This chapter is from the book
This chapter is from the book
This chapter is from the book
4.4 The Ideal Gas Law
We have just seen that the volume of a specified amount of a gas at constant pressure is proportional to the absolute temperature. In addition, we saw that the volume of a specified amount gas at a constant temperature is also inversely proportional to its pressure. We can correctly assume that pressure of a specified amount of gas at a constant volume is proportional to its absolute temperature. Let us also add the fact that the volume at constant pressure and temperature is also proportional to the amount of gas. Similarly, the pressure at constant volume and temperature is proportional to the amount of gas. Thus, these laws and relationships can be combined to give Equation 4.10.
Equation 4.10
Here, n is the number of moles of gas: Again, an absolute temperature must be used along with an absolute pressure.
Scientists and engineers have defined an ideal gas to be a gas with properties affected only by pressure and temperature. Thus, Equation 4.10 only needs a magical constant so that any one of its variables can be calculated if the other three are known. That constant is the ideal gas constant R and is used to form the Ideal Gas Law given by Equation 4.11.
Equation 4.11
Depending on the units of measure for the pressure, the volume, the number of moles, and the absolute temperature, some values for the ideal gas constant R are given in Table 4.1 and Appendix C for different units-of-measure systems.
Table 4.1 Values for R, the Ideal Gas Constant
System |
Value with Units of Measure |
SI |
8.314 (kPa)(m3)/(kg mol)(°K) |
cgs |
0.08206 (L)(atm)/(g mol)(°K) |
English |
0.7302 (ft3)(atm)/(lb mol)(°R) |
English |
10.73 (ft3)(psia)/lb-mol)(°R) |
Example 4.1 Volume of an Ideal Gas
Calculate the volume a one-pound mole (1.00 lbm mol) of an ideal gas occupies at the standard condition of 32°F and 1.00 atmosphere of pressure.
Solution
The information given in the statement of the problem simplifies this problem. We do not need to convert the gas’s mass to moles. However, we must convert the temperature in degrees Fahrenheit to the absolute temperature of degrees Rankine. That conversion is
T = 32° + 460° = 492°R
Rewriting the Ideal Gas Law given by Equation 4.11 to calculate the volume gives
Checking Table 4.1, we see that there is an ideal gas constant R for units of cubic feet, atmospheres, pound moles, and degrees Rankine. Substituting the values for the number of moles, the appropriate ideal gas constant, the absolute temperature, and the pressure gives
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The volume determined in the previous example is important and is usually memorized by engineers. It is the unit volume for 1 lb mol of gas at 32°F and 1 atm or 14.7 psia, which is usually expressed as 359 ft3/lb mol. Besides the standard volume of 1 lb mol at 32°F and 1 atmosphere, another standard unit volume primarily used by the natural gas industry is defined at 59°F and 1 atmosphere with the value of 379 ft3/lb mol.
Example 4.2 Volume of 11.0 Pounds of Methane at 80°F and 200 psig
Calculate the volume of 11.0 lbs of methane gas at 200 psig pressure and 80°F.
Solution
There are several steps involved in this calculation. First, we must determine the question, which is to calculate the volume of a quantity of gas at a given temperature and pressure. In a second step, after establishing a basis, we must convert the mass of methane that will be the basis into pound moles. Third, we must convert temperature in degrees Fahrenheit into absolute degrees Rankin and, fourth, convert pressure from psig into psia. Fifth, we must select the appropriate ideal gas constant and use it with a rewritten form of Equation 4.11 to determine the volume of 11.0 lbs of methane gas. Finally, we can substitute the values previously determined into the rewritten equation to calculate the volume.
Basis: 11.0 lb of methane gas
As outlined above, we must calculate the number of pound moles in 11.0 lbs of methane. The molecular weight of methane is about 16.04, which is composed of one atom of carbon with an atomic mass of 12.00 amu and 4 atoms of hydrogen with atomic masses of 1.01 amu:
Now, we must convert the temperature to the absolute scale of degrees Rankine:
T = 80° + 460° = 540°R
Just as we must use the absolute temperature, we must also use the absolute pressure. We were given the pressure in psig, which is pounds per square in gauge. Gauge pressure is taken relative to the pressure of the atmosphere, which would be 14.7 psia at sea level, and it will have lower pressures at higher elevations. The correction from gauge to absolute pressure at sea level under the standard condition of 1 atmosphere is:
p = 200 psig + 14.7 psi = 214.7 psia
The rewritten form of Equation 4.11 from Example 4.1 for volume is
Substituting the values for the number of moles, the appropriate ideal gas constant, the absolute temperature, and the absolute pressure gives
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Checking for dimensional consistency, we see that lb mol cancels with lb mol, atm cancels with atm, and °R cancels with °R, leaving only ft3, which is the appropriate unit for our answer of 18.5 cubic feet.
This result is expressed in only 3 digits, although it can be calculated to more. Why?
Hint: What was the number of significant figures in the mass of methane?
There are several ways that Equation 4.11 can be used. The most obvious is to “plug” all of the numbers into the equation each time there is a change and “turn the crank” to calculate a new volume or pressure, or whatever is unknown. However, most of the time not all of the variables are known for this method or it involves a lot of unnecessary arithmetic. In addition, it also involves looking up and confirming one of the gas constants previously described even if they have been memorized.
We could also go back to Boyle’s Law and Charles’s Law if we were changing only one variable, but due to thermodynamic effects we will discuss in the next chapter, the temperature of a system will usually change whenever the pressure changes.
So, what is the easiest way? Simply use Equation 4.11 to make a ratio of the variables before and after; this gives us Equation 4.12.
Equation 4.12
The subscripts 1 and 2 indicate the before and after states, respectively. Notice that the ideal gas constant R has been canceled because its ratio is unity. We can rewrite Equation 4.12 to calculate a new pressure when a given quantity of gas is compressed and it becomes Equation 4.13.
Equation 4.13
Alternatively, we can rewrite it to calculate the effect on the volume when a gas is compressed to a new pressure and temperature, as shown in Equation 4.14.
Equation 4.14
We could arrange Equations 4.13 and 4.14 to calculate the ratio of the pressures or volumes after and before. That would give us Equation 4.15.
Equation 4.15
We could also rewrite Equations 4.13 and 4.15 to give us Equation 4.16.
Equation 4.16
We can arrange Equation 4.12 as needed to calculate a final pressure, volume, or temperature given the before-and-after states of all of the other variables. Assuming that we understand the Ideal Gas Law and the “PVT” relationship between pressure, volume, and temperature, it is a lot easier to remember just one equation and rearrange it as necessary.
Example 4.3 Effect of Temperature on Pressure of a Batch Reactor
Calculate the expected pressure of a 1.748 ft3 batch reactor charged at 135 psig and heated from 95°F to 1,650°F.
Solution
Before plugging in a bunch of numbers, look at the problem. Here, we have an overstated problem. This is a sealed batch reactor. It may seem that we need to calculate the amount of gas charged to the reactor, but that is not the case. On the other hand, we may be in a laboratory where the reactor is weighed before and after it is charged. In that case, we could calculate the starting pressure of the gas. However, most times that number is measured and recorded by the experimenter.
Basis: Initial pressure of 135 psig
First, we convert pressure from psig to psia:
p1 = 135 psig + 14.7 psi = 149.7 psia
Next, we convert temperatures from degrees Fahrenheit into degrees Rankin:
T1 = 95°F + 460° = 555°R
and we obtain
T2 = 1650°F + 460° = 2110°R
Finally, we rewrite Equation 4.12 with V2 = V1 and n2 = n1 to give
Substituting values for p1, T1, and T2 with their units of measure gives
In this example, because there were only two units of measure, we did not use the dimensional equation form for the preceding calculation.
In addition to calculating absolute values for pressure, volume, and temperature, as stated above we can calculate ratios that may be more useful. This is demonstrated in Example 4.4.
Example 4.4 Change in Relative Volume
Calculate the relative change in volume when a gas at 165 psig and 95°F is compressed to 855 psig and then heated to 1,250°F.
Solution
Basis: Initial pressure and temperature of 165 psig and 95°F, respectively.
First, we convert pressures from psig to psia:
p1 = 165 psig + 14.7 psi = 179.7 psia
and
p2 = 855 psig + 14.7 psi = 869.7 psia
Next, we convert temperatures from °F to °R:
T1 = 95°F + 460° = 555°R
and
T2 = 1250°F + 460° = 1710°R
Finally, we rewrite Equation 4.12 with n2 = n1 canceling each other to give
Substituting values and dimensions yields
Note that all of the dimensions cancel out only leaving the ratio, which by definition is dimensionless.
Not only are relative changes affected by temperature and pressure, but they are also affected by a change in the number of moles of gas when a chemical reaction occurs. Recently, the construction of a number of ethylene crackers has been announced in the Gulf Coast area due to the abundance of natural gas that contains ethane. Those crackers use steam with the ethane in a complex set of chemical reactions that reduce the amount of solid carbon formed, consequently reducing the cracker’s “coking” caused by the deposition of elemental carbon. In those crackers, 1 mole of ethane will produce 1 mole of ethylene and 1 mole of hydrogen according to the chemistry of:
C2H6 → C2H4 + H2
An ethylene cracker is essentially a high-temperature furnace in which ethane flows through the piping, reaching temperatures on the order of 1,500°F in a matter of a few milliseconds. At that temperature, ethane starts to “pyrolyze” and form ethylene with a double bond as two hydrogen atoms are literally broken off. After exiting the furnace, it is quenched almost immediately lest it continue to break down, eventually becoming elemental carbon and molecular hydrogen.
Example 4.5 Effect on Velocity during Ethane Cracking
Approximate the effect on the velocity of ethane as it is cracked in a furnace to form ethylene when it is under 30 psig while flowing at 950 scf/min through a nominal 4-inch schedule-80 pipe entering the furnace at 350°F and exiting at 1,550°F. The ratio of steam to ethane is 1:10.
Solution
Basis: |
Ethane at 350°F |
Superfluous information: |
Initial pressure of 30 psig Flow rate of 950 scf/min Diameter of nominal 4-inch schedule-80 pipe |
For all practical purposes, the pressure drop in the pipe will be small as it passes through the furnace and we can set this as
p2 ≅ p1
The velocity of the gas is equal to the volumetric flow rate divided by the cross-sectional area of the pipe, which remains constant in the furnace. Thus, the effect on the velocity is essentially equivalent to the increase in the volume of the gas due to the change of temperature and the total number of moles of gas when the ethane is cracked. Therefore, the ratio of the velocities will be:
Using the Ideal Gas Law, we can write this relationship:
For our condition with p2 ≅ p1, the ratio of velocities is approximately
From the chemistry and the statement of the conditions, we have
n1 = Moles_ethane + Moles_steam
and
Moles_steam = 0.1 × Moles_ethane
and then
n1 = 1.1 × Moles_enthane
Assuming that 100% of the ethane is consumed when cracking to produce the ethylene, we have
n2 = Moles_ethylene + Moles_hydrogen + Moles_steam
and
Moles_ethylene = Moles_hydrogen = Moles_ethane_consumed
We then have
)
This gives us
When rewritten, it becomes
Adjusting the temperature to absolute degrees Rankine gives us:
T2 = 1550°F + 460° = 2010°R
T1 = 350°F + 460° = 810°R
From the data, we calculate the following:
Thus, we see that there will be an increase of over 4.7 times in the velocity of the gas as it passes through the pipe in the ethylene cracker furnace. This is a dramatic increase in velocity that had to be taken into account when the cracker was designed and operated.
The preceding example was actually a simple problem that has been made more complicated here. The simple solution can provide an estimate for Step 13 of the problem-solving technique in Chapter 1, “Introductory Concepts,” in which we judge our results. First, the cracking of ethane into ethylene with the hydrogen being given off doubles the number of moles of gas. Second, the increase of temperature, from 810°R (350°F) to 2,010°R (1,550°F), also more than doubles the volume. Thus, the volume of the gases would be about quadrupled, which approximates and confirms the 4.7 times increase in our previous calculation.
The exact increase in velocity is needed when designing the ethane cracker to insure that there will be sufficient residence time. Also, if the residence time is too great, the ethylene product will continue to crack to carbon and hydrogen gas. However, the actual calculations of the effect on cracking are much more difficult as the increase of the gas volume and thus the velocity is over the length of the pipe, while the temperature increase that more than doubles the volume occurs near the start of the pipe.