Page 192 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 12] GASES 181

Standard Conditions
According to the combined gas law, the volume of a given sample of gas can have any value, depending on its
temperature and pressure. To compare the quantities of gas present in two different samples, it is useful to adopt
a set of standard conditions of temperature and pressure. By universal agreement, the standard temperature is
chosen as 273 K (0 C), and the standard pressure is chosen as exactly 1 atm (760 torr). Together, these conditions
◦
are referred to as standard conditions or as standard temperature and pressure (STP). Many textbooks and
instructors find it convenient to use this short notation for this particular temperature and pressure.

EXAMPLE 12.10. A sample of gas occupies a volume of 1.88 L at 22 C and 0.979 atm pressure. What is the volume of
◦
this sample at STP?
Ans. P 1 = 0.979 atm P 2 = 1.00 atm
V 1 = 1.88 L V 2 = ?
T 1 = 22 C + 273 = 295 K T 2 = 273 K
◦
◦
P 1 V 1 P 2 V 2 (0.979 atm)(1.88 L) (1.00 atm)V 2
= = =
T 1 T 2 295 K 273 K
V 2 = 1.70 L

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
= k (a constant)
T
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
PV
= nR or PV = nRT
T
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.
EXAMPLE 12.11. How many moles of O 2 are present in a 0.500-L sample at 25 C and 1.09 atm?
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Ans. T = 25 C + 273 = 298 K
◦
PV (1.09 atm)(0.500 L)
n = = = 0.0223 mol O 2
RT [0.0821 L·atm/(mol·K)](298 K)
Note that T must be in kelvins.