<meta http-equiv="refresh" content="1; url=/nojavascript/"> Ideal Gas Law ( Read ) | Chemistry | CK-12 Foundation

# Ideal Gas Law

%
Progress
Practice Ideal Gas Law
Progress
%
Ideal Gas Law

Ideal Gas Law

The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro’s law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation:

$\frac{P_1 \times V_1}{T_1 \times n_1}=\frac{P_2 \times V_2}{T_2 \times n_2}$

As with the other gas laws, we can also say that  $\frac{\left(P \times V \right)}{\left(T \times n \right)}$ is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.

The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable  $R$ for the constant, the equation becomes:

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

The ideal gas law is conventionally rearranged to look this way, with the multiplication signs omitted:

$PV=nRT$

The variable  $R$ in the equation is called the ideal gas constant .

#### Sample Problem: Ideal Gas Law

What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19°C? Assume the oxygen is ideal.

Step 1: List the known quantities and plan the problem.

Known

• $P = 88.4 \ \text{kPa}$
• $T = 19^\circ \text{C} = 292 \text{ K}$
• $\text{mass} \ O_2 = 3.760 \text{ g}$
• $O_2 = 32.00 \text{ g/mol}$
• $R = 8.314 \text{ J/K} \cdot \text{mol}$

Unknown

• $V = ? \text{ L}$

In order to use the ideal gas law, the number of moles of O 2   $(n)$ must be found from the given mass and the molar mass. Then, use  $PV = nRT$ to solve for the volume of oxygen.

Step 2: Solve .

$3.760 \text{ g} \times \frac{1 \ \text{mol} \ O_2}{32.00 \text{ g} \ O_2}=0.1175 \ \text{mol} \ O_2$

Rearrange the ideal gas law and solve for $V$ .

$V=\frac{nRT}{P}=\frac{0.1175 \ \text{mol} \times 8.314 \text{ J/K} \cdot \text{mol} \times 292 \text{ K}}{88.4 \ \text{kPa}}=3.23 \text{ L } O_2$

The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume (22.4 L/mol) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for  $T$ and $P$ . Since a joule (J) = kPa • L, the units cancel correctly, leaving a volume in liters.