# A gas has a volume of "720.0 mL" at "20.0"^@"C" and "3.00 atm". What would the volume of the gas be at STP?

Feb 19, 2015

The final volume will be $\text{2039 mL}$.

#### Explanation:

STP means standard temperature and pressure. The current values for STP are 0^("o")"C" or $\text{273.15 K}$ and ${\text{10}}^{5}$ $\text{Pascals}$ $\left(\text{Pa}\right)$, usually given as $\text{100 kPa}$ to make it easier to work with. For the gas laws, the Celsius temperature must be converted to Kelvins by adding $273.15$ to the Celsius temperature.

This question can be answered using the equation for the combined gas law:

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

where:

${P}_{1}$ and ${P}_{2}$ are the initial and final pressures, ${V}_{1}$ and ${V}_{2}$ are the initial and final volumes, and ${T}_{1}$ and ${T}_{2}$ are the initial and final temperatures in Kelvins.

Known/Given:

P_1 = 3.00 color(red)cancel(color(black)("atm"))xx(101.325"kPa")/(1color(red)cancel(color(black)("atm")))="303.975 kPa"

${V}_{1} = 720.0 \text{mL}$

T_1 = 20.0^("o")"C" + 273.15 = 293.2"K"

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

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

Unknown:

${V}_{2}$

Solution: Rearrange the combined gas law so that ${V}_{2}$ is isolated, then solve for ${V}_{2}$

V_2 =(P_1V_1T_2)/(T_1P_2) = (303.975color(red)cancel(color(black)("kPa"))*720.0 "mL"*273.15 color(red)cancel(color(black)("K")))/(293.2 color(red)cancel(color(black)("K"))*100color(red)cancel(color(black)("kPa"))) = "2039 mL" (rounded to four significant figures)

Note:

If your teacher is still using $\text{1 atm}$ for standard pressure, substitute $\text{3.00 atm}$ for $\text{303.975 kPa}$, and $\text{1 atm}$ for $\text{100 kPa}$. The final volume will be $\text{2010 mL}$ rounded to three significant figures.