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Temperature as average molecular energy, pressure from collisions, internal energy, the ideal gas law and molecular speeds.
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By the end of this lesson you will be able to use the ideal gas law and connect temperature and pressure to the motion of molecules.
From Physics 1 you know kinetic energy $\tfrac{1}{2}mv^2$, that pressure is force per area, and that energy is conserved in a system you name. You also know from chemistry that matter is made of molecules and that a mole is $6.022 \times 10^{23}$ of them. This lesson connects the motion of molecules to the temperature and pressure you can measure.
| Term | What it means |
|---|---|
| Temperature | A measure of the average kinetic energy of a molecule: $\bar{K} = \tfrac{3}{2}k_BT$. |
| Absolute temperature | Temperature in kelvins, $T = t + 273$; zero at absolute zero. |
| Internal energy | The total energy of all the molecules in a system. |
| Heat | Energy transferred because of a temperature difference. |
| Pressure | Force per unit area, $P = F/A$, measured in pascals. |
| Ideal gas law | $PV = nRT$, with $R = 8.314$ J/(mol·K). |
| Boltzmann constant | $k_B = 1.381 \times 10^{-23}$ J/K, the gas constant per molecule. |
A gas is a vast number of molecules flying in random directions, colliding with each other and with the walls. Three measurable quantities come from that motion:
Together they obey the ideal gas law
$$PV = nRT,$$
which holds well for real gases at ordinary pressures and temperatures, when molecules are far apart compared with their size.
Another way: picture
Picture a room full of bouncing tennis balls. Make them bounce faster and they hit the walls harder and more often, so the push on the walls, the pressure, rises. Shrink the room and each wall gets hit more often, raising the pressure again. Add more balls and the same happens. The ideal gas law is exactly that bookkeeping.
Another way: steps
Celsius and Fahrenheit set their zeros at convenient temperatures, the freezing point of water or a cold winter day. The kelvin scale sets its zero where molecular motion would stop, at $-273.15$ °C, so a temperature in kelvins is directly proportional to the average molecular energy. Doubling the kelvin temperature doubles that energy.
Ratios and proportions in gas problems therefore need kelvins. Warming air from $27$ °C to $327$ °C does not multiply its molecular energy by twelve; it takes the gas from $300$ K to $600$ K, doubling it. Differences are the same on both scales, since a kelvin and a Celsius degree are the same size.
When a molecule bounces off a wall, its momentum perpendicular to the wall reverses, and the wall must supply an impulse to do that. Trillions of such impulses each second average into a steady force, and that force per area is the gas pressure. Air at sea level presses on everything with about $101$ kPa, or $14.7$ pounds per square inch.
Working the collisions through gives $PV = \tfrac{1}{3}Nm\overline{v^2}$, and comparing with $PV = Nk_BT$ shows that $\tfrac{1}{2}m\overline{v^2} = \tfrac{3}{2}k_BT$. That is how temperature and molecular energy are tied together, and why the pressure rises when a gas is heated in a rigid container.
The law $PV = nRT$ has four variables; given three, it gives the fourth. With pressure in pascals and volume in cubic meters, $R = 8.314$ J/(mol·K). A handy shortcut is that a joule per liter is exactly a kilopascal, so with volume in liters the law gives pressure in kPa directly.
When a fixed amount of gas changes state, $nR$ stays constant, so $P_1V_1/T_1 = P_2V_2/T_2$. Special cases have names: Boyle's law, $PV$ constant at fixed temperature; Charles's law, $V/T$ constant at fixed pressure; and Gay-Lussac's law, $P/T$ constant at fixed volume.
Since $\tfrac{1}{2}m\overline{v^2} = \tfrac{3}{2}k_BT$, the root-mean-square speed of a molecule is $v_{\text{rms}} = \sqrt{3k_BT/m} = \sqrt{3RT/M}$, where $M$ is the molar mass in kilograms per mole. Nitrogen at room temperature moves at about $517$ m/s, faster than a jet airliner; hydrogen, fourteen times lighter, moves nearly four times faster.
At the same temperature every gas has the same average kinetic energy per molecule, so lighter molecules must move faster. That is why Earth has lost almost all its hydrogen and helium to space: in the hot upper atmosphere, a fraction of light atoms exceed the $11.2$ km/s escape speed.
Checking an answer. Pressures of everyday gases are tens to hundreds of kPa. Molecular energies at room temperature are about $6 \times 10^{-21}$ J. Heating at fixed volume must raise the pressure; expansion at fixed temperature must lower it.
The ideal gas law assumes molecules are tiny compared with their spacing and exert no forces on each other except in brief collisions. For air at ordinary conditions, the molecules fill about a thousandth of the volume and the assumptions hold to better than one percent. At high pressure or near condensation, real gases deviate.
The combined law needs a fixed amount of gas: nothing leaks in or out. And temperature must be absolute, because the law's proportionality holds only when zero means zero molecular energy.
Internal energy is the sum over all molecules. For helium or argon, whose atoms can only move, $U = \tfrac{3}{2}nRT$. Molecules like nitrogen can also rotate, storing more energy at the same temperature: $U = \tfrac{5}{2}nRT$ for air.
Because internal energy is a total, it depends on how much material there is, while temperature does not. A swimming pool at $25$ °C holds vastly more internal energy than a cup of coffee at $80$ °C, yet heat flows from the coffee to the pool if they touch, because heat flows from higher to lower temperature, not from more energy to less.
When two objects touch, energy flows from the hotter to the colder until their temperatures are equal. At that point they are in thermal equilibrium, and no net heat flows. This is sometimes called the zeroth law of thermodynamics: two objects each in equilibrium with a third are in equilibrium with each other.
The zeroth law is what makes thermometers work. A thermometer placed in a gas comes to equilibrium with it, and its reading then tells the gas's temperature. Digital thermometers use resistors whose resistance changes with temperature; the principle is the same.
If a gas is cooled at fixed volume, its pressure falls in a straight line with temperature. Extend that line and it reaches zero pressure at $-273.15$ °C, whatever gas is used and however much of it there is. Nineteenth-century physicists took this as evidence of a lowest possible temperature, where molecular motion would be as small as it can be, and William Thomson, later Lord Kelvin, built his scale from it.
No real gas reaches that point; every gas condenses to a liquid first. But laboratories have come astonishingly close. At MIT and at the University of Colorado's JILA, physicists cooled clouds of atoms to less than a millionth of a kelvin in the 1990s, producing a new state of matter, the Bose-Einstein condensate, and winning the 2001 Nobel Prize. NASA's Cold Atom Laboratory on the International Space Station now makes such condensates in orbit, where they can float freely for longer.
The ideal gas law fails when molecules crowd together. At high pressure the molecules take up a noticeable fraction of the volume, and at low temperature their weak attractions pull them together, lowering the pressure below the ideal value. Near the point where a gas condenses, the law can be off by a large factor.
Johannes van der Waals added two corrections in 1873, one for the molecules' own volume and one for their attraction, and his equation predicts condensation itself. Engineers who design natural gas pipelines, which run at several megapascals, and the tanks of liquefied natural gas shipped from Gulf Coast terminals use such corrected equations. For air in a room, a tire or a balloon, the ideal gas law is accurate to within a percent, which is why it is the working tool of this course.
Twice a day, at about ninety sites across the United States, the National Weather Service launches balloons carrying radiosondes that radio back temperature, pressure and humidity as they climb. A balloon leaves the ground holding a couple of cubic meters of helium or hydrogen at $101$ kPa.
As it rises, the surrounding pressure falls faster than the temperature, so the gas expands. At $20$ kPa and $217$ K, the combined gas law gives a volume nearly four times larger; near $30$ km altitude, where the pressure is about $1$ kPa, the balloon is dozens of times its launch size and the rubber bursts. A small parachute brings the instrument down, and a label asks finders to mail it back.
Every fall, drivers across the northern United States see their tire-pressure warning lights come on during the first cold snap. The air in a tire is a fixed amount of gas in a nearly fixed volume, so its absolute pressure is proportional to its absolute temperature.
A tire set to $35$ psi gauge, $49.7$ psi absolute, at $70$ °F ($294$ K) drops to $45.5$ psi absolute at $20$ °F ($266$ K), a gauge reading of about $31$ psi. The rule of thumb that tires lose one psi for every ten degrees Fahrenheit comes straight from the gas law. Tire makers therefore recommend checking pressure when tires are cold, before driving warms them up.
A common mistake is to treat temperature as the amount of energy an object holds. Temperature is an average per molecule; internal energy is the total. A bathtub of warm water has a lower temperature than a cup of boiling water but far more internal energy, because it has so many more molecules.
A related error is to compute ratios with Celsius temperatures. Because $0$ °C is not zero energy, a gas at $20$ °C does not have twice the molecular energy of one at $10$ °C; in kelvins, $293$ against $283$, it has only $3.5$ percent more.
A $12$ L scuba tank holds air at $20.7$ MPa and $20$ °C. Convert to kelvins.
$T = 20 + 273 = 293\ \text{K}$
Absolute temperature.
Convert the pressure and volume to SI units.
$P = 2.07 \times 10^7\ \text{Pa}, \quad V = 0.012\ \text{m}^3$
Pascals and cubic meters.
Solve for the moles of air.
$n = \dfrac{PV}{RT} = \dfrac{2.07 \times 10^7 \times 0.012}{8.314 \times 293}$
From $PV = nRT$.
Evaluate the amount.
$n = 102\ \text{mol}$
About $3$ kg of air.
Find the volume at sea-level pressure.
$V = \dfrac{102 \times 8.314 \times 293}{101{,}000} = 2.46\ \text{m}^3$
Nearly $2500$ liters squeezed into $12$.
Find the average kinetic energy of an air molecule at $300$ K.
$\bar{K} = 1.5 \times 1.381 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\ \text{J}$
The same for every gas.
Find the mass of a nitrogen molecule.
$m = \dfrac{0.028}{6.022 \times 10^{23}} = 4.65 \times 10^{-26}\ \text{kg}$
Molar mass over Avogadro's number.
Solve for the mean square speed.
$\overline{v^2} = \dfrac{2 \times 6.21 \times 10^{-21}}{4.65 \times 10^{-26}} = 2.67 \times 10^5$
In m²/s².
Take the square root.
$v_{\text{rms}} = 517\ \text{m/s}$
Faster than sound.
Repeat for oxygen.
$v_{\text{rms}} = 517\sqrt{\tfrac{28}{32}} = 484\ \text{m/s}$
Slightly heavier, slightly slower.
Count the molecules in a liter of air.
$N = \dfrac{PV}{k_BT} = \dfrac{101{,}000 \times 0.001}{1.381 \times 10^{-23} \times 300} = 2.4 \times 10^{22}$
An enormous number.
A sealed can holds gas at $250$ kPa absolute at $22$ °C. Convert to kelvins.
$T_1 = 295\ \text{K}$
Absolute temperature.
The car interior reaches $65$ °C. Convert that too.
$T_2 = 338\ \text{K}$
Absolute temperature.
Identify what is fixed.
$V, n \ \text{fixed} \Rightarrow \dfrac{P}{T} \ \text{constant}$
Gay-Lussac's law.
Find the new pressure.
$P_2 = 250 \times \dfrac{338}{295} = 286\ \text{kPa}$
Up about $15$ percent.
Compare with the wrong Celsius ratio.
$250 \times \dfrac{65}{22} = 739\ \text{kPa}$
Far too high: Celsius ratios mislead.
Find the gauge pressure.
$286 - 101 = 185\ \text{kPa}$
What a gauge on the can would read.
Write the gas law for pressure.
$P = \dfrac{nRT}{V}$
Volume in liters gives kPa.
Substitute the values.
$P = \dfrac{1.0 \times 8.314 \times 300}{25}$
Numerator $2494$ J.
Evaluate the pressure.
A sealed gas is heated from $27$ °C to $627$ °C. By what factor does the average kinetic energy of its molecules increase?
Complete the worked solution: a car tire holds air at an absolute pressure of $300$ kPa at $10$ °C. After highway driving the air reaches $35$ °C. Treating the tire's volume as fixed, find both temperatures in kelvins and the new absolute pressure in kPa.
Convert the starting temperature.
$T_1 = 10 + 273 =$ x
Kelvins.
Convert the final temperature.
$T_2 = 35 + 273 =$ y
Kelvins.
Scale the pressure by the temperature ratio.
$P_2 = 300 \times \dfrac{T_2}{T_1} =$ p
Fixed volume and amount of air.
Explain why gauges read lower.
$\text{gauge} = \text{absolute} - 101\ \text{kPa}$
A gauge reads the excess over the atmosphere.
Match each quantity to what it describes in the molecular model of a gas.
| average kinetic energy of one molecule | total energy of all the molecules | energy transferred because of a temperature difference | force per area from molecular collisions | |
|---|---|---|---|---|
| temperature | ||||
| internal energy | ||||
| heat | ||||
| pressure |
A tank holds $3.0$ mol of helium at $320$ K in a volume of $80$ L. With $R = 8.314$ J/(mol·K) and $k_B = 1.381 \times 10^{-23}$ J/K, fill in the pressure in kPa, the average kinetic energy of one atom in units of $10^{-21}$ J, and the internal energy of the gas in J.
| value | |
|---|---|
| pressure (kPa) | |
| average energy per atom (10⁻²¹ J) | |
| internal energy (J) |
A rigid tank of volume $40$ L holds $4$ mol of an ideal gas. Using $R = 8.314$ J/(mol·K), write the pressure in kPa as a function of the absolute temperature $T$ in kelvins.
Answer:
What is the root-mean-square speed of hydrogen (H₂) molecules, molar mass $2$ g/mol, at $300$ K? Use $R = 8.314$ J/(mol·K) and give the answer in m/s.
Answer: m/s
The National Weather Service launches a sounding balloon holding $2.0$ m³ of helium at $101$ kPa and $288$ K. Aloft, the air pressure is $30$ kPa and the helium has cooled to $229$ K. Treating the rubber as putting no extra pressure on the gas, what is the balloon's volume, in m³?
Answer: m³
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A rigid tank of volume $10$ L holds $2$ mol of an ideal gas. Using $R = 8.314$ J/(mol·K), write the pressure in kPa as a function of the absolute temperature $T$ in kelvins.
Answer:
You can model a gas. Explain to someone why a bathtub of warm water holds more internal energy than a cup of boiling water.
21. Your turn: $1.0$ mol of gas at $300$ K occupies $25$ L. What is its pressure in kPa?, step 3
$P = 99.8\ \text{kPa}$
About one atmosphere.