Question

Boyle’s law states that the volume of a gas varies inversely with applied pressure. Suppose the pressure on 60 cubic meters of gas is raised from 1 atmosphere to 3 atmospheres. what new volume does the gas occupy?

Answer 1

$\bf \qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array}\\\\ -------------------------------$

$\bf \stackrel{\textit{the volume of a gas varies inversely with applied pressure}}{v=\cfrac{k}{p}} \\\\\\ \textit{we also know that } \begin{cases} v=60\\ p=1 \end{cases}\implies 60=\cfrac{k}{1}\implies 60=k \\\\\\ \boxed{v=\cfrac{60}{p}} \\\\\\ \stackrel{\textit{now if we apply 3 atmospheres, p=3, what is \underline{v}?}}{v=\cfrac{60}{3}}$

Answer 2

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:

$v=\dfrac{k}{p}\\\\i.e.\\\\pv=k$

where v is the volume of the gas and is the pressure applied.

This means that if firstly the volume and pressure exerted are:

$v_1\ and\ p_1$ respectively.

and afterwards the pressure applied and the volume of the gas is:

$p_2\ and\ v_2$ respectively.

Then by the inverse relation we have:

$p_1v_1=k\\\\and\\\\p_2v_2=k\\\\i.e.\\\\p_1v_1=p_2v_2$

From the data in the question we have:

$v_1=60\ m^3\ ,\ p_1=1\ atm\ ,\p_2=3\ atm$

Hence, we have from equation (1)

$1\times 60=3\times v_2\\\\i.e.\\\\60=3v_2\\\\i.e.\\\\v_2=\dfrac{60}{3}\\\\i.e.\\\\v_2=20\ m^3$