KEY2CHEM

Ideal Gas Law

Based on the relationship between the macroscopic properties of gases (as determined by kinetic molecular theory), the Ideal Gas Law (\(PV=nRT\)) combines the overall quantities with the universal gas constant (\(R= 0.08206 \frac{ L\cdot atm}{mol\cdot K}\)) to quantify measurements. The universal gas constant comes from measuring \(1\text{ mole}\) of ideal gas at standard temperature and pressure. Carefully note the units of the gas constant; unit conversions of the other variables may be required before using the Ideal Gas Law equation.


Example 1.

What is the pressure of a \(100.0\text{ mL}\) sample of \(0.029\text{ mol}\) gas at \(273\text{ K}\)?

A. \(65\text{ atm}\)

B. \(150\text{ atm}\)

C. \(6.5\text{ atm}\)

 

 

Solution 

C. \(6.5\text{ atm}\)

\(\begin{align} PV &= nRT \\ P &= \frac{nRT}{V} \\ P &= \frac{(0.029\text{ mol})(0.08206 \frac{L\cdot atm}{mol\cdot K})(273\text{ K})}{0.1000 \text{ L}} = 6.5\text{ atm}\end{align}\)


Example 2.

 

What is the volume of a \(0.65\text{ mol CO}_2\) sample at \(308\text{ K}\) and \(1.24\text{ atm}\)?

A. \(13\text{ L}\)

B. \(20.\text{ L}\)

C. \(0.075\text{ L}\)

 

 

 

Solution

A. \(13\text{ L}\)

\(\begin{align} PV &= nRT \\ V&= \frac{nRT}{P} \\ V &= \frac{(0.65\text{ mol})(0.08206 \frac{L\cdot atm}{mol\cdot K})(308\text{ K})}{1.24 \text{ atm}} = 13\text{ L}\end{align}\)


Example 3.

 

A \(0.250\text{ L}\) sample of ammonia (\(\require{mhchem}\ce{NH3}\)) gas exerts a pressure of \(1.10\text{ atm}\) at \(316\text{ K}\). What mass of ammonia is in the container?

A. \(0.0787\text{ g}\)

B. \(0.180\text{ g}\)

C. \(8.04\text{ g}\)

 

 

 

Solution

B. \(0.180\text{ g}\)

\(\begin{align} PV &= nRT \\ n&= \frac{PV}{RT} \\ n &= \frac{(1.10\text{ atm})(0.250\text{ L})}{(0.08206 \frac{L\cdot atm}{mol\cdot K})(316\text{ K})} = 0.0106\text{ mol}\times \frac{17.034\text{ g NH}_3}{1\text{ mol NH}_3} = 0.180\text{ g NH}_3\end{align}\)