F. Gases and Atmospheric Chemistry · Gases and Atmospheric ChemistryMeasuring Gases: Pressure, Temperature, STP and SATP
F2.1 F3.4 A1.12 A1.13About 65 minutes
Learning goals- I can define pressure and explain atmospheric pressure.
- I can convert between kPa, atm, mmHg and other pressure units.
- I can tell gauge pressure from absolute pressure.
- I can convert between degrees Celsius and kelvins and explain why gas laws need kelvins.
- I can state the conditions for STP and SATP.
Words to know
pressure
Force per unit area, P = F ÷ A; measured in pascals (Pa).
pascal (Pa)
The SI unit of pressure: 1 N/m². 1 kPa = 1000 Pa.
atmospheric pressure
The pressure exerted by the air; about 101.3 kPa at sea level.
barometer
An instrument for measuring atmospheric pressure.
gauge pressure
Pressure above atmospheric pressure, as read by a tire gauge.
absolute zero
0 K or −273.15 °C: the lowest possible temperature.
STP
Standard temperature and pressure: 0 °C and 101.325 kPa.
SATP
Standard ambient temperature and pressure: 25 °C and 100 kPa.
Reading the weather report
Listen to a Canadian weather forecast and you might hear: "The barometric pressure is 101.8 kilopascals and rising." Rising pressure often means clearer weather is coming; falling pressure often means a storm. But what exactly is being measured, and what is a kilopascal?
To use the gas laws, you need to measure four things about a gas: pressure (P), volume (V), temperature (T) and amount (n, in moles). This lesson is about measuring P and T correctly.
Pressure
Pressure is force per unit area:
P = F ÷ A
The SI unit is the pascal (Pa): 1 Pa = 1 N/m². A pascal is tiny (about the pressure of a sheet of paper lying on a table), so we usually use kilopascals (kPa). 1 kPa = 1000 Pa.
Atmospheric pressure is caused by the weight of the column of air above you, pushed on you by the collisions of air molecules. At sea level it averages about 101.3 kPa. That is huge: about 10 tonnes of force on every square metre. We don't get crushed because the fluids in our bodies push outward just as hard.
The barometer
In 1643, Evangelista Torricelli filled a long glass tube with mercury, turned it upside down in a dish of mercury, and saw the column drop until it was about 760 mm high. The air pushing down on the mercury in the dish held up the column. The space above the mercury is nearly a vacuum. When air pressure rises, the column rises; when it falls, the column falls. This is a mercury barometer, and it is why pressure is sometimes given in millimetres of mercury (mmHg).
A mercury barometer. Air pressure on the open dish supports a column of mercury about 760 mm high at sea level. Not to scale.| Unit | Equal to 1 standard atmosphere | Where you see it |
|---|
| kilopascal (kPa) | 101.325 kPa | SI unit; Canadian weather reports, chemistry |
| atmosphere (atm) | 1 atm | older chemistry, diving |
| millimetre of mercury (mmHg) | 760 mmHg | barometers, blood pressure |
| torr | 760 torr | lab vacuum gauges (1 torr = 1 mmHg) |
| bar | 1.01325 bar | weather maps (1 bar = 100 kPa) |
| pound per square inch (psi) | 14.7 psi | tire gauges, bike pumps |
Pressure unit conversions. Learn the first three by heart: 1 atm = 101.325 kPa = 760 mmHg.Pressure changes with altitude
The higher you go, the less air there is above you, so atmospheric pressure drops. A plastic water bottle capped tightly on a mountain top (or during a flight) looks crumpled when it is brought down to Toronto; a sealed bag of chips puffs up as you drive into the mountains.
Average air pressure falls as altitude increases (values from a standard model atmosphere). Near the top of Mount Logan, in Yukon, the air pressure is less than half of sea-level pressure.Gauge pressure versus absolute pressure
A tire gauge reads zero when it's not attached to anything, even though the atmosphere is pushing on it. It measures gauge pressure: the pressure above atmospheric pressure. The gas laws need absolute pressure:
P(absolute) = P(gauge) + P(atmospheric)
So a tire inflated to 220 kPa (gauge) actually contains air at about 220 + 101 = 321 kPa.
Temperature: why chemists use kelvins
The Celsius scale puts 0° at the freezing point of water. That's convenient for weather, but it's arbitrary: a gas at 20 °C does not have twice the kinetic energy of a gas at 10 °C.
The Kelvin scale starts at absolute zero (0 K = −273.15 °C), the temperature at which particles would have the lowest possible energy. Kelvin degrees are the same size as Celsius degrees, so:
T (K) = t (°C) + 273.15 (in this course, we usually use 273)
Notes: write K, not °K; a Kelvin temperature can never be negative; and in every gas-law calculation, temperatures must be in kelvins. Chemists write capital T for Kelvin temperature and lower-case t for Celsius.
The Celsius and Kelvin scales side by side. The degrees are the same size; the Kelvin scale is shifted so that zero is absolute zero.Standard conditions: STP and SATP
To compare gases fairly, chemists use reference conditions:
- STP (standard temperature and pressure): 0 °C (273.15 K) and 101.325 kPa
- SATP (standard ambient temperature and pressure): 25 °C (298.15 K) and 100 kPa
You will use both in the lessons on Avogadro's hypothesis and molar volume.
Worked exampleExample 1: Converting pressure units
A weather station in Thunder Bay reports 99.2 kPa. Express this in (a) atm and (b) mmHg.
- Use 1 atm = 101.325 kPa = 760 mmHg as conversion factors.
- (a) 99.2 kPa × (1 atm ÷ 101.325 kPa) = 0.979 atm
- (b) 99.2 kPa × (760 mmHg ÷ 101.325 kPa) = 744 mmHg
- Three significant figures in the data (99.2), so three in the answers.
(a) 0.979 atm (b) 744 mmHg
Worked exampleExample 2: Converting temperatures
Convert (a) −40 °C (a cold prairie morning) to kelvins; (b) 350 K to degrees Celsius.
- (a) T = t + 273 = −40 + 273 = 233 K
- (b) t = T − 273 = 350 − 273 = 77 °C
Worked exampleExample 3: Gauge and absolute pressure
A bike tire gauge reads 450 kPa on a day when the barometric pressure is 101 kPa. What is the absolute pressure in the tire, in kPa and in atm?
- P(absolute) = P(gauge) + P(atm) = 450 kPa + 101 kPa = 551 kPa
- 551 kPa × (1 atm ÷ 101.325 kPa) = 5.44 atm
551 kPa, or 5.44 atm. The air in the tire is at more than five times atmospheric pressure.
Hands-on activityThe incredible crushing atmosphere
You need: a grown-up, an empty, thin plastic water bottle (500 mL) with its cap, hot tap water, a sink, a sheet of paper and a ruler.
Safety: a grown-up handles the hot water; use hot tap water, never boiling water, and hold the bottle with a cloth if it feels too warm.
Part A
- Pour about 100 mL of hot tap water into the bottle. Put the cap on loosely and swirl for 20 s so the air inside warms up.
- Pour the water out, then immediately screw the cap on tightly.
- Stand the bottle in the sink and run cold water over it. Watch and listen.
What to look for: the bottle crumples inward. Cooling slows the gas particles inside (and some water vapour condenses), so the inside pressure drops. The atmosphere outside is now pushing harder than the gas inside.
Part B: calculate
- Measure your sheet of paper in metres (a letter-size sheet is about 0.216 m × 0.279 m).
- Find its area, A, in m².
- Calculate the force of the air on it: F = P × A, using P = 101 300 Pa.
- Divide by 9.8 to find the mass (kg) that would weigh as much.
What to look for: A ≈ 0.0603 m², F ≈ 101 300 × 0.0603 ≈ 6100 N, the weight of about 620 kg. The paper is not crushed because air pushes equally on both sides.
Just for funA barometer full of water instead of mercury would need a tube more than 10 m tall at sea level, about as high as a three-storey building. Mercury is about 13.6 times denser than water, which is why Torricelli's tube could fit on a lab bench.
Your turn
85.0 ÷ 101.325 = 0.839 atm
740 × (101.325 ÷ 760) = 98.7 kPa
4Convert 22 °C to kelvins.
5Convert 77 K (the boiling point of liquid nitrogen) to °C.
6State the conditions for STP and for SATP.
STP: 0 °C (273.15 K) and 101.325 kPa. SATP: 25 °C (298.15 K) and 100 kPa.
7A car tire gauge reads 240 kPa. Atmospheric pressure is 100 kPa. What is the absolute pressure?
More practice: warm-up
1What is the SI unit of pressure?
The pascal (Pa), which equals 1 N/m².
2How many kPa equal 1 atm?
3What is absolute zero in °C?
−273.15 °C (about −273 °C)
5Does air pressure increase or decrease as you climb a mountain?
More practice: on your own
1A force of 500 N acts on an area of 0.025 m². What is the pressure in kPa?
P = F ÷ A = 500 ÷ 0.025 = 20 000 Pa = 20 kPa
2Convert 32.0 psi (a typical car tire gauge reading) to kPa. (1 atm = 14.7 psi)
32.0 × (101.325 ÷ 14.7) = 221 kPa (gauge pressure)
3Explain why the temperature of a gas must be in kelvins for gas law calculations.
The gas laws use ratios and proportions. Kinetic energy (and so pressure and volume) is proportional to the Kelvin temperature, not the Celsius temperature. Celsius has an arbitrary zero and negative values, which would give wrong or impossible results.
4A student says that doubling the temperature from 10 °C to 20 °C doubles the average kinetic energy. Is the student right? Show why.
No. In kelvins, 10 °C = 283 K and 20 °C = 293 K. 293 ÷ 283 = 1.04, so the average kinetic energy rises by only about 4 %.
5A weather report gives 1013 hPa (hectopascals). Convert this to kPa and decide whether it is above or below standard pressure.
1 hPa = 100 Pa = 0.1 kPa, so 1013 hPa = 101.3 kPa. That is essentially standard pressure (101.325 kPa), very slightly below.
Challenge
1Mercury's density is 13.6 g/mL and water's is 1.00 g/mL. If a barometer used water, how tall would the column be when a mercury barometer reads 760 mm?
The column heights are inversely proportional to density: 760 mm × 13.6 = 10 336 mm ≈ 10.3 m of water.
2Your ears pop on a drive from Toronto (about 100 kPa) to a ski hill where the pressure is 90 kPa. What is the force difference on an eardrum with area 0.60 cm² (6.0 × 10⁻⁵ m²) before the pressure equalizes?
ΔP = 100 − 90 = 10 kPa = 10 000 Pa. F = ΔP × A = 10 000 × 6.0 × 10⁻⁵ = 0.60 N, pushing outward until the pressure behind the eardrum equalizes.
3A scuba tank gauge reads 2900 psi (gauge). Find the absolute pressure in kPa and atm. (1 atm = 14.7 psi = 101.325 kPa)
Absolute = 2900 + 14.7 = 2914.7 psi. In atm: 2914.7 ÷ 14.7 = 198 atm. In kPa: 198.3 × 101.325 ≈ 2.01 × 10⁴ kPa (about 20 100 kPa).
Check your understandingQuiz
1. Which is equal to 1 standard atmosphere?
2. What is 27 °C in kelvins?
4. A tire gauge reads 200 kPa where atmospheric pressure is 100 kPa. The absolute pressure is:
5. Why can a Kelvin temperature never be negative?
For parents and teachersBig idea (F2.1, F3.4, A1.12, A1.13): before using any gas law, students must express pressure in consistent units and temperature in kelvins. This lesson practises SI units, unit conversions and significant figures.
How to help: look at a weather app together (many give pressure in kPa) and convert it to atm and mmHg. Check tire pressure with a gauge and talk about gauge versus absolute pressure.
Common difficulties:
- Forgetting to convert °C to K. This is the single most common gas-law error. Make "K first!" a habit.
- Writing °K. The unit is simply K.
- Mixing up STP and SATP. STP is colder (0 °C) and has the "standard atmosphere" pressure.
- Using gauge pressure in gas law problems. Add atmospheric pressure first.
Safety: hot tap water only, handled by a grown-up.