# At 25°C, 3 M of a gas occupy a volume of 4.0 L at 1.9 kPa. If the volume decreases to 2.0 L, what will the new pressure be?

Mar 8, 2016

$\text{3.8 kPa}$

#### Explanation:

The important thing to realize here is that the temperature and number of moles of gas are kept constant, which should let you know that you can use the equation for Boyle's Law to find the new pressure of the gas.

Boyle's Law states that pressure and volume have an inverse relationship when temperature and number of moles of gas are kept constant.

In simple terms, when pressure increases, volume decreases, and when pressure decreases, volume increases.

This happens because as you increase the volume of the gas, you essentially increase the space in which the molecules are moving. Since their average kinetic speed is constant, i.e. since temperature is constant, the molecules will hit the walls of the container less frequently than they did in the smaller volume $\to$ pressure will decrease.

In your case, the volume of the gas decreased to half of its original value, which means that you can expect the pressure to double.

Mathematically, this is written as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} {V}_{1} = {P}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

${P}_{1}$, ${V}_{1}$ - the pressure and volume of a gas at an initial state
${P}_{2}$, ${V}_{2}$ - the pressure and volume of the gas at a final state

Rearrange the equation to solve for ${P}_{2}$

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

and plug in your values to get

P_2 = (4.0 color(red)(cancel(color(black)("L"))))/(2.0color(red)(cancel(color(black)("L")))) * "1.9 kPa" = color(green)(|bar(ul(color(white)(a/a)"3.8 kPa"color(white)(a/a)|)))