# Question 644b9

Jan 13, 2017

$\text{270 mL}$

#### Explanation:

The thing to remember about pressure and volume is that they have an inverse relationship when temperature and number of moles of gas are kept constant, as described by Boyle's Law.

In other words, when you keep the temperature of the gas and the amount of gas, i.e. the number of moles of gas, constant, increasing its pressure by a factor $f$ will cause its volume to decrease by the same factor $f$.

Similarly, decreasing its pressure by a factor $f$ will cause its volume to increase by the same factor $f$. Mathematically, you can say that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{P}_{1} \cdot {V}_{1} = {P}_{2} \cdot {V}_{2}}}}$

Here

• ${P}_{1}$ and ${P}_{2}$ represent the initial and the final pressure of the gas
• ${V}_{1}$ and ${V}_{2}$ represent the initial and final volume of the gas

In your case, you know that a sample of gas starts at a pressure ${P}_{1}$ and ends up at a pressure

${P}_{2} = 2 \times {P}_{1} \to$ the pressure is doubled

This means that you can expect the volume of the gas to decrease by a factor of $2$.

Rearrange the equation for Boyle's Law and plug in your values to confirm that this is the case

${V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1}$

V_2 = color(red)(cancel(color(black)(P_1)))/(2color(red)(cancel(color(black)(P_1)))) xx "540 mL" = color(darkgreen)(ul(color(black)("270 mL")))#

The answer is rounded to two sig figs, the number of sig figs you have for the initial volume of the gas.

As predicted, increasing the pressure by a factor of $2$ will cause the volume to decrease by a factor of $2$.