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Applications of the Ideal Gas Law

Determining the Molecular Weight (M) of an Unknown Gas

PV=nRT plug in n=mMPV=nRT\ plug\ in\ n=\frac{m}{M}
PV=mMRTPV=\frac{m}{M}RT
where, m=mass (g) and M=molar mass (g/mol)

Rearrange the equation to solve for M (molecular weight) of the unknown gas:

M=mRTPVM=\frac{mRT}{PV}

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Finding the Density of a Gas

The equation for density (𝛿) is:
δ=mV\boxed{\delta=\frac{m}{V}}

We can rearrange the ideal gas law to find density:

PV=nRT plug in n=mMPV=nRT\ plug\ in\ n=\frac{m}{M}

PV=mRTMPV=\frac{mRT}{M}
Solve for density:
mV=δ=MPRT\frac{m}{V}=\delta=\frac{MP}{RT}


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Standard Molar Volume

Recall: we talked about Avogadro's law where P and T are kept constant.
PV=nRT
V/n=RT/P where RT/P=constant
V/n=constant

Based on this, imagine if we were looking at two different gases in two different containers, each with the same P, T, and number of moles of gas particles.

We have made n (moles) a constant too. If that is the case, then the volume for all the gases should be the same, and they are!

Wize Concept
Specifically at STP (P=1atm and T=0°C or 273K), 1 mol of any gas occupies 22.4L! **Memorize this!**
STP=Standard Temperature and Pressure

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Photo by OpenStax / CC BY


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Example: Standard Molar Volume

In the following reaction, if you start with 4L H2(g) and 1.5L N2(g) at STP, what would the volume of each of the 3 gases be when the reaction is complete?

3H2(g) + N2(g) -> 2NH3(g)

@STP: T=273 K, P=1 atm (constant), 1 mol=22.4 L

We can convert the volumes into moles using 1 mol=22.4 L:
H2(g):
4L x (1 mol)22.4L =0.1786 moles4L\ x\ \frac{\left(1\ mol\right)}{22.4L}\ =0.1786\ moles

N2(g):
1.5L x (1 mol)22.4L=0.06696 moles1.5L\ x\ \frac{\left(1\ mol\right)}{22.4L}=0.06696\ moles

Now that we have moles of each reagent, we need to determine the limiting reagent.
To do this, divide each number of moles by the stoichiometric coefficient in the balanced equation:
H2(g):

0.1786 moles3 moles =0.05953\frac{0.1786\ moles}{3\ moles\ }=0.05953

N2(g):

0.06696 moles1 mole=0.06696\frac{0.06696\ moles}{1\ mole}=0.06696

The smallest number is for H2(g) so it is the limiting reagent!

Solve for moles of NH3(g) produced:

0.1786 moles H2(g) x 2 moles NH3(g)3 moles of H2(g)=0.1191 moles NH3(g)0.1786\ moles\ H_2\left(g\right)\ x\ \frac{2\ moles\ NH_3\left(g\right)}{3\ moles\ of\ H_2\left(g\right)}=0.1191\ moles\ NH_3\left(g\right)

Calculate the volume of NH3(g) using 1 mol =22.4 L

0.1191 moles NH3(g) x 22.4 L1 mol=2.6671 L 2.67L0.1191\ moles\ NH_3\left(g\right)\ x\ \frac{22.4\ L}{1\ mol}=2.6671\ L \ \boxed{\approx2.67L}

How many L of H2(g) are leftover?

0L of H2(g) leftover since it was the limiting reagent\boxed{\text{{0L of H2(g) leftover since it was the limiting reagent}}}
H2 was the limiting reagent and got completely used up!

We just need to find the L of N2(g) that are leftover.

First let's determine the moles leftover.

0.1786 moles H2(g) x (1 mol N2(g))3 moles H2(g)=0.05953 moles N2(g) used0.1786\ moles\ H_2\left(g\right)\ x\ \frac{\left(1\ mol\ N_2\left(g\right)\right)}{3\ moles\ H_2\left(g\right)}=0.05953\ moles\ N_2\left(g\right)\ used

We started with 0.06696 moles of N2(g)

0.06696 moles-0.05953 moles= 0.00743 moles N2(g) left over

Finally convert this into volume to determine the volume of N2(g) leftover:

0.00743 moles x 22.4L1 mol=0.166432 L  0.17 L0.00743\ moles\ x\ \frac{22.4L}{1\ mol}=0.166432\ L\ \ \boxed{\approx 0.17\ L}

Practice: Solving for Molar Mass of a Gas

The density of an unknown gas at 100oC and 746 torr is 1.994 g/L. What is the molecular mass of this unknown gas?

Practice: Determine the Gas

0.238 g of a gas were placed in a 250 mL vessel. If the gas is heated to 85 oC, the gas has a pressure of 7 atm. Which gas is in the vessel?

Practice: Solve for Molecular Weight

A 6.00 L flask is evacuated and weighed, 21.64g. The vessel is then filled with a gas at STP, after filling the vessel weighs 27.31g. What is the molecular weight of the gas?
Extra Practice