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# Ideal Gas Law

## Introduces law based on the Combined Gas Law and Avogadro's Law

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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:

P1×V1T1×n1=P2×V2T2×n2\begin{align*}\frac{P_1 \times V_1}{T_1 \times n_1}=\frac{P_2 \times V_2}{T_2 \times n_2}\end{align*}

As with the other gas laws, we can also say that (P×V)(T×n)\begin{align*}\frac{\left(P \times V \right)}{\left(T \times n \right)}\end{align*} 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\begin{align*}R\end{align*} for the constant, the equation becomes:

P×VT×n=R\begin{align*}\frac{P \times V}{T \times n}=R\end{align*}

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

PV=nRT\begin{align*}PV=nRT\end{align*}

The variable R\begin{align*}R\end{align*} 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 kPa\begin{align*}P = 88.4 \ \text{kPa}\end{align*}
• T=19C=292 K\begin{align*}T = 19^\circ \text{C} = 292 \text{ K}\end{align*}
• mass O2=3.760 g\begin{align*}\text{mass} \ O_2 = 3.760 \text{ g}\end{align*}
• O2=32.00 g/mol\begin{align*}O_2 = 32.00 \text{ g/mol}\end{align*}
• R=8.314 J/Kmol\begin{align*}R = 8.314 \text{ J/K} \cdot \text{mol}\end{align*}

Unknown

• V=? L\begin{align*}V = ? \text{ L}\end{align*}

In order to use the ideal gas law, the number of moles of O2 (n)\begin{align*}(n)\end{align*} must be found from the given mass and the molar mass. Then, use PV=nRT\begin{align*}PV = nRT\end{align*} to solve for the volume of oxygen.

Step 2: Solve.

3.760 g×1 mol O232.00 g O2=0.1175 mol O2\begin{align*}3.760 \text{ g} \times \frac{1 \ \text{mol} \ O_2}{32.00 \text{ g} \ O_2}=0.1175 \ \text{mol} \ O_2\end{align*}

Rearrange the ideal gas law and solve for V\begin{align*}V\end{align*}.

V=nRTP=0.1175 mol×8.314 J/Kmol×292 K88.4 kPa=3.23 L O2\begin{align*}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\end{align*}

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 \begin{align*}T\end{align*} and \begin{align*}P\end{align*}. Since a joule (J) = kPa • L, the units cancel correctly, leaving a volume in liters.

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