# If you have 16.8 L of gas at 1.0 atm, what is the new pressure if the volume is reduced to 15.0 L?

Mar 5, 2018

1.12 atm

#### Explanation:

I'm going to assume this is done at constant temperature.

We can use the combined gas law(?) :

$\frac{{P}_{1} \times {V}_{1}}{{T}_{1}} = \frac{{P}_{2} \times {V}_{2}}{{T}_{2}}$
Where P is the pressure, V is the volume and T is the temperature.

Units you plug into this equation can be varied (e.g ${m}^{3} , {\mathrm{dm}}^{3}$ for volume or $k P a , P a , a t m$ for pressure or degrees K or degrees C for temperature) but you are fine as long as you are consistent with the units you use on both sides. Now, let's assume this is done at STP (1 atm of pressure, 273 degrees Kelvin). For my units, I will use liters (${\mathrm{dm}}^{3}$) for volume, atm for pressure and Kelvin for temperature.

$\frac{1 \times 16.8}{273} = \frac{{P}_{2} \times 15}{273}$

${P}_{2} \approx 1.12 a t m$

Mar 5, 2018

$1.12 \setminus \text{atm}$

#### Explanation:

Using Boyle's law, it states that $P \propto \frac{1}{V}$, or ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, at a constant temperature.

Rearranging for ${V}_{2}$, we get

${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2$

Plugging in the given values,

V_2=(1 \ "atm"*16.8color(red)cancelcolor(black)"L")/(15color(red)cancelcolor(black)"L")=1.12 \ "atm"

So, the new pressure will be $1.12$ atmospheres.