas is:
[tex]p_2\ and\ v_2[/tex] respectively.
Then by the inverse relation we have:
[tex]p_1v_1=k\\\\and\\\\p_2v_2=k\\\\i.e.\\\\p_1v_1=p_2v_2[/tex]
From the data in the question we have:
[tex]v_1=60\ m^3\ ,\ p_1=1\ atm\ ,\p_2=3\ atm[/tex]
Hence, we have from equation (1)
[tex]1\times 60=3\times v_2\\\\i.e.\\\\60=3v_2\\\\i.e.\\\\v_2=\dfrac{60}{3}\\\\i.e.\\\\v_2=20\ m^3[/tex]
