# 9.5: The Ideal Gas Law

The three gas laws that we covered in Section 9.2, 9.3, 9.4 and 9.5 describe the effect of pressure, temperature and the number of moles of a gas on volume. The three independent gas laws are:

• Boyle’s law: ${\displaystyle V\propto {}^{1}\!\!\diagup \!\!{}_{P}\;}$
• Charles’s law: ${\displaystyle V\propto T{\text{ }}}$
• Avogadro’s law: ${\displaystyle V\propto n{\text{ }}}$

If volume (V) is proportion to each of these variables, it must also be proportional to their product:

$V\propto \frac{nT}{P}$

If we replace the proportionality symbol with a constant (let’s just choose R to represent our constant), we can re-write the equation as:

$V=R\left ( \frac{nT}{P}\right )$

or

$PV=nRT$

Example $$\PageIndex{1}$$:

The value of the proportionality constant R, can be calculated from the fact that exactly one mole of a gas at exactly 1 atm and at 0 ˚C (273 K) has a volume of 22.414 L.

Solution

Substituting in the equation:

$PV=nRT\; \; or\; \; R=\frac{PV}{nT}$

$R=\frac{(1\; atm)(22.414\; L)}{(1\; mole)(273\; K)}=0.082057\; L\; atm\; mol^{-1}\; K^{-1}$

The uncomfortable and somewhat obnoxious constant is called the universal gas constant, and you will need to know it (or look it up) whenever you solve problems using the combined ideal gas law.

Exercise $$\PageIndex{1}$$

What volume will 17.5 grams of N2 occupy at a pressure of 876 mm Hg and at 123 ˚C?

Many of the problems that you will encounter when dealing with the gas laws can be solved by simply using the “two-state” approach. Because R is a constant, we can equate an initial and a final state as:

$R=\frac{P_{1}V_{1}}{n_{1}T_{1}}\; \; for\; the\; initial\; state$

$R=\frac{P_{2}V_{2}}{n_{2}T_{2}}\; \; for\; the\; final\; state$

$\frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}}$

Using this equation, you can solve for multiple variables within a single problem.

Exercise $$\PageIndex{1}$$

A sample of oxygen occupies 17.5 L at 0.75 atm and 298 K. The temperature is raised to 303 K and the pressure is increased to 0.987 atm. What is the final volume of the sample?

If you noticed, we calculated the value of the proportionality constant R based on the fact that exactly one mole of a gas at exactly 1 atm and at 0 ˚C (273 K) has a volume of 22.414 L. This is one of the “magic numbers in chemistry; exactly one mole of any gas under these conditions will occupy a volume of 22.414 L. The conditions, 1 atm and 0 ˚C, are called standard temperature and pressure, or STP. The fact that all gases occupy this same molar volume can be rationalized by realizing that 99.999% of a gas is empty space, so it really doesn’t matter what’s in there, it all occupies the same volume. This realization is attributed to Amedeo Avogadro and Avogadro’s hypothesis, published in 1811, suggested that equal volumes of all gases at the same temperature and pressure contained the same number of molecules. This is the observation that led to the measurement of Avogadro’s number (6.0221415 × 1023), the number of things in a mole. The importance of the “magic number” of 22.414 L per mole (at STP) is that, when combined with the ideal gas laws, any volume of a gas can be easily converted into the number of moles of that gas.

Exercise $$\PageIndex{1}$$

1. A sample of methane has a volume of 17.5 L at 100.0 ˚C and 1.72 atm. How many moles of methane are in the sample?
2. A 0.0500 L sample of a gas has a pressure of 745 mm Hg at 26.4˚ C. The temperatureis now raised to 404.4 K and the volume is allowed to expand until a final pressure of 1.06 atm is reached. What is the final volume of the gas?
3. When 128.9 grams of cyclopropane (C3H6) are placed into an 8.00 L cylinder at 298 K, the pressure is observed to be 1.24 atm. A piston in the cylinder is now adjusted so that the volume is now 12.00 L and the pressure is 0.88 atm. What is the final temperature of the gas?

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