Gas Laws in Chemistry

Question

Calculate the density (in ^g/_L) of nitrogen dioxide gas at STP: 

_____________ ^g/_L 

Then calculate the density at 1.01 atm and 27.6 ˚C.

______________ ^g/_L 

Answer 1

Nitrogen dioxide = NO₂

molar mass NO₂ = 46.0 g / mole

AT STP, 1 mol of any ideal gas = 22.4 L. So assume we have 1 L and then we can calculate moles.

1 L x 1 mol / 22.4 L = 0.0446 mols NO₂

0.0446 mols NO₂ x 46.0 g / mol = 2.05 g / L = Density of NO₂ at STP

To find the density at 27.6ºC and 1.01 atm, we will use the ideal gas law

PV = nRT

P = pressure = 1.01 atm

V = volume in liters = 1.0 L (we can simply assume any volume we want)

n = mols NO₂ = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 27.6ºC + 273 = 300.6K

Solving for n, we have ...

n = PV/RT

n = (1.01)(1.0) / (0.0821)(300.6)

n = 0.0409 mols NO₂

Converting to g NO₂ we have ..

0.0409 mol NO₂ x 46 g / mol = 1.88g

Density = g / L

Density = 1.88 g / L

Answer 2

Hi Julia,

Gas laws can at first be difficult to understand, but with a little practice they get much easier!

To calculate the density of a gas we first need to take the ideal gas law:

PV = nRT (P= pressure, V = volume, n = moles, R = ideal gas constant, T = temperature (in Kelvin!))

And convert it to a more useful equation for this question which is (I will derive it at the bottom if you would like to see that):

D = PM/RT (D = density, P = pressure, M = molecular mass, R = ideal gas constant, T = temperature (in Kelvin!))

With this adjusted equation, it is just a matter of plugging in some values to calculate the density! STP stands for standard temperature and pressure which is 1 atm and 0°C. But, remember, we need temperature in Kelvin (K) and K = C + 273. So STP is also 1 atm and 273 K.

First question:

Calculate the density (in ^g/_L) of nitrogen dioxide gas at STP:

**use D = PM/RT.**

We are solving for D, P = 1atm (STP), M = molecular mass of nitrogen dioxide (NO₂) which is 46g/mol (14g/mol for Nitrogen, 16g/mol per O), R = constant = 0.0821 L·atm/K·mol, T = 273 K (STP).

So, the answer is 2.05 g/L (all other units cancel out and you get units of mass over volume, or, density!)

Second question:

Then calculate the density at 1.01 atm and 27.6 ˚C.

Same process, use D = PM/RT, just make sure to remember to covert °C to Kelvin (K = C + 273). So the Temp we will use in the equation is 300.6 K, and the Pressure is 1.01 atm, all other variables are the same.

So, the answer is 1.88 g/L.

Derivation of D = PM/RT (feel free to ignore!)

PV=nRT (ideal gas law), let's rearrange for n (moles):

n = PV/RT

n, or moles, is defined as mass (m) / molar mass (M). So, if we plug in for n, we get:

m/M = PV/RT

Now that we have mass and molar mass in the equation, let's remember the definition of density = mass/volume. Next, we rearrange to get mass (m) over volume (V):

m/V = PM/RT

Since m/V is equal to density (D), we now have the desired equation D=PM/RT!

Feel free to ask any questions!