Thermal Physics

**Introduction & Core Concept**

Assalam-o-Alaikum, my dear students and future leaders of Pakistan. Dr. Amir Hussain here, welcoming you to our lesson on Thermal Physics at SeekhoAsaan.com.

Imagine this: It’s a blistering hot June afternoon in Lahore. You've just parked your father's car, and you hear a distinct hissing sound. The tyre, which was perfectly fine this morning, has burst. Why? Or consider a pressure cooker in a Karachi kitchen, whistling away as it prepares a delicious daal. How does it cook the food so much faster than an open pot? The answers to these everyday Pakistani experiences lie not in magic, but in the elegant principles of Thermal Physics.

This topic is the bridge between two worlds. On one side, we have the macroscopic world we can see and measure: the pressure in a tyre, the volume of a balloon, the temperature of a cup of chai. On the other side is the microscopic world of countless, invisible atoms and molecules, zipping around at incredible speeds, colliding with each other and the walls of their container.

The big-picture mental model is this: The observable properties of a substance (like gas pressure) are simply the large-scale average effects of the chaotic, random behaviour of its microscopic particles.

Understanding this link is not just for passing your A Level exams. It is the foundation of thermodynamics, which is crucial for every mechanical engineer designing engines for WAPDA's power plants, every chemical engineer optimising processes in the Fauji Fertilizer Company, and every meteorologist forecasting the monsoon rains over the Indus River basin. So, let's build this bridge together, one concept at a time.

**Theoretical Foundation**

Let's lay down the bricks of our understanding, starting from the very core concepts and building up to the powerful equations that govern this field.

#### 1. Internal Energy (U)

Every object around you, from the steel in the Minar-e-Pakistan to the air in your room, contains energy. This total energy stored within a substance is its internal energy, denoted by the symbol `U`. It is the sum of the energies of all the individual particles (atoms or molecules) that make up the system.

This energy comes in two fundamental forms:

* Random Kinetic Energy (KE): This is the energy of motion. The particles in a substance are never still (unless at absolute zero, -273.15 °C). They move, they vibrate, they rotate. This motion is *random* – the particles are moving in all directions with a wide range of speeds. This is different from the *ordered* kinetic energy an object has when the entire object is moving (like a cricket ball flying towards the stumps). Internal energy is only concerned with the *random*, internal motion. Crucially, the average random kinetic energy of the particles is what we measure as temperature. Higher temperature means faster average particle motion.

* Random Potential Energy (PE): This is the energy stored in the bonds and forces between particles. Think of it like the energy stored in a stretched spring. In solids and liquids, particles are close together and are held by significant intermolecular forces (like electrostatic attractions). Work must be done to pull them apart, which means they possess potential energy due to their relative positions.

So, the full definition is: `U = ΣKE_random + ΣPE_random`

#### 2. The Ideal Gas: A Perfected Model

Real gases, like the natural gas from the Sui fields, are complex. Their molecules attract each other and take up space. To make the physics manageable, we create a simplified model called an ideal gas. It's a theoretical gas that obeys certain rules perfectly. Real gases behave very much like ideal gases under most conditions (i.e., at high temperatures and low pressures).

The assumptions of the kinetic theory for an ideal gas are fundamental and you *must* know them:

1. Large Number of Particles: The gas consists of a very large number of identical molecules, `N`.
2. Random Motion: The molecules move randomly and rapidly in all directions (Newtonian motion).
3. Negligible Volume of Molecules: The volume of the molecules themselves is negligible compared to the volume of the container they occupy. They are treated as point masses.
4. No Intermolecular Forces: There are no attractive or repulsive forces between the molecules, except during collisions. This means the potential energy component of the internal energy is zero.
5. Elastic Collisions: All collisions (between molecules and with the container walls) are perfectly elastic. This means kinetic energy and momentum are conserved.
6. Negligible Collision Time: The time taken for a collision is negligible compared to the time between collisions.

The most important consequence of these assumptions? For an ideal gas, since there are no intermolecular forces (Assumption 4), the potential energy is zero. Therefore, its internal energy consists *entirely* of the random kinetic energy of its molecules.

For an ideal gas: `U = ΣKE_random`

Since temperature is a measure of the average KE of the particles, this leads to a profound conclusion for ideal gases: The internal energy of a fixed amount of an ideal gas depends *only* on its temperature. `U ∝ T`. If you increase its temperature, you increase its internal energy. It doesn't matter if you change its pressure or volume.

#### 3. The Ideal Gas Equation

This equation links the macroscopic properties of an ideal gas. It's one of the most important formulae in this topic.

`pV = nRT`

* `p` = pressure of the gas (in Pascals, Pa)

* `V` = volume of the gas (in cubic metres, m³)

* `n` = number of moles of the gas (in mol)

* `R` = the universal gas constant (8.31 J K⁻¹ mol⁻¹)

* `T` = absolute temperature of the gas (in Kelvin, K)

There is an alternative form that works with the number of molecules instead of moles:

`pV = NkT`

* `N` = number of molecules

* `k` = the Boltzmann constant (1.38 x 10⁻²³ J K⁻¹).

The two constants are related by Avogadro's constant, `N_A` (6.02 x 10²³ mol⁻¹): `R = N_A * k`. The Boltzmann constant is essentially the gas constant *per molecule*, while R is the gas constant *per mole*.

A crucial note on temperature: You **must** use Kelvin for all gas law calculations. The Kelvin scale is the absolute temperature scale, where 0 K represents the theoretical point of zero internal energy (no particle motion). To convert from Celsius (°C) to Kelvin (K): `K = °C + 273.15` (for A-Level, using 273 is usually sufficient).

#### 4. The Kinetic Theory Derivation: The Heart of the Matter

Now, we will derive the equation that links the microscopic world of molecules to the macroscopic world of pressure. This derivation is a classic and demonstrates the power of physical modelling.

Imagine a single molecule of mass `m` inside a perfect cube-shaped box of side length `L`. Its velocity `c` can be split into components in the x, y, and z directions: `c_x`, `c_y`, `c_z`.

1. Collision with a Wall: Let's focus on the molecule's motion in the x-direction. It travels towards the right-hand wall with velocity `c_x`. Its initial momentum is `m*c_x`.
2. Change in Momentum: It collides elastically with the wall and bounces back with velocity `-c_x`. Its final momentum is `-m*c_x`. The change in momentum (`Δp`) is `final momentum - initial momentum = (-mc_x) - (mc_x) = -2mc_x`. By Newton's third law, the momentum imparted *to the wall* is `+2mc_x`.
3. Time Between Collisions: To hit the same wall again, the molecule must travel to the opposite wall (a distance `L`) and back again (another distance `L`). The total distance is `2L`. The time taken (`Δt`) is `distance / speed = 2L / c_x`.
4. Force Exerted by One Molecule: According to Newton's second law, force is the rate of change of momentum.

`Force (F_1) = Δp / Δt = (2mc_x) / (2L / c_x) = mc_x² / L`.

This is the average force exerted by *one* molecule on that one wall.

1. Pressure from One Molecule: Pressure is force per unit area (`P = F/A`). The area of the wall is `L²`.

`Pressure (P_1) = F_1 / A = (mc_x² / L) / L² = mc_x² / L³`.

Since `L³` is the volume `V` of the box, `P_1 = mc_x² / V`.

1. Generalising to N Molecules: Now, let's consider all `N` molecules in the box. They all have different x-velocities. The total pressure `p` will be the sum of the pressures from each molecule.

`p = (m/V) * (c_{x1}² + c_{x2}² + ... + c_{xN}²)`.

We can simplify this using the concept of the mean (average) square velocity in the x-direction, ``.

` = (c_{x1}² + c_{x2}² + ... + c_{xN}²) / N`.

So, `(c_{x1}² + c_{x2}² + ... + c_{xN}²) = N * `.

Substituting this back, we get: `p = (m/V) * N * = (Nm/V) * `.

1. Connecting to 3D Motion: The molecules are moving randomly in three dimensions. The total speed `c` is related to its components by Pythagoras' theorem in 3D: `c² = c_x² + c_y² + c_z²`.

Taking the average for all molecules: ` = + + `.

Since the motion is random, there is no preferred direction. Therefore, the average motion in each direction is the same: ` = = `.

This means we can write ` = 3 * `, or the crucial step: ` = (1/3) * `.

1. The Final Equation: Substitute this final piece into our pressure equation:

`p = (Nm/V) * (1/3) * `.

Rearranging this gives the famous Kinetic Theory Equation:

`pV = (1/3) * N * m * `

Here, `` is the mean square speed of the molecules. Its square root is called the root-mean-square speed (c_rms).

#### 5. The Grand Unification: Linking Energy and Temperature

Now we perform the most beautiful step. We have two equations for `pV`:

1. From the ideal gas law: `pV = NkT`
2. From kinetic theory: `pV = (1/3) * N * m * `

Let's equate them:

`NkT = (1/3) * N * m * `

The `N` on both sides cancels out. Now, let's rearrange to make it look like the kinetic energy formula (`(1/2)mv²`).

Multiply both sides by 3: `3kT = m`

Divide both sides by 2: `(3/2)kT = (1/2)m`

This is a monumental result. The right-hand side, `(1/2)m`, is the average translational kinetic energy of a single gas molecule. The left-hand side contains the Boltzmann constant `k` and the absolute temperature `T`.

This equation tells us, mathematically, that the absolute temperature of an ideal gas is directly proportional to the average random translational kinetic energy of its molecules. This is the microscopic definition of temperature and the ultimate goal of our derivation.

**Key Definitions & Formulae**

Here is a summary of the essential tools for your problem-solving toolkit.

| Concept | Formula | Symbol Definitions | Units |

| :--- | :--- | :--- | :--- |

| Ideal Gas Law | `pV = nRT` | `p`: pressure
`V`: volume
`n`: number of moles
`R`: universal gas constant (8.31)
`T`: absolute temperature | Pa
m³
mol
J K⁻¹ mol⁻¹
K |

| Ideal Gas Law (Molecular) | `pV = NkT` | `p`: pressure
`V`: volume
`N`: number of molecules
`k`: Boltzmann constant (1.38x10⁻²³)
`T`: absolute temperature | Pa
m³
(dimensionless)
J K⁻¹
K |

| Kinetic Theory Equation | `pV = (1/3)N m ` | `p`: pressure
`V`: volume
`N`: number of molecules
`m`: mass of one molecule
``: mean square speed | Pa
m³
(dimensionless)
kg
m² s⁻² |

| Mean KE of a Molecule | `E_k = (1/2)m = (3/2)kT` | `E_k`: average translational KE
`m`: mass of one molecule
``: mean square speed
`k`: Boltzmann constant
`T`: absolute temperature | J
kg
m² s⁻²
J K⁻¹
K |

| Internal Energy (Ideal Gas) | `U = (3/2)NkT` or `U = (3/2)nRT` | `U`: total internal energy
*(for a monatomic ideal gas)* | J |

Dimensional Analysis Check: Let's verify that `kT` has units of energy (Joules).

Units of `k` are J K⁻¹. Units of `T` are K.

So, units of `kT` are (J K⁻¹) * (K) = J. It works perfectly.

**Worked Examples**

Theory is one thing, but applying it is where true understanding is forged. Let's solve some problems with a Pakistani flavour.

Example 1: The Rickshaw's CNG Tank

A rickshaw driver in Karachi gets his CNG tank filled. The tank has a volume of 25 litres (0.025 m³) and is filled with methane (molar mass = 16 g/mol) to a pressure of 2.0 x 10⁷ Pa. The temperature on that day is a scorching 40 °C.

(a) Calculate the number of moles (`n`) of methane in the tank.

(b) Calculate the mass of the methane gas.

Solution:

(a) Finding the number of moles

* Step 1: Identify the correct formula. We have p, V, T and need to find n. The ideal gas law `pV = nRT` is the perfect tool.

* Step 2: Convert all units to SI.

* `p = 2.0 x 10⁷ Pa` (already in SI)

* `V = 0.025 m³` (already in SI)

* `T = 40 °C`. We MUST convert to Kelvin. `T = 40 + 273 = 313 K`.

* `R = 8.31 J K⁻¹ mol⁻¹` (a constant)

* Step 3: Rearrange and solve for n.

`pV = nRT => n = pV / RT`

`n = (2.0 x 10⁷ Pa * 0.025 m³) / (8.31 J K⁻¹ mol⁻¹ * 313 K)`

`n = 500,000 / 2601.03`

`n = 192.2 mol`

* Step 4: State the answer with appropriate significant figures. The input values have 2 s.f., so our answer should too.

`n ≈ 190 mol`

(b) Finding the mass

* Step 1: Recall the relationship between moles, mass, and molar mass.

`moles (n) = mass (m_total) / molar mass (M)`

* Step 2: Use the molar mass given. Molar mass of methane is 16 g/mol, which is 0.016 kg/mol in SI units.

* Step 3: Rearrange and solve for mass.

`m_total = n * M`

`m_total = 192.2 mol * 0.016 kg/mol` (using the unrounded value for n for better accuracy)

`m_total = 3.075 kg`

* Step 4: State the final answer.

The mass of the methane gas is approximately `3.1 kg`.

Example 2: Speed of Air Molecules at a Cricket Match

During a day/night cricket match at Gaddafi Stadium in Lahore, the air temperature is a pleasant 27 °C. Air is approximately 80% nitrogen. Calculate the root-mean-square speed (c_rms) of a nitrogen molecule (`N₂`) under these conditions. (Molar mass of nitrogen = 28 g/mol; Avogadro's constant `N_A` = 6.02 x 10²³ mol⁻¹)

Solution:

* Step 1: Find the mass of a single nitrogen molecule (`m`).

`m = Molar Mass / Avogadro's Constant`

First, convert molar mass to kg/mol: `28 g/mol = 0.028 kg/mol`.

`m = 0.028 kg/mol / 6.02 x 10²³ mol⁻¹ = 4.65 x 10⁻²⁶ kg`.

* Step 2: Relate temperature to kinetic energy. We use the key equation:

`(1/2)m = (3/2)kT`

* Step 3: Convert temperature to Kelvin.

`T = 27 °C + 273 = 300 K`.

* Step 4: Rearrange the formula to solve for ``.

The (1/2) on both sides cancels: `m = 3kT`

` = 3kT / m`

* Step 5: Substitute the values and calculate ``.

` = (3 * 1.38 x 10⁻²³ J K⁻¹ * 300 K) / (4.65 x 10⁻²⁶ kg)`

` = (1.242 x 10⁻²⁰) / (4.65 x 10⁻²⁶)`

` = 267,096 m² s⁻²`

* Step 6: Calculate the r.m.s. speed. The r.m.s. speed is the square root of the mean square speed.

`c_rms = √ = √267,096`

`c_rms = 516.8 m/s`

* Step 7: Final answer with sig figs.

`c_rms ≈ 520 m/s`.

This is incredibly fast – over 1800 km/h! This is the typical speed of the molecules that make up the air we breathe.

Example 3: Internal Energy of Helium

A PTCL weather balloon is filled with 50 moles of helium gas (which behaves as a monatomic ideal gas). Before release from the ground station in Quetta, its temperature is 15 °C. What is the total internal energy of the helium in the balloon?

Solution:

* Step 1: Identify the core concept. For a monatomic ideal gas, internal energy `U` is the sum of the kinetic energies of all its atoms.

`U = N * (average KE per atom) = N * (3/2)kT`.

* Step 2: Choose the most convenient formula. We are given moles (`n`), not number of molecules (`N`). We can use the relationship `N = n * N_A` and `R = N_A * k`.

`U = (n * N_A) * (3/2)kT = n * (3/2) * (N_A * k) * T = (3/2)nRT`. This is a very useful form.

* Step 3: Convert temperature to Kelvin.

`T = 15 °C + 273 = 288 K`.

* Step 4: Substitute the values and calculate U.

`U = (3/2) * 50 mol * 8.31 J K⁻¹ mol⁻¹ * 288 K`

`U = 1.5 * 50 * 8.31 * 288`

`U = 179,496 J`

* Step 5: Final answer.

The total internal energy of the helium is approximately `1.8 x 10⁵ J` or `180 kJ`.

**Visual Mental Models**

Sometimes, a picture is worth a thousand equations. Use these mental models to visualise the concepts.

1. The Beehive in a Box:

Imagine a glass box filled with angry bees (our molecules). They are moving chaotically, in every direction. The temperature is like the average "agitation" or speed of the bees. The pressure is the constant, collective force of the bees hitting the walls. If you heat the box (increase temperature), the bees become more agitated, fly faster, and hit the walls harder and more often, increasing the pressure.

+-----------------+

| . . -> . |

| . <- . . |

| . . . | <-- Molecules moving randomly,

| . -> . <- . | exerting pressure on walls.

| . . . |

+-----------------+

1. The p-V Graph:

This is a vital tool in thermodynamics. For a fixed amount of gas at a constant temperature (an *isothermal* process), the ideal gas law `pV = nRT` tells us `p ∝ 1/V`. This creates a hyperbolic curve on a pressure-volume graph. Higher temperature curves (isotherms) are further away from the origin because for the same volume, the pressure will be higher.

^ Pressure (p)

|

| T₂ > T₁

| /

| / .-- T₂

| / .----

| / .--- T₁

| +----------------> Volume (V)

1. The Maxwell-Boltzmann Distribution Curve:

This graph shows that not all molecules in a gas have the same speed. Some are slow, some are fast, and most are somewhere in the middle. The peak of the curve represents the most probable speed. As you increase the temperature:

* The curve flattens and spreads out, meaning there is a wider range of speeds.

* The peak (most probable speed) and the average speed shift to the right (higher speeds).

^ Number of Molecules

|

| /| \

| / | \ /| \

| / | \ / | \

| / | \_____/ | \___

|/____|_____\__/____|_____\__> Speed

T₁ T₂ (T₂ > T₁)

This is a more sophisticated (A* level) model, showing that temperature is about the *average* KE, but individual molecular energies vary greatly.

**Common Mistakes & Misconceptions**

Many students stumble on the same points. Let's clear them up before you make these mistakes.

1. Confusing Heat and Temperature.

* Mistake: "This cup of chai has a lot of heat."

* Why it's wrong: Temperature is a measure of the average internal kinetic energy of particles (a property of the system). Heat is the energy *transferred* from a hotter object to a colder one. The chai has a high temperature and possesses high internal energy. It *transfers* heat to the cooler air.

* Correct thinking: Temperature is a state; heat is a process of energy transfer.

1. Forgetting to Convert to Kelvin.

* Mistake: Using `T = 20 °C` in the `pV = nRT` equation.

* Why it's wrong: The ideal gas law is based on an absolute scale where pressure and volume would theoretically be zero at 0 K. Using Celsius, which has an arbitrary zero point, will give completely wrong answers. A change from 10°C to 20°C is not a doubling of temperature in the physical sense.

* Correct thinking: ALWAYS convert temperatures to Kelvin (K = °C + 273) before they touch any gas law formula.

1. Misunderstanding Internal Energy of an Ideal Gas.

* Mistake: "If I compress an ideal gas isothermally (constant temperature), its internal energy decreases because the particles are closer."

* Why it's wrong: For an ideal gas, we assume there are *no intermolecular forces*. Therefore, the distance between particles is irrelevant to its energy. The internal energy of an ideal gas depends *only* on its temperature.

* Correct thinking: If temperature `T` is constant, internal energy `U` is constant. Period. `ΔT = 0` implies `ΔU = 0` for a fixed amount of an ideal gas.

1. Pressure is Caused by Molecules Hitting Each Other.

* Mistake: "The pressure comes from all the molecules colliding with each other."

* Why it's wrong: While molecules do collide with each other, this simply redistributes energy among them. The macroscopic pressure we measure is the result of countless molecules colliding with the walls of the container, imparting momentum and thus exerting a force.

* Correct thinking: Pressure = Force on walls / Area of walls.

1. Root-Mean-Square Speed Confusion.

* Mistake: Calculating the average speed `(c₁ + c₂ + ...)/N` and using that in the kinetic theory equation.

* Why it's wrong: The derivation shows that pressure is related to the *square* of the velocity (`mc_x²/L`). Therefore, we must average the *squares* of the speeds, giving the mean square speed ``. The `c_rms` (`√`) is a type of statistical average that correctly relates to the kinetic energy of the gas. It's not the same as the simple mean speed.

* Correct thinking: Energy is proportional to `c²`, so we must average `c²`, not `c`.

**Exam Technique & Mark Scheme Tips**

Let's talk about how to impress the Cambridge examiners and secure those top marks.

1. Master the Command Words:

* State: Give a concise term or a short sentence. E.g., "State two assumptions of the kinetic theory." You would write: "1. No intermolecular forces. 2. Volume of molecules is negligible." No explanation needed.

* Describe: Give a step-by-step account. E.g., "Describe the motion of molecules in an ideal gas." You would write: "Molecules move rapidly and randomly in straight lines between collisions. They collide elastically with each other and the container walls."

* Explain: This requires a reason. Use words like "because," "therefore," "as a result of." E.g., "Explain, using the kinetic theory, why pressure increases with temperature at constant volume." A good answer: "An increase in temperature means the average kinetic energy of the molecules increases (1). Therefore, molecules move faster and collide with the container walls more frequently and more forcefully (1). Since pressure is the force per unit area on the walls, the pressure increases (1)." (Note the marks for each point).

1. Show All Your Working: In calculation questions, marks are awarded for the method, not just the final answer.

* Write the formula you are using (`pV = nRT`).

* Show the substitution of values, including units.

* Give the final answer with the correct unit and to an appropriate number of significant figures (usually 2 or 3, matching the data in the question).

1. Assumptions are Free Marks: Questions asking for the assumptions of the kinetic theory are very common. Memorise them perfectly. Don't be vague. "Molecules are small" is not as good as "The volume of the molecules is negligible compared to the volume of the container."

1. The Derivation Steps: If asked to derive `pV = (1/3)Nm`, examiners look for key logical steps. Marks are often allocated for:

* Stating the change in momentum for a single collision (`2mc_x`).

* Calculating the time between collisions on the same wall (`2L/c_x`).

* Combining these to find the force from one molecule.

* The crucial averaging step: relating `` to `` using ` = (1/3)`.

1. Watch Out for Traps:

* Units: Pressure in kPa? Volume in cm³ or litres? Molar mass in g/mol? Always convert to base SI units (Pa, m³, kg/mol) before calculating.

* Diatomic vs. Monatomic: The formula `E_k = (3/2)kT` is for the *translational* kinetic energy. For simple monatomic gases like Helium or Neon, this is the total internal energy. Diatomic gases like Nitrogen (`N₂`) can also have rotational kinetic energy, but for the A-Level syllabus, you primarily focus on the translational part and the monatomic ideal gas model.

**Memory Tricks & Mnemonics**

Your brain is a muscle; let's give it some tools to lift these heavy concepts.

1. Ideal Gas Assumptions (PV = nRT): Think of a perfect, ideal student named "RANDI".

* Random motion

* Attractions are zero (no intermolecular forces)

* Negligible volume

* Duration of collision is negligible

* Ideal (Elastic) collisions

1. Ideal Gas Law Formula:

* `pV = nRT` -> "Pakistani Vehicles = need Really good Tyres"

1. Kinetic Energy Formula:

* `E_k = (3/2)kT` -> "Energy in a 3D world is 3-over-2 kT". This links the 3 dimensions of space to the numerator 3.

1. The Two Gas Constants:

* `R` is the "Regal" or "Royal" constant – it's big (8.31) and works with moles, which are large quantities.

* `k` is the "kursi" (chair) constant – it's tiny (10⁻²³) and works with individual molecules, which sit on their own little chairs.

**Pakistan & Everyday Connections**

Let's bring the physics from the textbook right into your daily life here in Pakistan.

1. The Pressure Cooker Principle: Why does your mother's pressure cooker make daal or paye so quickly? It's a direct application of the ideal gas law. By sealing the lid, the volume `V` is kept constant. As the stove provides heat, the temperature `T` of the water vapour (steam) inside increases. Since `p = (nR/V)T`, pressure `p` is directly proportional to the absolute temperature `T`. This high pressure (up to twice the atmospheric pressure) raises the boiling point of water from 100°C to about 120°C. The food is now cooking at a much higher temperature, so chemical reactions happen faster, and the food gets tender in a fraction of the time.

1. Karachi's Sea Breeze: This is a magnificent example of thermal physics at work. During the day, the land (sand, concrete) heats up much faster than the Arabian Sea. The air above the land gets hot, its molecules gain kinetic energy, move faster, and spread out, making the air less dense. This hot, less dense air rises. To fill the low-pressure area created, the cooler, denser air from over the sea moves inland. This is the famous Karachi sea breeze that provides relief in the afternoon. At night, the process reverses as the land cools faster than the sea.

1. Punctures and Aerosol Cans: Have you ever noticed that when air rushes out of a punctured tyre, the escaping air feels cold? This is because the gas is expanding rapidly and doing work on the surrounding atmosphere. This work requires energy, which is taken from the internal energy of the gas itself. As the internal energy `U` drops, the temperature `T` also drops (`U ∝ T`). The same principle applies when you use a deodorant spray can – the can gets noticeably colder as the pressurised gas expands.

**Practice Problems**

Test your newfound knowledge with these exam-style questions.

Question 1 (Knowledge):

State four of the fundamental assumptions of the kinetic theory of ideal gases.

* Answer Outline: List any four from the "RANDI" mnemonic or the list in the theory section. Be precise. E.g., (1) No intermolecular forces of attraction or repulsion. (2) The volume of the molecules is negligible compared to the container volume. (3) Collisions are perfectly elastic. (4) Motion is random.

Question 2 (Calculation):

A car tyre on a Suzuki Alto has a volume of 0.030 m³ and is inflated to a gauge pressure of 210 kPa in Sialkot, where the morning temperature is 17 °C. Atmospheric pressure is 100 kPa. By the afternoon, the temperature rises to 37 °C. Calculate the new gauge pressure of the tyre, assuming its volume remains constant.

* Answer Outline:

1. Calculate the initial absolute pressure: `P₁ = Gauge + Atmospheric = 210 + 100 = 310 kPa`.
2. Convert temperatures to Kelvin: `T₁ = 17 + 273 = 290 K`, `T₂ = 37 + 273 = 310 K`.
3. Use the pressure law (a form of `pV=nRT`): `P₁/T₁ = P₂/T₂`.
4. Solve for `P₂`: `P₂ = P₁ * (T₂/T₁) = 310 kPa * (310/290) = 331.4 kPa`.
5. Convert back to gauge pressure: `Gauge P₂ = Absolute P₂ - Atmospheric = 331.4 - 100 = 231.4 kPa`.
6. Final answer to 2 s.f.: `230 kPa`.

Question 3 (Explanation):

A sealed container of an ideal gas is heated. Use the kinetic theory to explain, in terms of molecular motion, the change in the pressure of the gas.

* Answer Outline:

1. Heating increases the internal energy of the gas, which for an ideal gas is the total kinetic energy of the molecules.
2. This means the average speed of the molecules increases.
3. As a result, the molecules collide with the container walls more frequently.
4. Furthermore, the change in momentum during each collision is greater (as they are faster).
5. Since force is the rate of change of momentum, the average force on the walls increases.
6. As pressure is force per unit area, the pressure increases.

Question 4 (Application & Calculation):

Oxygen (`O₂`, molar mass 32 g/mol) and Helium (`He`, molar mass 4 g/mol) gases are mixed together in a container at thermal equilibrium.

(a) State and explain which gas molecules have a higher average kinetic energy.

(b) Calculate the ratio of the r.m.s. speed of a helium molecule to that of an oxygen molecule.

* Answer Outline:

(a) "Thermal equilibrium" means they are at the same temperature `T`. Since the average kinetic energy of a molecule is given by `(3/2)kT`, it depends *only* on temperature. Therefore, both types of molecules have the same average kinetic energy.

(b) We know `(1/2)m` is the same for both. So, `(1/2)m_He * = (1/2)m_O₂ * `.

This simplifies to `m_He / m_O₂ = / `.

The ratio of r.m.s. speeds is `c_rms(He) / c_rms(O₂) = √() / √() = √(m_O₂ / m_He)`.

The ratio of molecular masses is the same as the ratio of molar masses: `32 / 4 = 8`.

So, the ratio of speeds is `√8 ≈ 2.83`. Helium molecules move, on average, 2.83 times faster than oxygen molecules at the same temperature.