CoreChem:Dalton's Law of Partial Pressures
From ChemPRIME
The ideal gas law can also be rearranged to show that the pressure of a gas is proportional to the amount of gas:
Thus the factor RT/V may be used to interconvert amount of substance and pressure in a container of specified volume and temperature.
Equation (1) is also useful in dealing with the situation where two or more gases are confined in the same container (i.e., the same volume). Suppose, for example, that we had 0.010 mol of a gas in a 250-ml container at a temperature of 32°C. The pressure would be
Now suppose we filled the same container with 0.004 mol H_{2}(g) at the same temperature. The pressure would be
If we put 0.006 mol N_{2} in the container,
Now suppose we put both the 0.004 mol H_{2} and the 0.006 mol N_{2} into the same flask together. What would the pressure be? Since the ideal gas law does not depend on which gas we have but only on the amount of any gas, the pressure of the (0.004 + 0.006) mol, or 0.010 mol, would be exactly what we got in our first calculation. But this is just the sum of the pressure that H_{2} would exert if it occupied the container alone plus the pressure of N_{2} if it were the only gas present. That is,
P_{total} = p_{H2} + p_{N2}
We have just worked out an example of Dalton’s law of partial pressures (named for John Dalton, its discoverer). This law states that in a mixture of two or more gases, the total pressure is the sum of the partial pressures of all the components. The partial pressure of a gas is the pressure that gas would exert if it occupied the container by itself. Partial pressure is represented by a lowercase letter p.
TABLE 1 Vapor Pressure of Water as a Function of Temperature.
Temperature (in °C) | Vapor Pressure/mmHg | Vapor Pressure/kPa |
0 | 4.6 | 0.61 |
5 | 6.5 | 0.87 |
10 | 9.2 | 1.23 |
15 | 12.8 | 1.71 |
20 | 17.5 | 2.33 |
25 | 23.8 | 3.17 |
30 | 31.8 | 4.24 |
50 | 92.5 | 12.33 |
70 | 233.7 | 31.16 |
75 | 289.1 | 38.63 |
80 | 355.1 | 47.34 |
85 | 433.6 | 57.81 |
90 | 525.8 | 70.10 |
95 | 633.9 | 84.51 |
100 | 760.0 | 101.32 |
EXAMPLE 1 Assume 0.321 g zinc metal is allowed to react with excess hydrochloric acid (an aqueous solution of HCl gas) according to the equation
Zn(s) + 2HCl(aq) → ZnCl_{2}(aq) + H_{2}(g)
The resulting hydrogen gas is collected over water at 25°C, while the barometric pressure is 745.4 mmHg. What volume of wet hydrogen will be collected?
Solution From Table 1 we find that at 25°C the vapor pressure of water is 23.8 mmHg. Accordingly p_{H2} = p_{total}– p_{H2O} = 754 mmHg – 23.8 mmHg = 721.6 mmHg. This must be converted to units compatible R:
The road map for this problem is
Thus