MathConnections
CH 3 CHCH 2 CH3CH2CH2CH2
5.3 History and Application of the Gas Law
pounds per square inch. None of these should cause you particular diffi culty as long as you are careful that any calculations you do properly account for the units used.
Now, we might ask how gas pressure plays a role in air pollution. Gases natu- rally flow from higher pressure to lower pressure, and pollutants are often pro- duced in a high-pressure environment, such as in the cylinder of an auto engine or a reaction vessel in an industrial processing plant. When gases are released from these high-pressure environments, they disperse rapidly into the surrounding air. The pres- sure difference helps speed the spread of the pollutants.
do not change. (R is a universal constant, so it never changes!) Then rearrange the gas law equation to collect all of those terms onto the same side, leaving those that do change on the other side. If pressure and volume change while the number of moles and the temperature remain constant, for example, we’d get
PV = nRT = constant
Because the product nRT remains constant, we know that PV must also be con- stant. This tells us that the product of P times V must be the same before and after the change, so we can write
P1V1= P2V2 (5.3)
Here the subscripts ‘1’ and ‘2’ indicate the conditions before and after the change takes place. Now if we know any three of those four quantities, we can solve for the missing one, as seen in the following example.
E X A M P L E P RO B L E M 5 .1
In addition to ongoing sources, such as car engines, pollutants can also be introduced into the air through isolated incidents like the rupture of a gas storage tank. Most gases are stored and transported at high pressures. A common laboratory cylinder of methane, for example, has a volume of 49.0 L and is fi lled to a pressure of 154 atm.
Suppose that all of the CH4 from this cylinder is released and expands until its pressure falls to 1.00 atm. What volume would the CH4 occupy?
Strategy The problem involves a change in the conditions, here pressure and vol- ume, of a sample of gas. We will want to use the ideal gas law because it is our model that relates the various properties of a gas to one another. So we begin by assuming that the gas will obey the ideal gas law under both the initial and fi nal conditions. We will further assume that temperature is constant because we are not given any data to suggest otherwise. Because we are working with the same sample of gas throughout, we also know that the number of moles (n) is constant. Collect the constant terms together and work toward a solution.
Solution In this case, n and T are constant, and R is always constant. So, PV = nRT = constant
Therefore,
P1V1= P2V2 Solving this for the fi nal volume (V2) gives
V2= P1V1
— P2
We know all three terms on the right: P1= 154 atm, V1= 49.0 L, and P2= 1.00 atm.
Insert those above and solve:
V2= (154 atm)(49.0 L)
— (1.00 atm) = 7550 L
Analyze Your Answer Our result is in liters, which is an appropriate unit for volume.
The actual value is much larger than the initial volume, so we might ask if this seems sen- sible. Looking at the initial and fi nal conditions, we have lowered the pressure by a large factor, and this should lead to a much larger volume. So our answer seems plausible.
Discussion Because of the high initial pressure, the assumption that the gas behaves ideally may not be very accurate. We’ll examine that assumption later in Example 5.8.
In Section 5.6, we will look at gases that do not behave ideally.
In Section 5.6, we will look at gases that do not behave ideally.
Check Your Understanding A small sample of gas is generated in a laboratory.
In a 125-mL vessel, it exerts 115 torr of pressure. Then a valve is opened and the gas expands until its volume is 175 mL. If the temperature is constant throughout this process, what is the new pressure of the gas?
In other situations, we might fi nd that V and T vary, whereas n and P are constant.
That would lead us to write the following relationship — V
T = — nR
P = constant This in turn leads to Equation 5.4:
V1
— T1
= V2
— T2
(5.4)
E X A M P L E P RO B L E M 5 . 2
A balloon is fi lled with helium, and its volume is 2.2 L at 298 K. The balloon is then dunked into a thermos bottle containing liquid nitrogen. When the helium in the balloon has cooled to the temperature of the liquid nitrogen (77 K), what will the volume of the balloon be?
Strategy We should start by recognizing this as a change in conditions for a gas and realize that we will want to use the ideal gas law. So we assume that the gas will behave ideally under both the initial and fi nal conditions. We will further assume that the pres- sure is constant since we are not given any data to suggest otherwise. And since the bal- loon is tied off throughout, we can also assume that n is constant. Collect the constant terms together and work toward a solution.
Solution As in the discussion above, we have — V
T = — nR
P = constant So
V1
— T1
= V2
— T2
Solve this for the fi nal volume (V2):
V2= V1T2
— T1
Because the three terms on the right are known, we can just insert values and solve:
V2= (2.2 L)(77 K)
— (298 K) = 0.57 L
Analyze Your Answer We have decreased the temperature from 298 K to 77 K, which means that it is smaller by a factor of nearly 4. This should lead to a volume that is also smaller by a factor of nearly 4. So our answer seems likely to be correct.
Check Your Understanding The balloon in the example above will burst if its volume exceeds 2.3 L. At what temperature would you expect the balloon to burst?
For any change in the conditions of a gas sample, we should always be able to em- ploy this same idea by separating those parameters that are constant from those that
are changing. Because you are very unlikely to forget the ideal gas law, this approach is almost certainly more reliable than attempting to remember a list of equations cov- ering all the possible combinations of variables.
Units and the Ideal Gas Law
Suppose that we had expressed the temperature in °C rather than in kelvins in the above example? It should take you only a little work to convince yourself that the equation we used will not hold true. Changing the temperature of a gas from 10 to 20°C, for example, will not double its volume. The ideal gas law itself is only valid when we use an absolute temperature scale where T = 0 corresponds to absolute zero temperature. To see why this must be so, notice that putting T = 0 into the gas law implies that either P or V or both must also go to zero, meaning that there must be no gas present. This simply can’t be correct for T = 0°C or T = 0°F; that’s obvious because the winter temperature routinely reaches these levels in many areas, and yet the air in the atmosphere obviously does not vanish. Absolute zero (0 K), on the other hand, is a very special temperature, and it is rea- sonable that there might be peculiar consequences if we actually cooled a gas to this level.
In contrast to the temperature, which must be expressed on an absolute scale, we can use any convenient units for pressure and volume, so long as we then reconcile the units on those quantities with those for the gas constant, R. Why this difference?
The common temperature scales are fundamentally different from other kinds of units because converting among them involves not only multiplying by a conversion factor but also adding or subtracting. This, in turn, results directly from the fact that differ- ent temperature scales (K, °C, or °F ) have different zero points. In all the other quan- tities that we work with, the meaning of “zero” is the same no matter what units we use. If you say the mass of a sample is zero, for example, it doesn’t really matter much what units you measured in—there’s still nothing there. But for temperature, the dif- ferences among 0°C, 0°F, and 0 K are obvious. In some engineering fi elds, you may encounter the Rankine temperature scale (°R). This is also an absolute scale, meaning that 0°R = 0 K. But one degree on the Rankine scale is the same size as 1°F. Because they are both absolute scales, Rankine and Kelvin temperatures can be interconverted by using a multiplicative unit factor: 1°R = 1.8 K.
If you look at a table like the one in Appendix B of this book, you will fi nd values for R in several combinations of units.
R = 0.08206 L atm mol−1 K−1
= 8.314 J mol−1 K−1
= 62.37 L torr mol−1 K−1
Students often ask how to decide which of these values is “the right R” to use in a given situation. The answer is that they are all really the same quantity, just expressed in different units. (If you are curious, you can do unit conversions to verify this for yourself.) If pressure is given in atm and volume in L, then it will be convenient to use R as 0.08206 L atm mol−1 K−1, for example. But if pressures are given in torr, it might be easier to insert 62.37 L torr mol−1 K−1 for R. Or, if you always prefer to use the same value of R, you could simply convert pressures from any other units into atm before doing any calculations. It doesn’t matter which set of units we choose, as long as we are careful to reconcile the units in all of our calculations.
E X A M P L E P RO B L E M 5 . 3
A sample of CO2 gas has a volume of 575 cm3 at 752 torr and 72°F. What is the mass of carbon dioxide in this sample?
Strategy We are trying to fi nd the mass of a sample of gas. If we knew how many moles of CO were in the sample, we could easily use the molar mass to fi nd the mass
Although absolute zero is a fairly abstract concept, temperatures well below 1 K can be achieved routinely in the laboratory.
Although absolute zero is a fairly abstract concept, temperatures well below 1 K can be achieved routinely in the laboratory.
No degree sign is used when writing Kelvin temperatures.
No degree sign is used when writing Kelvin temperatures.
of the sample. And since the problem involves a gas, we can attempt to use the ideal gas law to fi nd the number of moles. We are given values for V, P, and T, so we will be able to solve for n. Some of the units given are unusual, so we will need to be careful to handle them properly.
Solution We can begin by converting some of the given information into units more commonly used for gases.
T = (72 − 32) × 5/9 = 22°C = 295 K V = 575 cm3= 575 mL = 0.575 L
For P, we will use the value in torr and choose the corresponding value for R. We could also convert from torr to atm, of course.
n = — PV
RT = (752 torr)(0.575 L)
—— (62.37 L torr mol−1 K−1)(295 K) = 0.0235 mol The gas in our sample is CO2, with a molar mass of 44.0 g/mol, so
0.0235 mol × 44.0 g
— mol = 1.03 g CO2
Analyze Your Answer The temperature and pressure are near typical room con- ditions, and the volume is a little more than a half liter. So this should correspond to a small balloon fi lled with CO2. One gram is a pretty small mass but seems reasonable for the gas in a small balloon.
Check Your Understanding Find the volume in ft3 of 1.0 mole of an ideal gas at −25°C and 710 torr.