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}