# How would you calculate the density, in grams per liter, of a gas at STP if 3.78 L of the gas at 28.8 Celsius and 765.6 mm Hg weighs 0.791 g?

Mar 5, 2017

The density of the gas at STP is $\text{0.227 g/L}$.

#### Explanation:

This question involves the combined gas law . The equation for the combined gas law is:

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

The density of a substance changes with change in temperature. This is because the volume increases as temperature increases and decreases as the temperature decreases. The mass is not affected by temperature or volume. We need to determine the volume at STP in order to determine the density at STP.

The strategy is to use the combined gas law, using the values given in the question as the first set of variables ${V}_{1}$, ${P}_{1}$, and ${T}_{1}$, and STP as the second set of variables ${V}_{2}$, ${P}_{2}$, and ${T}_{2}$.
STP is $\text{0"^@"C}$, or $\text{273.15 K}$, and $\text{100 kPa}$ . For the gas laws we use Kelvins. Pressure will have to be converted from mmHg to kPa, and the Celsius temperature will have to be converted to Kelvins.

Write what you know.

${V}_{1} = \text{3.78 L}$
P_1=765.6 color(red)(cancel(color(black)("mmHg")))xx"1 kPa"/7.5006183 color(red)(cancel(color(black)("mmHg")))="102.058 kPa"
${T}_{1} = \text{28.8"^@"C"+273.15="302.0 K}$
${P}_{2} = \text{100 kPa}$
${T}_{2} = \text{273.15 K}$

Write what you don't know: ${V}_{2}$ (volume at STP)

Solution
Rearrange the equation to isolate ${V}_{2}$. Substitute the values into the equation and solve.

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

V_2=(3.78"L"xx102.058color(red)(cancel(color(black)("kPa")))xx273.15color(red)(cancel(color(black)("K"))))/(302.0color(red)(cancel(color(black)("K")))xx100color(red)cancel(color(black)("kPa")))="3.4892 L"

Density at STP

$\text{Density"="mass"/"volume}$

$\text{Density"=(0.791 color(white)(.)"g")/(3.4892 color(white)(.)"L")="0.227 g/L}$ (rounded to three significant figures)