Chapter 14
Expansion of Gases:
The thermal expansion of a gas involves 3 variables: volume, temperature, and pressure. The pressure of a gas, in a closed container is the result of the collision of its molecules on the walls of that container. It is important to note that the kinetic energy of each gas molecule depends on its temperature only. Recall the definition of temperature: " the temperature of an object is a result of the vibrations of its atoms and molecules. In a gas, molecules are free to move and bounce repeatedly against each other as well as their container's walls. In each collision, a gas molecule transfers some momentum to its container's walls. Gas pressure is the result of such momentum transfers. The faster they move, the greater the number of collisions per second and the greater impulse per collision they impart to the container's walls causing a higher pressure. For a fixed volume, if the temperature of a gas increases (by heating), its pressure increases as well. This is simply because of increased kinetic energy of gas molecules that cause more number of collisions per second and therefore increased pressure.
One important formula to know is the formula for the average kinetic energy of a number of gas molecules that are at a given temperature.
The average K.E. of gas molecules is a function of temperature only. The formula is
(K.E.)avg. = (3/2) kT
where T is the absolute temperature in Kelvin scale and k is called the " Boltzman's constant " with a value of k = 1.38x10^{-23} J/K.
By the number of gas molecules, we do not mean 1000 or even 1000,000 molecules. Most often we mean much more than 10^{24} molecules.
The formula for kinetic energy on the other hand is K.E. = (1/2)MV^{2} where V is the average speed of gas molecules that are at a given temperature.
According to above formula, since at a given temperature, the average K.E. of gas molecules is constant, a gas molecule that has a greater mass oscillates slower, and a gas molecule that has a smaller mass oscillates faster. The following example clarifies this concept.
Example 1: Calculate the average K.E. of air molecules at 27.0^{o}C. Also, calculate the average speed of its constituents: mainly oxygen molecules and nitrogen molecules. Note that 1 mole of O_{2} = 32.0 grams and 1 mole of N_{2} = 28.0 grams. By one mole of O_{2}, we mean 6.02x10^{23 }molecules of O_{2}. By one mole of N_{2}, we mean 6.02x10^{23 }molecules of N_{2}.
Solution:
K.E. = (3/2)kT = (3/2)(1.38x10^{-23}J/K)(27+273)K = 6.21x10^{-21} J/molecule.
This means that every gas molecule at this temperature, on the average, has this energy whether it is a single O_{2} molecule or N_{2} molecule.
For each O_{2} molecule, we may write: K.E. = (1/2)MV^{2} and solve for V.
6.21x10^{-21} J = (1/2)[ 32.0x10^{-3}kg/6.02x10^{23}]V^{2} ; V = 483m/s.
Note that the bracket calculates the mass of each O_{2} molecule in kg.
For each N_{2} molecule:
6.21x10^{-21} J = (1/2)[ 28.0x10^{-3}kg/6.02x10^{23}]V^{2} ; V = 517m/s.
Expansion of Gases: Perfect Gas Law:
If a gas fulfills two conditions, it is called a " perfect gas" or an " ideal gas " and its expansion follows the perfect gas law:
PV = nRT
where P is the gas absolute pressure (pressure with respect to vacuum), V is its volume (the volume of its container), n is the number of moles of gas in the container, R is the Universal gas constant, R =8.314 [J/(mole K)], and T is the gas absolute temperature in Kelvin.
The two conditions for a gas to be ideal or obey this equation are:
1) The gas pressure should not exceed about 8 atmospheres.
2) The gas must be superheated (gas temperature sufficiently above its boiling point) at the operating pressure and volume.
The Unit of " PV ":
Note that the product " PV " has dimensionally the unit of "energy." In SI, the unit of "P" is N/m^{2} and the unit of volume " V " is m^{3}. On this basis, the unit of the product " PV " becomes Nm or Joule. The " Joule " that appears in R = 8.314J /(mole K) is for this reason.
Example 2: A 0.400m^{3} tank contains nitrogen at 27^{o}C. The pressure gauge on it reads 3.75 atmosphere. Find (a) the number of moles of gas in the tank, and (b) its mass in kg.
Solution: (a)
P_{abs.} = P_{gauge} + 1atm. = 4.75 atm. Also, T_{abs.} = 27^{o}C + 273 = 300K.
PV = nRT ; n = PV/[RT] ; Write the following with horiz. fraction bars.
n = (4.75x101,000Pa)(0.400m^{3}) /{[8.314J/(mole K)]300K} = 76.9moles.
(b) M = (76.9 moles)(28.0 grams /mole) = 2150 grams = 2.15 kg.
Example 3: A 0.770m^{3} hydrogen tank contains 0.446 kg of hydrogen at 127 ^{o}C. The pressure gage on it is not working. What pressure should the gauge show? Each mole of H_{2 }is 2.00grams.
Solution: n = (0.446x10^{3} grams) / (2.00 grams / mole) = 223moles.
PV = nRT ; P = (nRT) / V ; Use horiz. fraction bars when solving.
P = (223 moles)[ 8.314 J/(mole K)](127+273)K /(0.770m^{3}).
P_{abs} = 963,000 Pascals.
P_{gauge} = P_{abs}- atm = 963,000Pa - 101,000Pa = 862,000Pa (about 8.6 atm.)
Equation of State:
Equation PV = nRT is also called the "equation of state." The reason is that for a certain amount of a gas, i.e., a fixed mass, the number of moles is fixed. A change in any of the variables: P , V , or T, or any two of them, results in a change in one or the other two. Regardless of the changes, PV = nRT holds true for any state that the gas is in, as long as the two conditions of a perfect gas are maintained. That's why it is called the equation of state. A gas is considered to be ideal if its temperature is quite above its boiling point and its pressure is under about 8 atmospheres. These two conditions must be met in any state that the gas is in, in order for this equation to be valid.
Now suppose that a fixed mass of a gas is in state 1: P_{1}, V_{1}, and T_{1}. We can write P_{1}V_{1} = nRT_{1}. If the gas goes through a certain change and ends up in state 2: P_{2}, V_{2}, and T_{2}, the equation of state for it becomes P_{2}V_{2} = nRT_{2}.
Dividing the 2nd equation by the 1st one results in: (P_{2}V_{2}) /(P_{1}V_{1}) = (nRT_{2}) /(nRT_{1}).
Simplifying yields: (P_{2}V_{2}) / (P_{1}V_{1}) = T_{2} / T_{1}.
This equation simplifies the solution to many problems. Besides its general form shown above, it has 3 other forms: one for constant pressure, one for constant temperature, and one for constant volume.
Example 4: 1632 grams of oxygen is at 2.80 atm. of gauge pressure and a temperature of 127^{o}C. Find (a) its volume. It is then compressed to 6.60 atm. of gauge pressure while cooled down to 27^{o}C. Find (b) its new volume.
Solution: n = (1632/32.0)moles = 51.0moles; (a) P_{1}V_{1} = nRT_{1} ; V_{1} = nRT_{1}/P_{1} ;
V_{1} = (51.0moles)[(8.314 J/(mole K)](127+273)K /(3.80x101,000)Pa.
V_{1} = 0.442m^{3}.
(b) (P_{2}V_{2})/(P_{1}V_{1}) = T_{2 }/T_{1} ; (7.6atm)(V_{2}) /[(3.8atm)(0.442m^{3})] = 300K/400K
Use horizontal fraction bars. V_{2} = 0.166m^{3}.
Constant Pressure (Isobar) Processes:
A process in which the pressure of an ideal gas does not change is called an " isobar process." Const. pressure means P_{2}=P_{1}. Equation (P_{2}V_{2}) / (P_{1}V_{1}) = T_{2 }/ T_{1} becomes: V_{2}/V_{1} = T_{2}/T_{1}.
Example 5: A piston-cylinder mechanism as shown below may be used to keep a constant pressure. The pressure on the gas under the piston is 0 gauge plus the extra pressure that the weight generates. Let the piston's radius be 10.0cm and the weight 475N, and suppose that the position of the piston at 77^{o}C is 25.0cm from the bottom of the cylinder. Find its position when the system is heated and the temperature is 127^{o}C.
Solution: V_{1} = πr^{2}h_{1} = π(10.0)^{2}(25.0) = 2500π V_{2} = πr^{2}h_{2} = π(10.0)^{2} h_{2 }= (100π)h_{2} . P_{2}= P_{1}=Const. (Won't change, no need to calculate). T_{1} = 77 ^{o}C + 273^{o}C = 350 K. T_{2} = 127^{o}C + 273^{o}C = 400 K. Using V_{2}/V_{1} = T_{2}/T_{1} results in: 100πh_{2}/(2500π) = 400/350 or, h_{2 } = 28.6cm. |
Note: As the gas under the piston expands, it pushes the piston up, but the gas pressure remains constant. |
Constant Temperature (Isothermal) Processes:
A process in which the temperature of a gas does not change is called an "isothermal process." Constant temperature means T_{2}=T_{1}. Equation (P_{2}V_{2}) / (P_{1}V_{1}) = T_{2 }/ T_{1} becomes: (P_{2}V_{2})/(P_{1}V_{1}) = 1. Cross-multiplication results in: P_{2}V_{2} = P_{1}V_{1}.
Example 6: A piston cylinder system has an initial volume of 420cm^{3} and the air in it is at a pressure of 3.00 atmospheres as its gauge shows. The gas is compressed to a volume of 140cm^{3} by pushing the piston. The generated heat is removed by enough cooling such that the temperature remains constant. Find the final pressure of the gas.
Solution: Since T = constant, P_{2}V_{2} = P_{1}V_{1 }; P_{2}(140cm^{3}) = (4.00atm)(420cm^{3}) ; (P_{2})_{abs.} = 12.0 atm. ; (P_{2})_{gauge} = 11.0 atm.
Constant Volume (Isometric) Processes:
A process in which the volume of an ideal gas does not change is called an " isometric process." Constant volume means V_{2}=V_{1}. Equation (P_{2}V_{2}) / (P_{1}V_{1}) = T_{2}/T_{1} becomes: P_{2}/P_{1} = T_{2}/T_{1}. Rigid gas cylinders have constant volumes.
Example 7: A 15.0 liter gas cylinder contains helium at 7^{o}C and 11.0atm of gauge pressure. It is warmed up to 147^{o}C. Find its new gauge pressure.
Solution: P_{2}/P_{1} = T_{2}/T_{1} ; P_{2} = P_{1 }(T_{2}/T_{1 }) ; (P_{2})_{gauge} = 17.0 atm.
Note: If your solution resulted in 16.5atm and you rounded it to 17 atmospheres, it is wrong. The answer is exactly 17.0 atm. without rounding.
Chapter 14 Test Yourself 1:
1) The temperature of a gas is the result of the (a) average K.E. of its atoms or molecules (b) average P.E. of its atoms and molecules (c) net momentum its atoms and molecules have. click here.
2) The net momentum of gas molecules in a container is (a) equal to the net K.E. of the atoms or molecules (b) zero (c) neither a nor b.
3) The pressure of a gas in its container (a) is the result of the average momentum transfer of its molecules to the walls of the container (b) depends on the temperature of the gas that in turn depends on the average K.E. of the gas molecules (c) both a & b. click here.
4) In a head-on collision of two equal and rigid masses, (a) the same-mass molecules exchange velocities (b) if one molecule is initially at rest, it will move at the velocity of the colliding molecule, and the colliding molecule comes to stop. (c) both a & b.
5) According to the "Kinetic Theory of Gases", the average K.E. of gas molecules is a function of (a) pressure only (b) volume only (c) temperature only. click here.
6) The average K.E. of gas molecules in a container that is at temperature T is equal to (a) kT (b) 1/2 kT (c) 3/2 kT where k is the Boltzman's constant.
7) The Boltzman's constant (k) has a value of (a) 1.38x10^{-23}/K (b) 1.38x10^{-23}J (c) 1.38x10^{-23}^{ }J/K where K stands for Kelvin temperature. click here.
8) The formula (K.E.)_{avg.} = (3/2)kT is applicable to (a) all gases (b) mono-atomic gases only (c) diatomic gases only.
9) The energy of a gas in a container due to its temperature is because of the energy it constantly receives from (a) the surroundings via heat transfer (b) the pressure from its container (c) the gravitational field of the Earth. click here.
10) The ratio of the mass of an average oxygen molecule to that of an average hydrogen molecule is (a) 32 (b) 16 (c) 8.
11) The (K.E.)_{avg.} of a gas molecule is on one hand (3/2)kT and on the other hand is (a) Mv^{2} (b) (1/2)Mv^{2} (c) Mv.
12) At a given temperature, if the average speed of oxygen molecules is 480m/s, the average speed of hydrogen molecules is (a)120m/s (b)1920m/s (c)960m/s. click here.
13) A gas is treated as a perfect gas if its pressure and temperature are respectively: (a) less than 8atm and under boiling point (b) more than 8atm, at boiling point (c) less then 8atm, quite above boiling point.
14) When the temperature of a gas is quite above its boiling point, the molecules are quite energetic and bounce around, and therefore do not stick together to condensate into liquid phase. This is a good reason for a gas to be a perfect gas and follow the perfect gas formula. (a) True (b) False. click here.
15) When the pressure of a gas is under about 8atm, the pressure is not too high to keep the molecules closer together to where the molecular attraction forces between the molecules become significant. Since no provision is made in this form of perfect gas law (PV=nRT) for molecular attraction; therefore, the formula is valid for pressures under about 8 atmospheres. (a) True (b) False
16) In the formula PV = nRT, (a) P is absolute pressure only (b) T is absolute temperature only (c) both P and T are absolute quantities. click here.
17) In the formula PV = nRT, the value of R is 8.314 J/(mole K) in (a) SI units only (b) all systems (c) American systems only.
18) The product PV has unit of (a) force (b) energy (c) power.
19) In SI, since R = 8.314 J/(mole K), the unit of PV is (a) lb-ft (b) Joule (c)watt. click here.
20) The gauge on a gas tank shows the pressure as 2.5atm. The absolute pressure is (a) 1.5atm (b) 3.5atm (c) also 2.5atm.
21) The gas pressure in a tank is (340kPa)_{gauge}. The absolute pressure is (a) 440kPa (b) 240kPa (c) also 140kPa.
22) The absolute pressure of a gas tank is 44.7psi. Its gauge should show (a) 59.4psi (b) 43.7psi (c) 30.0psi.
23) Avogadro number is (a) 6.02x10^{23}moles (b) 6.02x10^{23} molecules/mole (c) 6.02x10^{23} grams.
24) 6.02x10^{23} molecules of O_{2} have a mass of (a)32.0grams (b)16.0grams (c)8.0grams. click here.
25) 6.02x10^{23} molecules of N_{2} have a mass of (a)34.0grams (b)18.0grams (c)28.0grams.
26) One mole of N_{2} has a mass of (a) 34.0grams (b)18.0grams (c)28.0grams.
27) One mole of O_{2} has a mass of (a) 32.0grams (b)16.0grams (c)8.0grams. click here.
Problem: Let us select 1 mole of a perfect gas, any perfect gas, hydrogen, nitrogen, helium, etc., and put it in a container that can have a variable volume. Also, let us create STP (Standard Pressure and Temperature) for it, that is 1atm of absolute pressure and 0^{o}C. In Metric units, the standard pressure and temperature are: 101,000Pa and 273K. Answer the following:
28) Plugging these values: n = 1 mole, P_{abs} = 101,000Pa, T = 273K in the perfect gas formula PV = nRT , and solving for volume, yields: (a) V = 0.0224m^{3} (a) V = 22.4 Liter (c) both a & b. Note that: 1m^{3} = 1000 liter.
29) We may say that: one mole of any perfect gas at STP ( 1atm of absolute pressure, or 0 gauge pressure, and 0^{o}C, or 273^{o}K) occupies the same volume of 22.4 liter. (a) True (b) False click here.
30) In an isothermal process, (a) T_{2} = T_{1} (b) P_{2}V_{2} = P_{1}V_{1} (c) both a & b.
31) In an isometric process, (a) P_{2} = P_{1} (b) V_{2} = V_{1} (c) P_{2}V_{2} = P_{1}V_{1}.
32) In an isometric process, (a) V_{2} = V_{1} (b) P_{2 }/ P_{1} = T_{2 }/ T_{1} (c) both a & b.
33) In an isobar process, (a) P_{2} = P_{1} (b) V_{2} = V_{1} (c) P_{2}V_{2} = P_{1}V_{1}.
34) In an isobar process, (a) P_{2} = P_{1}. (b) V_{2} / V_{1} = T_{2 }/ T_{1}. (c) both a & b. click here.
35) At constant pressure, if the absolute temperature of a gas is doubled, its volume (a) doubles (b) triples (c) becomes half of what it was.
36) At constant volume, if the absolute temperature of a gas is doubled, its absolute pressure (a) doubles (b) triples (c) becomes half of what it was. click here.
37) At constant temperature, if the volume of a gas is tripled by expansion, its pressure (a) doubles (b) triples (c) becomes 1/3 of what it was.
38) At constant temperature, if the pressure of a gas quadrupled by compression, its volume (a) doubles (b) triples (c) becomes 1/4 of what it was.
39) At constant temperature, when the pressure of a gas increases, its volume (a) increases (b) decreases (c) does not change. click here.
40) The trick to create constant pressure for a gas is to (a) place it in a tank with fixed boundaries (b) place it in a piston-cylinder system in which the piston can move easily without allowing the gas to escape (c) neither a nor b.
41) To keep an ideal gas at a constant temperature, (a) its volume must be kept constant regardless of its pressure (b) its pressure must be kept constant regardless of its volume (c) heat must be supplied to the gas or removed from it to keep it at the desired temperature.
42) Constant volume for a gas means keeping it (a) in a rigid closed cylinder (b) keep it under a piston-cylinder system such that the piston cannot move (c) both a & b. click here.
43) A certain amount of gas in a closed rigid cylinder loses pressure if it is (a) cooled down (b) warmed up (c) both a & b.
44) If there is a certain amount of gas under a piston-cylinder system and the piston is pulled up such that the gas volume is increased, the gas temperature (a) goes up (b) goes down (c) remains unchanged (d) insufficient information. click here.
Problems:
1) Calculate (a) the mass (in kg) of each CO2 molecule as well as each He molecule knowing that their molar weights are 44.0 gr/mole and 4.00 gr/mole, respectively. Find (b) the average K.E. of each at 57.0^{o}C. Find (c) the average speed of each at 57.0^{o}C. The Avogadro Number is 6.02x10^{23} atoms/mole.
2) A 0.200m^{3} tank contains CO_{2} at 77^{o}C. The pressure gauge on it reads 7.75 atmospheres. Find (a) the number of moles of gas in the tank, and (b) the gas mass in kg. Each mole of CO_{2} has a mass of 44.0grams. R = 8.314 J/(mole K).
3) A 555-liter tank contains 1.20 kg of helium at -73.0 ^{o}C. What pressure should the gauge on it show (a) in kPa and (b) in psi? Each mole of He_{ }is 4.00grams. 1m^{3} = 1000 liter.
4) A tank contains 2.800 kg of nitrogen at a pressure of 3.80 atmosphere as its gauge shows. It is kept at a temperature of 27^{o}C. Find (a) its volume in liters. It is then kept in another room for several hours at a temperature of 77^{o}C. Find (b) its new pressure. Neglect the very small change in the volume of the tank due to thermal expansion and treat the tank volume as a constant.
5) A piston cylinder system has an initial volume of 960cm^{3} and the air in it is at zero gauge pressure. (a) If it is taken to outer space while keeping its volume and temperature constant, what pressure will its gauge show? Here on Earth, if the gas is compressed to a volume of 160cm^{3} by pushing on its piston while keeping its temperature constant by cooling, find (b) its final gauge pressure.
6) A 36.0 liter metal cylinder contains nitrogen at 127^{o}C and 3.00atm of gauge pressure. It is warmed up to 177^{o}C. Find (a) its new gauge pressure. (b) how much gas does it contain?
Answers:
1) 7.31x10^{-26}kg , 6.64x10^{-27}kg , 6.83x10^{-21}J, 432m/s, 1430m/s
2) 60.7moles, 2.67kg 3) 798kPa, 116psi 4) 513 liters, 4.6 atm.
5) 1.0atm, 5.0atm 6) 3.50atm, 123grams