THE IDEAL GAS LAW

12.7. THE IDEAL GAS LAW

All the gas laws described so far apply only to a given sample of gas. If a gas is produced during a chemical reaction or some of the gas under study escapes during processing, these gas laws are not used because the gas sample has only one temperature and one pressure. The ideal gas law works (at least approximately) for any sample of gas. Consider a given sample of gas, for which

PV

( a constant)

If we increase the number of moles of gas at constant pressure and temperature, the volume must also increase. Thus, we can conclude that the constant k can be regarded as a product of two constants, one of which represents the number of moles of gas. We then get

where P, V , and T have their usual meanings; n is the number of moles of gas molecules; and R is a new constant that is valid for any sample of any gas. This equation is known as the ideal gas law. You should remember the following value of R; values of R in other units will be introduced later.

R= 0.0821 L·atm/(mol·K) ( Note that there is a zero after the decimal point.) In the simplest ideal gas law problems, values for three of the four variables are given and you are asked

to calculate the value of the fourth. As usual with the gas laws, the temperature must be given as an absolute temperature, in kelvins. The units of P and V are most conveniently given in atmospheres and liters, respectively, because the units of R with the value given above are in terms of these units. If other units are given for pressure or volume, convert each to atmospheres or liters, respectively.

C and 1.09 atm? Ans.

EXAMPLE 12.11. How many moles of O 2 are present in a 0.500-L sample at 25 ◦

1.09 atm)(0.500 L)

[0.0821 L·atm/(mol·K)](298 K) 2 Note that T must be in kelvins.

n=

= 0.0223 mol O

RT

EXAMPLE 12.12. What is the volume of 1.00 mol of gas at STP?

n RT

1.00 mol)(0.0821 L·atm/mol·K)(273 K) =

The volume of 1.00 mol of gas at STP is called the molar volume of a gas. This value should be memorized.

C and 775 torr? Ans.

EXAMPLE 12.13. How many moles of SO 2 are present in a 765-mL sample at 37 ◦

Since R is defined in terms of liters and atmospheres, the pressure and volume are first converted to those units. Temperature as usual is given in kelvins. First, change the pressure to atmospheres and the volume to liters:

1.02 atm)(0.765 L)

[0.0821 L·atm/(mol·K)](310 K)

Alternately, we can do the entire calculation, including the conversions, with one equation:

PV

( 775 torr)(1.00 atm/760 torr)(765 mL)(1 L/1000 mL)

[0.0821 L·atm/(mol·K)](310 K)

EXAMPLE 12.14. At what temperature will 0.0750 mol of CO 2 occupy 2.75 L at 1.11 atm?

PV

1.11 atm)(2.75 L)

[0.0821 L·atm/(mol·K)](0.0750 mol)

EXAMPLE 12.15. What use was made of the information about the chemical identity of the gas in Examples 12.13 and 12.14?

Ans. None. The ideal gas law works no matter what gas is being used.

EXAMPLE 12.16. What volume will 7.00 g of Cl 2 occupy at STP?

Ans. The value of n is not given explicitly in the problem, but the mass is given, from which we can calculate the number of moles:

7.00 g Cl

2 2 = 71.0 g Cl 0.0986 mol Cl 2

n RT

= 0.0986 mol)[0.0821 L·atm/(mol·K)](273 K) =

The identity of the gas is important here to determine the number of moles.

EXAMPLE 12.17. (a) How many moles of O atoms were present in Example 12.11? (b) How many moles of He atoms are present in 465 mL at 25 ◦

C and 1.00 atm?

Ans. (a)

0.0223 mol O 2 = 0.0446 mol O atoms

1 mol O 2

Note that n in the ideal gas equation in the example refers to moles of O 2 molecules.

PV

1.00 atm)(0.465 L)

(b)

n=

= 0.0190 mol He

RT

[0.0821 L·atm/(mol·K)](298 K)

The He molecules are individual He atoms, and thus there are 0.0190 mol He atoms present. As soon as you recognize P, V, and T data in a problem, you can calculate n. (In a complicated problem, if you

are given P, V, and T, but you do not see how to do the whole problem, first calculate n and then see what you can do with it.)

EXAMPLE 12.18. If 4.58 g of a gas occupies 3.33 L at 27 ◦ C and 808 torr, what is the molar mass of the gas? Ans.

If you do not see at first how to solve this problem to completion, at least you can recognize that P, V, and T data are given. First calculate the number of moles of gas present:

1.06 atm)(3.33 L)

[0.0821 L·atm/(mol·K)](300 K)

We now know the mass of the gas and the number of moles. That is enough to calculate the molar mass: 4.58 g

32.0 g/mol

0.143 mol

EXAMPLE 12.19. What volume is occupied by the oxygen liberated by heating 0.250 g of KClO 3 until it completely decomposes to KCl and oxygen? The gas is collected at STP.

Ans. Although the temperature and pressure of the gas are given, the number of moles of gas is not. Can we get it somewhere? The chemical reaction of KClO 3 yields the oxygen, and the rules of stoichiometry (Chap. 10) may be used to calculate the number of moles of gas. Note that the number of moles of KClO 3 is not used in the ideal gas law equation.

2 KClO heat 3 −→

2 KCl + 3 O 2

3 mol O

0.250 g KClO

3 = 122 g KClO 0.00307 mol O 2

3 2 mol KClO 3

Now that we know the number of moles, the pressure, and the temperature of O 2 , we can calculate its volume:

n RT

= 0.00307 mol)[0.0821 L·atm/(mol·K)](273 K) = 0.0688 L

V=

1.00 atm