# What is the volume of an ideal gas at STP, if its volume is "2.85 L" at 14^@ "C" and "450 mm Hg"?

Aug 10, 2017

The volume at STP is $\approx \text{1200 L} .$

#### Explanation:

This is an example of the combined gas law, which combines Boyle's, Charles', and Gay-Lussac's laws. It shows the relationship between the pressure, volume, and temperature when the quantity of ideal gas is constant.

The equation to use is:

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

https://en.wikipedia.org/wiki/Gas_laws

Housekeeping Issues

Current STP

Standard temperature is $\text{0"^@"C}$ or $\text{273.15 K}$, and standard pressure after 1982 is ${10}^{5}$ $\text{Pa}$, usually written as $\text{100 kPa}$ for easier use.

The Kelvin temperature scale must be used in gas problems. To convert temperature in degrees Celsius to Kelvins, add $273.15$ to the Celsius temperature.

${14}^{\circ} \text{C"+273.15="287.15 K}$

The pressure in $\text{mmHg}$ must be converted to $\text{kPa}$.

$\text{1 mmHg}$ $=$ $\text{101.325 kPa}$

450color(red)cancel(color(black)("mmHg"))xx(101.325"kPa")/(1color(red)cancel(color(black)("mmHg")))="45600 kPa"

I'm giving the pressure to three sig figs to reduce rounding errors.

Organize the data:

Known

${P}_{1} = \text{45600 kPa}$

${V}_{1} = \text{2.85 L}$

${T}_{1} = \text{287.15 K}$

${P}_{2} = \text{100 kPa}$

${T}_{2} = \text{273.15 K}$

Unknown

${V}_{2}$

Solution

Rearrange the combined gas law equation to isolate ${V}_{2}$. Insert the data and solve.

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

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

V_2=((45600color(red)cancel(color(black)("kPa")))xx(2.85"L")xx(273.15color(red)cancel(color(black)("K"))))/((287.15color(red)cancel(color(black)("K")))xx(100color(red)cancel(color(black)("kPa"))))="1200 L" (rounded to two significant figures)