14.4. Gas Mixtures and Molecular Speeds http://www.ck12.org
TABLE14.4:(continued)
Temperature (°C) Vapor Pressure (mmHg) Temperature (°C) Vapor Pressure (mmHg)
15 12.79 55 118.04
20 17.54 60 149.38
25 23.76 65 187.54
30 31.82 70 233.7
35 42.18Sample Problem 14.9: Gas Collected by Water Displacement
A certain experiment generates 2.58 L of hydrogen gas, which is collected over water. The temperature is 20°C and
the atmospheric pressure is 98.60 kPa. Find the volume that the dry hydrogen would occupy at STP.
Step 1: List the known quantities and plan the problem.
Known
- VTotal= 2.58 L
- T = 20°C = 293 K
- PTotal= 98.60 kPa = 739.7 mmHg
Unknown
- VH 2 at STP =? L
The atmospheric pressure is converted from kPa to mmHg in order to match units with the table above (Table14.4).
The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of
the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined
gas law.
Step 2: Solve.
PH 2 =PTotal−PH 2 O= 739 .7 mmHg− 17 .54 mmHg= 722 .2 mmHgNow the combined gas law is used, solving for V 2 , the volume of hydrogen at STP.
V 2 =
P 1 ×V 1 ×T 2
P 2 ×T 1
=
722 .2 mmHg× 2 .58 L×273 K
760 mmHg×293 K= 2 .28 L H 2
Step 3: Think about your result.
If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be
2.28 L. This is less than the actual collected volume because some of that is water vapor. The conversion to its dry
volume at STP is useful for stoichiometry purposes.
Practice Problem- 845 mL of oxygen gas is collected over water at 25°C. The atmospheric pressure is 754.3 mmHg. Calculate
the volume of the dry oxygen gas at STP.