SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 5

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:

The Rate of Effusion of a Gas Is Inversely Proportional to the Square Root of its Molecular Mass.

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

        

Let A be H2 and B be O2.

        

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).

Charles’s Law 

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:

If the Pressure Remains Constant, the Volume of a Gas Varies Directly as the Absolute Temperature. Then

Initial  = Final  at constant pressure or 

Plots of versus for representative gases.

TIP 

Charles’s Law  at constant pressure is a direct proportion.

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 is held constant.

Convert temperatures:

     20°C = 20° + 273 = 293 K
      0°C = 0° + 273 = 273 K

TIP 

Assume dry gases unless otherwise stated.

 

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

        

Or you can use the formula and substitute known values:

        

TIP 

STP = standard temperature of 273K standard pressure of 760 mm Hg or 1 atmosphere (atm) or 105 pascals.

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

         

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:

If the Temperature Remains Constant, the Volume of a Gas Varies Inversely as the Pressure Changes. Then

P1V1 = P2V2 at a constant temperature
or
PV = k

TIP 

Boyle’s Law

P1V1 = P2V2 temperature is held constant

Graphic relationship—Boyle’s Law

TIP 

Volume vs. pressure for a gas at constant temperature. This is an inverse proportion. As the pressure increases by 2, the volume drops by 1/2.

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

Or you can use the formula:

       

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

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:

TIP 

In the combined gas law, all subscripts are 1s on the left and all subscripts are 2s on the right.

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 be  Also, since the temperature is going from 30°C (303 K) to 0°C (273 K), the volume should decrease; this fraction must be  So

Or you can use the formula:

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?

Using the combined gas law, 

and solving for 



V2 = 26.3 mL

Pressure Versus Temperature (Gay-Lussac’s Law)

At Constant Volume, the Pressure of a Given Mass of Gas Varies Directly with the Absolute Temperature. Then

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 . So

Or you can use the formula:

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?

Using the relationship of: 

and solving for 



     P2 = 1.30 atm

Dalton’s Law of Partial Pressures

When a Gas Is Made Up of a Mixture of Different Gases, the Total Pressure of the Mixture Is Equal to the Sum of the Partial Pressures of the Components; That Is, the Partial Pressure of the Gas Would Be the Pressure of the Individual Gas If It Alone Occupied the Volume. The Formula Is:

Ptotal = Pgas 1 + Pgas 2 + Pgas 3 + …

TIP 

Know Dalton’s Law of Partial Pressures.

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 × 760. = 494 mm pressure (N2)
0.150 × 760. = 114 mm pressure (O2)
0.200 × 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 × 1.0 atm = 0.650 atm (N2)
0.150 × 1.0 atm = 0.150 atm (O2)
0.200 × 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 contributes 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.

TIP 

When a gas is collected over water, subtract the water vapor pressure at the given temperature from the atmospheric pressure to find the partial pressure of the gas. Pgas = Patm - PH2O

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 than 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.

Figure 21. Same Pressure Exerted on Both Liquids

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:

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 pressure, 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

PV, and retain their usual meanings, but stands for the number of moles of the gas and represents the ideal gas constant.

TIP 

Know how to use the ideal gas law, PV nRT

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 is constant. Therefore, PV k. At the initial set of conditions of a problem, P1V 1 = a constant (k). At the second set of conditions, the terms on the right side of the equation 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:

In Charles’s Law the number of moles and the pressure are constant. Substituting for the constant term, , we have

The expression relating two sets of conditions can be written as

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.

TIP 

Know the ideal gas constant

PV nRT 

Rearranging the equation to solve for gives

Substituting the known values on the right, we have

Calculating R, we get

Once 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

The molar mass of hydrogen is 2.00 g/mol, so the number of moles in this problem would be

Substituting the known values, we have

TIP 

Remember to use appropriate units: moles (mol) liters (L) atmosphere (atm)

Calculating the value, we get

= 4.66 atm

Another use of the ideal gas law is to find the number of moles of a gas when PT, and 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 gives

Changing temperature to kelvins and pressure to atmospheres gives

= 0° + 273 = 273 K

Substituting in the equation, we have

= 0.0085 mol of nitrogen gas

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.

TIP 

Least deviations occur at low pressures and high temperatures.
High deviations occur at high pressures and very low temperatures.

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 conditions 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.