SAT Chemistry Gases and the Gas Laws - Gas Laws and Related Problems Graham’s Law of Effusion (Diffusion)

SAT Chemistry Gases and the Gas Laws - Gas Laws and Related Problems Graham’s Law of Effusion (Diffusion)

This law relates the rate at which a gas diffuses (or effuses) to the type of molecule in the gas. It can be expressed as follows:
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(143-2)
Hydrogen, with the lowest molecular mass, can diffuse more rapidly than other gases under similar conditions.
Type Problem ________________________________________
Compare the rate of diffusion of hydrogen to that of oxygen under similar conditions.
The formula is
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(143-3)
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(144-1)

Therefore hydrogen diffuses four times as fast as oxygen.
In dealing with the gas laws, a student must know what is meant by standard conditions of temperature and pressure (abbreviated as STP). The standard pressure is defined as the height of mercury that can be held in an evacuated tube by 1 atmosphere of pressure (14.7 lb/in.2). This is usually expressed as 760 millimeters of Hg or 101.3 pascals. Standard temperature is defined as 273 Kelvin or absolute (which corresponds to 0° Celsius).
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(144-2)
Jacques Charles, a French chemist of the early nineteenth century, discovered that, when a gas under constant pressure is heated from 0°C to 1°C, it expands 1/273 of its volume. It contracts this amount when the temperature is dropped 1 degree to -1°C. Charles reasoned that, if a gas at 0°C was cooled to -273°C (actually found to be -273.15°C), its volume would be zero. Actually, all gases are converted into liquids before this temperature is reached. By using the Kelvin scale to rid the problem of negative numbers, we can state Charles’s Law as follows:
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(144-3)
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(144-4)

Graphic relationship - Charles’s Law. The dashed lines represent extrapolation of the data into regions where the gas would become liquid or solid. Extrapolation shows that each gas, if it remained gaseous, would reach zero volume at 0 K or -273°C.
Type Problem _____________________________________

The volume of a gas at 20°C is 500. mL. Find its volume at standard temperature if pressure held constant
Convert temperatures:
20°C = 20° + 273 = 293 K
0°C = 0° + 273 = 273 K
If you know that cooling a gas decreases its volume, then you know that 500. mL will have to be multiplied by a fraction (made up of the Kelvin temperatures) that has a smaller numer­ator than the denominator. So
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(145-1)
Another Example _____________________________________
A sample of gas occupies 24 L at 175.0 K. What volume would the gas occupy at 400.0 K?
The temperature of the gas is increased. Charles’s Law predicts that the gas volume will also increase. So
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(145-2)
The final volume has increased as predicted.

Boyle’s Law (PV = k)
Robert Boyle, a seventeenth century English scientist, found that the volume of a gas decreases when the pressure on it is increased, and vice versa, when the temperature is held constant. Boyle’s Law can be stated as follows:
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(145-3)
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(146-1)

Type Problem _______________________________________
 Given the volume of a gas as 200. mL at 1.05 atm pressure, calculate the volume of the same gas at 1.01 atm. Temperature is held constant.
If you know that this decrease in pressure will cause an increase in the volume, then you know 200. mL must be multiplied by a fraction (made up of the two pressures) that has a larger numerator than the denominator. So
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(146-2)

Another Example ____________________________________
The gas in a balloon has a volume of 7.5 L at 100. kPa. The balloon is released into the atmo­sphere, and the gas in it expands to a volume of 11. L. Assuming a constant temperature, what is the pressure on the balloon at the new volume?
The volume of the gas has increased. Boyle’s Law predicts that the gas pressure will decrease. So
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(146-3)
The final pressure has decreased as predicted.

Combined Gas Law
This is a combination of the two preceding gas laws. The formula is as follows:
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(147-1)
Type Problem _____________________________________________

The volume of a gas at 780. mm pressure and 30.°C is 500. mL What volume would the gas occupy at STP?
You again can use reasoning to determine the kind of fractions the temperatures and pressures must be to arrive at your answer. Since the pressure is going from 780. mm to 760. mm, the volume should increase. The fraction must then be780/760 Also, since the temperature is going from 300C (303 K) to OOC (273 K), the volume should decrease; this fraction must be
273/303. So
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(147-2)
Another Example ____________________________________

The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm?
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(147-3)
Pressure Versus Temperature (Gay-Lussac’s Law)
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(148-1)
Type Problem _________________________________________

A steel tank contains a gas at 27°C and a pressure of 12.0 atms. Determine the gas pressure when the tank is heated to 100.°C.
Reasoning that an increase in temperature will cause an increase in pressure at constant volume, you know the pressure must be multiplied by a fraction that has a larger numerator than denominator. The fraction must be 373 K/300. K So
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(148-2)
Another Example _________________________________________

At 120.°C, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 205°C, assuming constant volume?
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(149-1)

Dalton’s Law of Partial Pressures
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(149-2)

Type Problem _________________________________________
A mixture of gases at 760. mm Hg pressure contains 65.0% nitrogen, 15.0% oxygen, and 20.0% carbon dioxide by volume. What is the partial pressure of each gas?
0. 650 x 760. = 494 mm pressure (N2)
0.150 x 760. = 114 mm pressure (O2)
0.200 x 760. = 152 mm pressure (CO2)
If the pressure was given as 1.0 atm, you would substitute 1.0 atm for 760. mm Hg. The answers would be:
. 0.650 x 1.0 atm = 0.650 atm (N2)
0.150 x 1.0 atm = 0.150 atm (O2)
0.200 x 1.0 atm = 0.200 atm (CO2)

Corrections of Pressure
CORRECTION OF PRESSURE WHEN A GAS IS COLLECTED OVER WATER. When a gas is collected over a volatile liquid, such as water, some of the water vapor is present in the gas and con¬tributes to the total pressure. Assuming that the gas is saturated with water vapor at the given temperature, you can find the partial pressure due to the water vapor in a table of such water vapor values. This vapor pressure, which depends only on the temperature, must be subtracted from the total pressure to find the partial pressure of the gas being measured.

CORRECTION OF DIFFERENCE IN THE HEIGHT OF THE FLUID. When gases are collected in eudi­ometers (glass tubes closed at one end), it is not always possible to get the level of the liquid inside the tube to equal the level on the outside. This deviation of levels must be taken into account when determining the pressure of the enclosed gas. There are then two possibilities:
(1) When the level inside is higher than the level outside the tube, the pressure on the inside is less, by the height of fluid in excess, than the outside pressure. If the fluid is mercury, you simply subtract the difference from the outside pressure reading (also in height of mercury and in the same units) to get the corrected pressure of the gas. If the fluid is water, you must first convert the difference to an equivalent height of mercury by dividing the difference by 13.6 (since mercury is 13.6 times as heavy as water, the height expressed in terms of Hg will be 1 /13.6 the height of water). This is shown pictorially in Figure 21. Again, care must be taken that this equivalent height of mercury is in the same units as the expression for the outside pressure before it is subtracted to obtain the corrected pressure for the gas in the eudiometer.
(2) When the level inside is lower them the level outside the tube, a correction must be added to the outside pressure. If the difference in height between the inside and the outside is expressed in terms of water, you must take 1/13.6 of this quantity to correct it to millimeters of mercury. This quantity is then added to the expression of the outside pressure, which must also be in millimeters of mercury. If the tube contains mercury, then the difference between the inside and outside levels is merely added to the outside pressure to get the corrected pressure for the enclosed gas.
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(150-1)

Type Problem _____________________________________
Hydrogen gas was collected in a eudiometer tube over water. It was impossible to level the outside water with that in the tube, so the water level inside the tube was 40.8 mm higher than that outside. The barometric pressure was 730. mm of Hg. The water vapor pressure at the room temperature of 29°C was found in a handbook to be 30.0 mm. What is the pressure of the dry hydrogen?

STEP 1 To find the true pressure of the gas, we must first subtract the water-level difference expressed in mm of Hg:
40.8/13.6 = 3.00 mm of Hg
Then 730. mm - 3.00 mm = 727 mm total pressure of gases in the eudiometer

STEP 2 Correcting for the partial pressure due to water vapor in the hydrogen, we subtract the vapor pressure (30.0 mm) from 727 mm and get our answer: 697 mm.

Ideal Gas Law
The preceding laws do not include the relationship of number of moles of a gas to the pres-sure, volume, and temperature of the gas. A law derived from the Kinetic-Molecular Theory relates these variables. It is called the Ideal Gas Law and is expressed as
PV = nRT
P, V, and T retain their usual meanings, but n stands for the number of moles of the gas and R represents the ideal gas constant.
Boyle’s Law and Charles’s Law are actually derived from the Ideal Gas Law. Boyle’s Law applies when the number of moles and the temperature of the gas are constant. Then in PV = nRT, the number of moles, n, is constant; the gas constant (R) remains the same; and by definition Tis constant. Therefore, PV= k. At the initial set of conditions of a problem, P1V1 = a. constant (fc). At the second set of conditions, the terms on the right side of the equa¬tion are equal to the same constant, so P1V1 = P2V2 This matches the Boyle’s Law equation introduced earlier.
The same can also be done with Charles’s Law, because PV = nRT can be expressed with the variables on the left and the constants on the right:
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(151-1)
In Charles’s Law the number of moles and the pressure are constant. Substituting k for the constant term, nR/P, we have
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(151-2)
To use the Ideal Gas Law in the form PV = nRT, the gas constant, R, must be determined. This can be done mathematically as shown in the following example.
One mole of oxygen gas was collected in the laboratory at a temperature of 24.0°C and a pressure of exactly 1 atmosphere. The volume was 24.38 liters. Find the value of R.
PV = nRT
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(151-3)
Once R is known, the Ideal Gas Law can be used to find any of the variables, given the other three.
For example, calculate the pressure, at 16.0°C, of 1.00 gram of hydrogen gas occupying 2.54 liters.
Rearranging the equation to solve for P, we get
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(152-1)
The molar mass of hydrogen is 2.00 g/mol, so the number of moles in this problem would be
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(152-2)
Calculating the value, we get
P = 4.66 atm
Another use of the ideal gas law is to find the number of moles of a gas when P, T, and V are known.
For example, how many moles of nitrogen gas are in 0.38 liter of gas at 0°C and 0.50 atm pressure?
Rearranging the equation to solve for n gives
sat-chemistry-gases-and-the-gas-laws-gas-laws-and-related-problems-grahams-law-of-effusion-diffusion-(152-3)
Ideal Gas Deviations
In the use of the gas laws, we have assumed that the gases involved were “ideal” gases. This means that the molecules of the gas were not taking up space in the gas volume and that no intermolecular forces of attraction were serving to pull the molecules closer together. You will find that a gas behaves like an ideal gas at low pressures and high temperatures, which move the molecules as far as possible from conditions that would cause condensation. In general,
pressures below a few atmospheres will cause most gases to exhibit sufficiently ideal properties for the application of the gas laws with a reliability of a few percent or better.
If, however, high pressures are used, the molecules will be forced into closer proximity with
each other as the volume decreases until the attractive force between molecules becomes a
factor. This factor decreases the volume, and therefore the PV values at high pressure condi-
tions will be less than those predicted by the Ideal Gas Law, where PV remains a constant.
Examining what occurs at very low temperatures creates a similar situation. Again, the
molecules, because they have slowed down at low temperatures, come into closer proximity with each other and begin to feel the attractive force between them. This tends to make the
gas volume smaller and, therefore, causes the PV to be lower than that expected in the ideal
gas situation. Thus, under conditions of very high pressures and low temperatures, deviations from the expected results of the Ideal Gas Law will occur.

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