# Question #ccbc8

Jul 31, 2016

$R = 0.082 \left(\text{atm" * "dm"^3)/("mol" * "K}\right)$

#### Explanation:

Your starting point here will be the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Here you have

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant
$T$ - the absolute temperature of the gas

Rearrange the above equation to solve for $R$

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

Now, notice that the problem doesn't provide you with the number of moles of nitrogen gas present in your sample. This means that you're going to have to improvise a bit.

As you know, the number of moles present in a sample of a given compound is equal to the mass of the sample divided by the molar mass of the compound.

If you take $m$ to be the mass and ${M}_{M}$ the molar mass of nitrogen gas, you can say that

$n = \frac{m}{M} _ M$

Plug this into the above equation to get

$R = \frac{P V}{\frac{m}{M} _ M \cdot T} = \frac{P V}{m \cdot T} \cdot {M}_{M}$

Now, the problem tells you that the gas has a density of ${\text{1.250 g dm}}^{- 3}$. As you know, density is defined by mass divided by volume.

You already used $m$ as the mass of the sample, so you can say that you have

$\rho = \frac{m}{V} \implies \frac{V}{m} = \frac{1}{\rho}$

Plug this into the equation to get

$R = \frac{P}{T} \cdot \frac{1}{\rho} \cdot {M}_{M}$

Nitrogen gas, ${\text{N}}_{2}$, has a molar mass of ${\text{28.0134 g mol}}^{- 1}$. Plug in the values given to you to calculate the value of $R$ -- do not forget to convert the temperature from degrees Celsius to Kelvin

$R = {\text{1 atm"/((273.15 + 0)"K") * 1/(1.250 color(red)(cancel(color(black)("g"))) "dm"^(-3)) * 28.0134 color(red)(cancel(color(black)("g"))) "mol}}^{- 1}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{R = 0.082 \left(\text{atm" * "dm"^3)/("mol" * "K}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs, but don't forget that you only have one sig fig for the temperature and pressure of the sample.

The actual value of $R$ listed using these units is

$R \approx 0.0820575 \left(\text{atm" * "dm"^3)/("mol" * "K}\right)$

so this is an excellent result.

http://www.cpp.edu/~lllee/gasconstant.pdf