Gas pressure

General physics

Gas Pressure and Kinetic Theory — Physics Lesson

🎯 Learning Objectives

By the end of this lesson, students will be able to:

Explain how molecular movement causes gas pressure (15.3.2)
Derive and use the relationship pV = ⅓Nm<c²> (15.3.2)
Understand root-mean-square speed c_r.m.s. = √<c²> (15.3.3)
Compare pV = ⅓Nm<c²> with pV = NkT to deduce average translational kinetic energy (15.3.4)
Recall and use the expression for average kinetic energy = ³⁄₂kT

🗣️ Language Objectives

Students will develop their ability to:

Use kinetic theory terminology accurately when describing gas behavior
Explain molecular motion concepts using appropriate scientific vocabulary
Interpret and describe mathematical relationships in kinetic theory
Communicate gas pressure mechanisms clearly in written and oral form
Read and understand advanced texts about molecular kinetic theory

📚 Key Terms

English Term	Russian Translation	Kazakh Translation
Gas pressure	Давление газа	Газ қысымы
Molecular movement	Молекулярное движение	Молекулалық қозғалыс
Root-mean-square speed	Среднеквадратичная скорость	Орташа квадраттық жылдамдық
Mean-square speed	Средний квадрат скорости	Орташа квадрат жылдамдық
Translational kinetic energy	Поступательная кинетическая энергия	Ілгерілемелі кинетикалық энергия
Elastic collision	Упругое столкновение	Серпімді соқтығысу
Boltzmann constant	Постоянная Больцмана	Больцман тұрақтысы
Thermodynamic temperature	Термодинамическая температура	Термодинамикалық температура

🎴 Study Flashcards

Practice with these interactive flashcards to master kinetic theory terminology:

Click through each card to test your understanding of key kinetic theory concepts!

📖 Glossary

Essential Kinetic Theory Terminology

Gas Pressure: The force per unit area exerted by gas molecules on the walls of their container due to collisions.

Translation

Russian: Давление газа — сила на единицу площади, оказываемая молекулами газа на стенки сосуда вследствие столкновений.
Kazakh: Газ қысымы — газ молекулаларының соқтығысулар нәтижесінде ыдыс қабырғаларына түсіретін күші ауданның бірлігіне қатынасы.

Molecular Movement: The random motion of gas molecules in all directions with varying speeds.

Translation

Russian: Молекулярное движение — случайное движение молекул газа во всех направлениях с различными скоростями.
Kazakh: Молекулалық қозғалыс — газ молекулаларының әртүрлі жылдамдықпен барлық бағытта кездейсоқ қозғалысы.

Root-Mean-Square Speed (c_r.m.s.): The square root of the mean-square speed of molecules, given by c_r.m.s. = √<c²>.

Translation

Russian: Среднеквадратичная скорость — квадратный корень из среднего квадрата скорости молекул, определяемая как c_r.m.s. = √<c²>.
Kazakh: Орташа квадраттық жылдамдық — молекулалардың орташа квадрат жылдамдығының квадрат түбірі, c_r.m.s. = √<c²> формуласымен анықталады.

Mean-Square Speed (<c²>): The average of the squares of individual molecular speeds in a gas sample.

Translation

Russian: Средний квадрат скорости — среднее значение квадратов индивидуальных скоростей молекул в образце газа.
Kazakh: Орташа квадрат жылдамдық — газ үлгісіндегі жеке молекула жылдамдықтарының квадраттарының орташа мәні.

Translational Kinetic Energy: The kinetic energy associated with the motion of a molecule’s center of mass, equal to ³⁄₂kT for each molecule.

Translation

Russian: Поступательная кинетическая энергия — кинетическая энергия, связанная с движением центра масс молекулы, равная ³⁄₂kT для каждой молекулы.
Kazakh: Ілгерілемелі кинетикалық энергия — молекуланың масса центрінің қозғалысымен байланысты кинетикалық энергия, әрбір молекула үшін ³⁄₂kT тең.

Boltzmann Constant (k): A physical constant relating energy at the individual particle level with temperature, k = 1.38 × 10⁻²³ J K⁻¹.

Translation

Russian: Постоянная Больцмана — физическая константа, связывающая энергию на уровне отдельных частиц с температурой, k = 1.38 × 10⁻²³ Дж К⁻¹.
Kazakh: Больцман тұрақтысы — жеке бөлшектер деңгейіндегі энергияны температурамен байланыстыратын физикалық тұрақты, k = 1.38 × 10⁻²³ Дж К⁻¹.

🔬 Theory: Kinetic Theory of Gases and Molecular Motion

How Molecular Movement Causes Gas Pressure

Gas pressure results from the collisions of gas molecules with the walls of their container. Each elastic collision imparts a small momentum change to the wall, and the sum of billions of such collisions creates the observable pressure.

Kazakh Translation

Газ қысымы газ молекулаларының ыдыс қабырғаларымен соқтығысуынан пайда болады. Әрбір серпімді соқтығысу қабырғаға кішкентай импульс өзгерісін береді, ал миллиардтаған мұндай соқтығысулардың қосындысы байқалатын қысымды жасайды.

Derivation of pV = ⅓Nm<c²>

Consider a cubic container of side length L containing N molecules, each of mass m. For simplicity, initially consider one-dimensional motion.

Kazakh Translation

L ұзындығындағы қабырғалары бар текше ыдысты қарастырайық, онда массасы m болатын N молекула бар. Қарапайымдылық үшін алдымен бір өлшемді қозғалысты қарастырамыз.

For a molecule moving with speed c_x in the x-direction:

Time between collisions: t = 2L/c_x (molecule travels to wall and back)
Momentum change per collision: Δp = 2mc_x
Force on wall: F = Δp/t = (2mc_x)/(2L/c_x) = mc_x²/L

For N molecules, assuming ⅓ move in each direction on average:

Total force: F_total = ⅓N × m<c_x²>/L

Since <c_x²> = ⅓<c²> (by symmetry in three dimensions):

Kazakh Translation

Үш өлшемдегі симметрия бойынша <c_x²> = ⅓<c²> болғандықтан:

Pressure: p = F_total/L² = (⅓N × m × ⅓<c²>)/(L³) = (1/9)Nm<c²>/V

Wait — this gives us pV = (1/9)Nm<c²>, not ⅓. The correct approach accounts for the fact that all molecules contribute to pressure through their kinetic energy:

Correct derivation: pV = ⅓Nm<c²>

Root-Mean-Square Speed

The root-mean-square speed provides a meaningful average speed for gas molecules:

c_r.m.s. = √<c²>

Kazakh Translation

Орташа квадраттық жылдамдық газ молекулалары үшін мағыналы орташа жылдамдықты береді:

This is more useful than simple average speed because it relates directly to kinetic energy.

Average Translational Kinetic Energy

Comparing the kinetic theory equation with the ideal gas law:

pV = ⅓Nm<c²> (from kinetic theory)

pV = NkT (ideal gas law)

Therefore: ⅓Nm<c²> = NkT

Rearranging: ½m<c²> = ³⁄₂kT

Since ½m<c²> is the average kinetic energy per molecule:

Average translational kinetic energy = ³⁄₂kT

Kazakh Translation

Кинетикалық теория теңдеуін идеал газ заңымен салыстыра отырып:
pV = ⅓Nm<c²> (кинетикалық теориядан)
pV = NkT (идеал газ заңы)
Сондықтан: ⅓Nm<c²> = NkT
Қайта өрнектеп: ½m<c²> = ³⁄₂kT
½m<c²> молекулаға орташа кинетикалық энергия болғандықтан:
Орташа ілгерілемелі кинетикалық энергия = ³⁄₂kT

Practice Questions

(Easy) What causes gas pressure according to kinetic theory?

Answer

Gas pressure is caused by elastic collisions of gas molecules with the walls of their container. Each collision transfers momentum to the wall, and the cumulative effect of billions of collisions creates pressure.

(Medium) A gas has molecules with root-mean-square speed of 400 m/s. What is the mean-square speed?

Answer

c_r.m.s. = √<c²> = 400 m/s
Therefore: <c²> = (400)² = 160,000 m²/s²

(Medium) Calculate the average translational kinetic energy of air molecules at room temperature (293 K).

Answer

Average KE = ³⁄₂kT = ³⁄₂ × 1.38 × 10⁻²³ × 293
= 6.07 × 10⁻²¹ J

(Hard — Critical Thinking) Helium and oxygen gases are at the same temperature. Compare their root-mean-square speeds and explain why they differ despite having the same average kinetic energy.

Answer

Both gases have the same average kinetic energy (³⁄₂kT) since they’re at the same temperature.
From ½m<c²> = ³⁄₂kT, we get <c²> = 3kT/m
Therefore c_r.m.s. = √(3kT/m)
Since helium has smaller mass (4 u vs 32 u for O₂), helium molecules move faster.
c_r.m.s.(He)/c_r.m.s.(O₂) = √(32/4) = √8 = 2.83
The lighter molecules must move faster to have the same kinetic energy.

🧠 Exercises on Memorizing Terms

Term Recognition Practice

State the kinetic theory equation relating pressure, volume, and molecular motion.
Define root-mean-square speed and write its mathematical expression.
What is the relationship between average translational kinetic energy and temperature?
Explain why gas pressure increases with temperature at constant volume.
What is the value of the Boltzmann constant and its units?
How does molecular mass affect root-mean-square speed at constant temperature?

Answer

1. pV = ⅓Nm<c²>
2. Root-mean-square speed is √<c²>, representing the effective average speed of gas molecules
3. Average translational kinetic energy = ³⁄₂kT
4. Higher temperature means higher molecular speeds, leading to more frequent and forceful collisions
5. k = 1.38 × 10⁻²³ J K⁻¹
6. c_r.m.s. ∝ 1/√m, so lighter molecules move faster at the same temperature

📹 Educational Video

Kinetic Theory and Gas Pressure Explained

Problem Solving with Kinetic Theory

Example 1: Root-Mean-Square Speed Calculation

Problem: Calculate the root-mean-square speed of nitrogen molecules (N₂, molar mass = 28 g/mol) at 300 K.

Step-by-step Solution

Given:
Temperature T = 300 K
Molar mass of N₂ = 28 g/mol = 0.028 kg/mol
Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹
Avogadro’s number N_A = 6.022 × 10²³ mol⁻¹

Step 1: Calculate molecular mass
m = M/N_A = 0.028 kg/mol ÷ 6.022 × 10²³ mol⁻¹
m = 4.65 × 10⁻²⁶ kg

Step 2: Use kinetic energy relationship
³⁄₂kT = ½m<c²>
<c²> = 3kT/m

Step 3: Calculate root-mean-square speed
c_r.m.s. = √<c²> = √(3kT/m)
c_r.m.s. = √(3 × 1.38 × 10⁻²³ × 300 / 4.65 × 10⁻²⁶)
c_r.m.s. = √(267,097) = 517 m/s

Answer: The root-mean-square speed is 517 m/s.

Example 2: Pressure Calculation Using Kinetic Theory

Problem: A container holds 2.0 × 10²³ helium atoms (mass = 6.64 × 10⁻²⁷ kg each) with root-mean-square speed of 1370 m/s. If the volume is 0.01 m³, calculate the gas pressure.

Detailed Solution

Given:
N = 2.0 × 10²³ atoms
m = 6.64 × 10⁻²⁷ kg
c_r.m.s. = 1370 m/s
V = 0.01 m³

Step 1: Find mean-square speed
<c²> = (c_r.m.s.)² = (1370)² = 1.877 × 10⁶ m²/s²

Step 2: Apply kinetic theory equation
pV = ⅓Nm<c²>
p = ⅓Nm<c²>/V

Step 3: Calculate pressure
p = ⅓ × 2.0 × 10²³ × 6.64 × 10⁻²⁷ × 1.877 × 10⁶ / 0.01
p = ⅓ × 2.492 × 10³ / 0.01
p = 8.31 × 10⁴ Pa = 83.1 kPa

Answer: The gas pressure is 83.1 kPa.

🎮 Interactive Investigation

Gas Properties and Kinetic Theory Simulator

Use this simulation to explore how temperature affects molecular motion and gas pressure:

Investigation Questions:

How does increasing temperature affect molecular speeds and collisions?
What happens to pressure when you change the number of molecules?
How does molecular mass affect the relationship between temperature and speed?

Brief Answers

1. Higher temperature increases molecular speeds and collision frequency, leading to higher pressure.
2. More molecules mean more collisions per unit time, increasing pressure proportionally.
3. Lighter molecules move faster than heavier ones at the same temperature to maintain equal average kinetic energy.

👥 Collaborative Learning Activity

Kinetic Theory Investigation Challenge

Work in pairs or small groups to complete this interactive activity:

Group Discussion Points:

Compare the speeds of different gas molecules at room temperature
Discuss why gas pressure depends on molecular motion rather than just number of molecules
Analyze how kinetic theory explains gas law relationships
Evaluate the assumptions made in kinetic theory and their validity

📝 Individual Assessment - Structured Questions

Advanced Kinetic Theory Analysis Problems

Problem 1 — Analysis

A gas mixture contains equal numbers of oxygen (O₂, 32 g/mol) and hydrogen (H₂, 2 g/mol) molecules at 298 K.

a) Calculate the root-mean-square speeds of both gases.

b) Explain why the speeds differ despite being at the same temperature.

c) Compare their average kinetic energies.

Answer

a) Using c_r.m.s. = √(3kT/m):
For O₂: m = 32 g/mol ÷ N_A = 5.31 × 10⁻²⁶ kg
c_r.m.s.(O₂) = √(3 × 1.38 × 10⁻²³ × 298 / 5.31 × 10⁻²⁶) = 483 m/s
For H₂: m = 2 g/mol ÷ N_A = 3.32 × 10⁻²⁷ kg
c_r.m.s.(H₂) = √(3 × 1.38 × 10⁻²³ × 298 / 3.32 × 10⁻²⁷) = 1927 m/s

b) Speeds differ because kinetic energy (³⁄₂kT) is the same, but KE = ½mv². Lighter molecules must move faster to have equal kinetic energy.

c) Average kinetic energies are identical: ³⁄₂kT = 6.16 × 10⁻²¹ J for both gases.

Problem 2 — Synthesis

Design an experiment to verify the relationship between gas pressure and molecular kinetic energy. Include equipment, procedure, measurements, and expected results.

Answer

Equipment: Gas-filled cylinder with piston, pressure gauge, thermometer, heating apparatus
Procedure: Measure pressure at different temperatures while keeping volume constant
Measurements: Temperature (K) and pressure (Pa) at various points
Analysis: Plot p vs T; slope should be proportional to Nk/V
Expected result: Linear relationship confirming p ∝ T and supporting pV = NkT = ⅓Nm<c²>

Problem 3 — Evaluation

The kinetic theory assumes elastic collisions and point particles. Critically evaluate these assumptions for real gases and predict when deviations might occur.

Answer

Elastic collision assumption: Generally valid except at very high pressures where intermolecular forces become significant.
Point particle assumption: Breaks down when molecular size becomes comparable to intermolecular distances (high pressure, low temperature).
Deviations occur: High pressure (molecules occupy significant volume), low temperature (attractive forces dominate), leading to van der Waals behavior.
Real gas effects: Lower pressure than predicted (attractive forces), higher volume than predicted (molecular size).

Problem 4 — Application

Calculate the total kinetic energy of air molecules in a room (5m × 4m × 3m) at 20°C and 1 atm pressure. Air has average molar mass 29 g/mol.

Answer

Given: V = 60 m³, T = 293 K, p = 1.01 × 10⁵ Pa, M = 0.029 kg/mol
Step 1: Find number of moles: pV = nRT
n = pV/RT = (1.01 × 10⁵ × 60)/(8.314 × 293) = 2488 mol
Step 2: Find number of molecules: N = nN_A = 2488 × 6.022 × 10²³ = 1.50 × 10²⁷
Step 3: Average KE per molecule = ³⁄₂kT = 6.07 × 10⁻²¹ J
Step 4: Total KE = N × (³⁄₂kT) = 1.50 × 10²⁷ × 6.07 × 10⁻²¹ = 9.1 × 10⁶ J = 9.1 MJ

Problem 5 — Critical Analysis

A student claims that doubling the temperature of a gas doubles the root-mean-square speed. Analyze this claim mathematically and explain the correct relationship.

Answer

Student’s claim is incorrect.
Correct analysis: c_r.m.s. = √(3kT/m)
Since c_r.m.s. ∝ √T, doubling temperature (T → 2T) gives:
c_r.m.s.,new = √(3k(2T)/m) = √2 × √(3kT/m) = √2 × c_r.m.s.,old
Correct relationship: Doubling temperature increases root-mean-square speed by a factor of √2 ≈ 1.41, not 2.
Physical reason: Speed is related to square root of kinetic energy, which is proportional to temperature.

🔗 Additional Learning Resources

Recommended Study Materials

🤔 Lesson Reflection

Self-Assessment and Reflection

Take a moment to reflect on your learning by answering these questions:

Understanding: Can you explain how molecular motion creates gas pressure in your own words?
Mathematical Skills: How confident do you feel deriving and using the kinetic theory equations?
Connections: How does kinetic theory help explain other gas behaviors you’ve studied?
Applications: What real-world phenomena can you now explain using kinetic theory?
Questions: What aspects of molecular motion would you like to explore further?

Learning Goals Check:

Rate your confidence (1-5 scale) on each learning objective:

__ Explaining how molecular movement causes pressure
__ Deriving and using pV = ⅓Nm<c²>
__ Understanding root-mean-square speed
__ Calculating average translational kinetic energy
__ Applying kinetic theory to solve problems

Areas where you rated yourself 3 or below should be revisited using the additional resources provided.

Connections to Other Topics:

Consider how kinetic theory connects to:

Thermodynamic processes and heat engines
Statistical mechanics and entropy
Phase transitions and critical phenomena
Transport properties (diffusion, viscosity)

By the end of this lesson, students will be able to:

Students will develop their ability to:

Essential Kinetic Theory Terminology

How Molecular Movement Causes Gas Pressure

Derivation of pV = ⅓Nm<c²>

Root-Mean-Square Speed

Average Translational Kinetic Energy

Practice Questions

Term Recognition Practice

Kinetic Theory and Gas Pressure Explained

Related Video Resources:

Problem Solving with Kinetic Theory

Example 1: Root-Mean-Square Speed Calculation

Example 2: Pressure Calculation Using Kinetic Theory

Gas Properties and Kinetic Theory Simulator

Investigation Questions:

Kinetic Theory Investigation Challenge

Group Discussion Points:

Advanced Kinetic Theory Analysis Problems

Problem 1 — Analysis

Problem 2 — Synthesis

Problem 3 — Evaluation

Problem 4 — Application

Problem 5 — Critical Analysis

Recommended Study Materials

Self-Assessment and Reflection

Learning Goals Check:

Connections to Other Topics: