# What is the temperature in Celsius inside a sealed 10.0 L flask which contains 14.2 g of nitrogen gas at 2.0 atm?

Jan 17, 2016

$t = {210}^{\circ} \text{C}$

#### Explanation:

Your strategy here will be to

• use the ideal gas law equation to find the temperature of the gas in Kelvin
• use the known conversion factor to go from Kelvin to degrees Celsius

The ideal gas law equation looks like this

$\textcolor{b l u e}{P V = n R T} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas, always expressed in Kelvin!

Notice that the problem does not provide you with the number of moles of gas, but that it does give you the mass of the sample.

To get the number of moles that are present in that $\text{14.2-g}$ sample of nitrogen gas, ${\text{N}}_{2}$, use the compound's molar mass.

14.2 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(28.0134color(red)(cancel(color(black)("g")))) = "0.5069 moles N"_2

Rearrange the ideal gas law equation and solve for $T$ - notice that all the values given to you are expressed in units that match those used for the universal gas constant!

$P V = n R T \implies T = \frac{P V}{n R}$

This will give you

T = (2.0 color(red)(cancel(color(black)("atm"))) * 10.0color(red)(cancel(color(black)("L"))))/(0.5069color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K")) = "480.6 K"

Now, the following relationship exists between temperature expressed in degrees Celsius and temperature expressed in Kelvin

$\textcolor{b l u e}{\left[\text{^@"C"] = ["K}\right] - 273.15}$

In your case, the temperature of the gas in degrees Celsius will be

$t = 480.6 - 273.15 = {207.45}^{\circ} \text{C}$

Rounded to two sig figs, the number of sig figs you have for the pressure of the gas, the answer will be

$t = \textcolor{g r e e n}{{210}^{\circ} \text{C}}$