# The Ideal Gas Equation

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 13:23

Elsewhere, we can learn quantitatively the effects of pressure and thermodynamic temperature on gas volume, we can return to the relation between volume and amount of substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula.. Avogadro's law tells us that at constant P and T, the volume of a gas is directly proportional to the amount of gas. Boyle's law says that volume is inversely proportional to pressure, and Charles' indicates that volume is directly proportional to temperature. These three laws may all be applied at once if we write

$V\propto n\text{ }\times \text{ }\frac{\text{1}}{P}\text{ }\times \text{ }T\text{ (1)}$

or, introducing a constant of proportionality R,

$V=R\text{ }\frac{nT}{P}\text{ (2)}$

Equation (2) applies to all gases at low pressures and high temperatures and is a very good approximation under nearly all conditions. The value of R, the gas constantA proportionality constant between the product of the pressure and volume of a gas and the product of the chemical amount (moles) and temperature of the gas., is independent of the kind of gas, the temperature, or the pressure. To calculate R, we rearrange Eq. (2):

$R=\frac{PV}{nT}\text{ (3)}$

and substitute appropriate values of P, V , n, and T. From Table 9.1 we saw that the molar volumes of several gases at 0°C (273.15 K) and 1 atmAbbreviation for atmosphere, a unit of pressure equal to 101.325 kPa or 760 mmHg. (101.3 kPa) were close to 22.4 liters (22.4 dm3). Substituting into Eq. (3),

$R=\frac{\text{1 atm }\times \text{ 22}\text{.4 liters}}{\text{1 mol }\times \text{ 273}\text{.15 K}}=\text{0}\text{.0820}\frac{\text{liter atm}}{\text{mol K}}$

If we use SI unitsThe international system of units (Syst&egrave;me International d'Unit&eacute;) based on seven fundamental units: meter, kilogram, second, ampere, kelvin, candela, mole. for pressure and volume, as well as for amount of substance and temperature,

$R=\frac{\text{101}\text{.3 kPa }\times \text{ 22}\text{.4 dm}^{\text{3}}}{\text{1 mol }\times \text{ 273}\text{.15 K}}=\text{8}\text{.31}\frac{\text{kPa dm}^{\text{3}}}{\text{mol K}}=\text{8}\text{.31 J mol}^{-\text{1}}\text{ K}^{-\text{1}}$

Thus the gas constant has units of energyA system's capacity to do work. divided by amount of substance and thermodynamic temperature.

Equation (2) is usually rearranged by multiplying both sides by P, so that it reads

PV = nRT      (4)

This is called the ideal gas equationAn equation which gives a simple relationship among the pressure, volume, temperature, and chemical amount (moles) of a well behaved gas: PV, = nRT. or the ideal gasA hypothetical gas for which the relationship among the pressure, volume, temperature, and chemical amount (moles) can be described by simple proportionalities summarized by the ideal gas equation, PV = nRT. law. With the ideal gas equation we can convert from volume of a gas to amount of substance (provided that P and T are known). This is very useful since the volume, pressure, and temperature of a gas are easier to measure than massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object., and because knowledge of the molar massThe mass of a mole of substance; the same as molecular weight for molecular substances. is unnecessary.

EXAMPLE 1 Calculate the amount of gas in a 100-cm³ sample at a temperature of 300 K and a pressure of 750 mmHg.

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. Either of the two values of the gas constant may be used, so long as the units P, V, and T are adjusted properly. Using R = 0.0820 liter atm mol–1 K–1,

\begin{align} & V=\text{100 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 liter}}{\text{1000 cm}^{\text{3}}}\text{ }=\text{0}\text{.100 liter} \ & P=\text{750 mmHg }\times \text{ }\frac{\text{1 atm}}{\text{760 mmHg}}\text{ }=\text{0}\text{.987 atm} \ & T=\text{300 K} \ \end{align}

Rearranging Eq. (4) and substituting,

$n=\frac{PV}{RT}=\frac{\text{0}\text{.987 atm }\times \text{ 0}\text{.100 liter}}{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 300 K}}=\text{4}\text{.01 }\times \text{ 10}^{-\text{3}}\text{ mol}$

Essentially the same calculations are required when SI units are used:

\begin{align} & V=\text{100 cm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 dm}}{\text{10 cm}} \right)^{\text{3}}\text{ }=\text{0}\text{.100 dm}^{\text{3}} \ & P=\text{750 mmHg }\times \text{ }\frac{\text{101}\text{.3 kPa}}{\text{760 mmHg}}\text{ }=\text{100}\text{.0 kPa} \ & T=\text{300 K} \ & n=\frac{PV}{RT}=\frac{\text{100}\text{.0 kPa }\times \text{ 0}\text{.100 dm}^{\text{3}}}{\text{8}\text{.31 J mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 300K}}=\text{4}\text{.01 }\times \text{ 10}^{-\text{3}}\text{ mol} \ \end{align}

Note: Since 1 kPa dm3 = 1 J, the units cancel as shown. For this reason it is convenient to use cubic decimeters as the unit of volume when kilopascals are used as the unit of pressure in the ideal gas equation.

The ideal gas law enables us to find the amount of substance, provided we can measure V, P, and T. If we can also determine the mass of a gas, it is possible to calculate molar mass (and molecular weightThe mass of one mole of molecules of a substance; the molar mass of a molecular substance.). One way to do this is vaporize a volatile liquidA state of matter in which the atomic-scale particles remain close together but are able to change their positions so that the matter takes the shape of its container (one which has a low boilingThe process of a liquid becoming vapor in which bubbles of vapor form beneath the surface of the liquid; at the boiling temperature the vapor pressure of the liquid equals the pressure of the gas in contact with the liquid. temperature) so that it fills a flask of known volume. When the flask is cooled, the vapor condenses to a liquid and can easily be weighed.

EXAMPLE 2 The empirical formulaThe chemical formula of a substance written using the smallest possible integer subscripts that reflect the elemental composition. of benzene is CH. When heated to 100°C in a flask whose volume was 247.2 ml, a sample of benzene vaporized and drove all air from the flask. When the benzene was condensed to a liquid, its mass was found to be 0.616 g. The barometric pressure was 742 mmHg. Calculate (a) the molar mass and (b) the molecular formulaThe chemical formula of a substance written using the subscripts that reflect the number of each kind of atom present in a molecule of the substance. For example, the simplest formula for ethane is CH3, but the molecular formula is C2H6 because there are 2 C atoms and 6 H atoms in a molecule of ethane. of benzene.

Solution

a) Molar mass is mass divided by amount of substance. The latter quantity can be obtained from the volume, temperature, and pressure of benzene vapor:

\begin{align} & V=\text{247}\text{.2 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 liter}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }=\text{0}\text{.2472 liter} \ & T=(\text{273}\text{.15 + 100) K}=\text{373 K} \ & P=\text{742 mmHg }\times \text{ }\frac{\text{1}\text{.00 atm}}{\text{760 mmHg}}\text{ }=\text{0}\text{.976 atm} \ & n=\frac{PV}{RT}=\frac{\text{0}\text{.976 atm }\times \text{ 0}\text{.2472 liter}}{\text{0}\text{.0820 liter atm mol}^{-\text{1}}\text{ K}^{-\text{1}}\text{ }\times \text{ 373 K}}=\text{7}\text{.89 }\times \text{ 10}^{-\text{3}}\text{ mol} \ \end{align}

The molar mass is

$M=\frac{m}{n}=\frac{\text{0}\text{.616 g}}{\text{7}\text{.89 }\times \text{ 10}^{-\text{3}}\text{ mol}}=\text{78}\text{.1 g mol}^{-\text{1}}$

b) The empirical formula CH would imply a molar mass of (12.01 + 1.008) g mol–1, or 13.02 g mol–1. The experimentally determined molar mass is 6 times larger:

$\frac{\text{78}\text{.1 g mol}^{-\text{1}}}{\text{13}\text{.02 g mol}^{-\text{1}}}=\text{6}\text{.00}$

and so the molecular formula must be C6H6.