Ideal Gas Law pV = nRT, Boyle's Law, Root Mean Square Speed & Kinetic Theory Assumptions
Thermal physics links macroscopic properties of matter, temperature T, pressure p, volume V, and internal energy U - directly to the microscopic random kinetic and potential energies of atoms and molecules. While heating a solid or liquid without a state change raises average molecular kinetic energy (Q = mcΔT), heating at the melting or boiling point breaks intermolecular bonds to increase molecular potential energy at constant temperature (Q = ml). For an ideal gas (where intermolecular potential energy is zero), combining Boyle's, Charles's, and the Pressure Laws yields pV = nRT = NkT, which kinetic theory derives from Newton's laws of momentum change during elastic wall collisions as pV = ⅓Nm(c_rms)². In particular, equating these two expressions proves that the mean translational kinetic energy of a gas molecule depends solely on absolute Kelvin temperature: ½m(c_rms)² = (3/2)kT.
- Internal Energy, Specific Heat Capacity (Q = mcΔT) & Specific Latent Heat (Q = ml)
- Internal energy (U) of a body is the sum of the randomly distributed kinetic energies and potential energies of all its particles. Absolute Zero (0 K = -273.15°C) is the minimum possible temperature at which particles have minimum internal energy (zero kinetic energy). 1. Temperature change without phase change (kinetic energy changes, potential energy constant): Q = mcΔT, where c is specific heat capacity (energy needed to raise 1 kg by 1 K, in J kg⁻¹ K⁻¹); for electrical heating over time t: VIt = mcΔT. 2. Phase change at constant temperature (potential energy increases as bonds break, kinetic energy and T stay constant): Q = ml, where l is specific latent heat of fusion (solid ↔ liquid) or vaporisation (liquid ↔ gas) in J kg⁻¹.
- Empirical Gas Laws (Boyle's, Charles's, Pressure Law) & Absolute Zero
- For a fixed mass of ideal gas: 1. Boyle's Law (constant T): pressure is inversely proportional to volume (p ∝ 1/V, or p₁V₁ = p₂V₂; p vs 1/V is a straight line through the origin; work done at constant pressure W = pΔV). 2. Charles's Law (constant p): volume is directly proportional to absolute temperature T in Kelvin (V ∝ T, or V₁/T₁ = V₂/T₂). 3. Pressure Law (constant V): pressure is directly proportional to absolute temperature T in Kelvin (p ∝ T, or p₁/T₁ = p₂/T₂). Plotting p (or V) against temperature in °C gives a straight line whose x-intercept extrapolates back to Absolute Zero at -273°C (0 K) where gas pressure/volume reaches zero.
- The ideal gas equation: pV = nRT = NkT
- Combining the three gas laws gives two equivalent forms of the Ideal Gas Equation: 1. Molar form: pV = nRT, where p is pressure in Pascals (Pa = N m⁻²), V is volume in m³, n = m / M_r is number of moles, R = 8.31 J K⁻¹ mol⁻¹ is the molar gas constant, and T is absolute temperature in Kelvin (K = °C + 273). 2. Molecular form: pV = NkT, where N = n × N_A is the total number of gas molecules and k = R / N_A = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant.
Ideal Gas Law pV = nRT, Boyle's Law, Root Mean Square Speed & Kinetic Theory Assumptions: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Key Concepts & Exam Definitions (6)
- Internal Energy, Specific Heat Capacity (Q = mcΔT) & Specific Latent Heat (Q = ml)
- Internal energy (U) of a body is the sum of the randomly distributed kinetic energies and potential energies of all its particles. Absolute Zero (0 K = -273.15°C) is the minimum possible temperature at which particles have minimum internal energy (zero kinetic energy). 1. Temperature change without phase change (kinetic energy changes, potential energy constant): Q = mcΔT, where c is specific heat capacity (energy needed to raise 1 kg by 1 K, in J kg⁻¹ K⁻¹); for electrical heating over time t: VIt = mcΔT. 2. Phase change at constant temperature (potential energy increases as bonds break, kinetic energy and T stay constant): Q = ml, where l is specific latent heat of fusion (solid ↔ liquid) or vaporisation (liquid ↔ gas) in J kg⁻¹.
- Empirical Gas Laws (Boyle's, Charles's, Pressure Law) & Absolute Zero
- For a fixed mass of ideal gas: 1. Boyle's Law (constant T): pressure is inversely proportional to volume (p ∝ 1/V, or p₁V₁ = p₂V₂; p vs 1/V is a straight line through the origin; work done at constant pressure W = pΔV). 2. Charles's Law (constant p): volume is directly proportional to absolute temperature T in Kelvin (V ∝ T, or V₁/T₁ = V₂/T₂). 3. Pressure Law (constant V): pressure is directly proportional to absolute temperature T in Kelvin (p ∝ T, or p₁/T₁ = p₂/T₂). Plotting p (or V) against temperature in °C gives a straight line whose x-intercept extrapolates back to Absolute Zero at -273°C (0 K) where gas pressure/volume reaches zero.
- The ideal gas equation: pV = nRT = NkT
- Combining the three gas laws gives two equivalent forms of the Ideal Gas Equation: 1. Molar form: pV = nRT, where p is pressure in Pascals (Pa = N m⁻²), V is volume in m³, n = m / M_r is number of moles, R = 8.31 J K⁻¹ mol⁻¹ is the molar gas constant, and T is absolute temperature in Kelvin (K = °C + 273). 2. Molecular form: pV = NkT, where N = n × N_A is the total number of gas molecules and k = R / N_A = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant.
- Kinetic Theory Equation (pV = ⅓Nm(c_rms)²) & Derivation Steps
- Relates macroscopic pV to microscopic molecular motion: pV = ⅓Nm(c_rms)² (or p = ⅓ρ(c_rms)² since density ρ = Nm / V), where m is the mass of one gas molecule in kg, and c_rms = √[(c₁² + c₂² + . + c_N²) / N] is the root-mean-square speed. Derivation for a cube of side L: 1. One molecule of mass m and x-velocity u collides elastically with wall of area L²: momentum change Δp = (-mu) - (+mu) = -2mu (or magnitude 2mu). 2. Time between collisions with the same wall is Δt = 2L / u. 3. By Newton's 2nd Law, average force on molecule = Δp / Δt = 2mu ÷ (2L / u) = mu² / L (equal and opposite force on wall by Newton's 3rd Law). 4. Pressure from N molecules = Force / Area = Nm(u_mean²) / L³ = Nm(u_mean²) / V. 5. By 3D Pythagoras (c² = u² + v² + w²) and isotropic motion (u_mean² = v_mean² = w_mean² = ⅓(c_rms)²), pV = ⅓Nm(c_rms)².
- Five Assumptions of the Kinetic Theory of an Ideal Gas
- Memorise at least four of the five classic exam assumptions: 1. Gas molecules move in continuous, rapid, random motion. 2. The total volume of the gas molecules themselves is negligible compared to the volume of the container. 3. There are NO intermolecular forces of attraction or repulsion between molecules except during collisions (so all internal energy of an ideal gas is purely translational kinetic energy; zero potential energy). 4. All collisions between molecules and with container walls are perfectly elastic (no loss of kinetic energy). 5. The duration of a collision is negligible compared to the time between collisions.
- Average molecular kinetic energy: ½m(c_rms)² = (3/2)kT = (3RT) / (2N_A)
- Equating pV = NkT and pV = ⅓Nm(c_rms)² cancels N and gives: ½m(c_rms)² = (3/2)kT = (3RT) / (2N_A). Key implications: 1. The mean translational kinetic energy of a gas molecule is directly proportional to absolute temperature T in Kelvin only (at the same temperature, a heavy Xenon atom and a light Helium atom have the exact same average kinetic energy, though Helium has a much higher c_rms). 2. Total internal energy of n moles of an ideal monatomic gas is U = N × (3/2)kT = (3/2)nRT = (3/2)pV.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Why does the temperature of boiling water remain constant at 100°C even though thermal energy is continuously supplied?
Explain using Newton's laws of motion how a gas at constant volume exerts a higher pressure on the walls of its container when heated.
State four assumptions of the kinetic theory model of an ideal gas.
Under what conditions of temperature and pressure do real gases deviate from ideal gas behaviour, and why?
A gas mixture at 300 K contains Hydrogen (M_r = 2) and Oxygen (M_r = 32). Compare the mean kinetic energy and root-mean-square speed (c_rms) of the H₂ and O₂ molecules.
What is Brownian motion, what is observed under a microscope with smoke particles in air, and what does it prove?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
An electric kettle of power 2.40 kW contains 0.850 kg of water at an initial temperature of 18.0°C. Assuming 92.0% of the electrical energy is transferred to the water, calculate the time taken for the water to reach its boiling point of 100.0°C, and the additional time required to boil away 0.150 kg of the water into steam at 100.0°C. (Specific heat capacity of water c = 4200 J kg⁻¹ K⁻¹; specific latent heat of vaporisation of water l_v = 2.26 × 10⁶ J kg⁻¹). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate useful thermal power delivered to the water: P_useful = 0.920 × 2400 W = 2208 W [1 mark]. 2. Thermal energy needed to heat 0.850 kg of water from 18.0°C to 100.0°C (ΔT = 82.0 K): Q₁ = mcΔT = 0.850 × 4200 × 82.0 = 2.927 × 10⁵ J [1 mark]. 3. Time to reach 100.0°C: t₁ = Q₁ ÷ P_useful = (2.927 × 10⁵ J) ÷ 2208 W = 133 s [1 mark]. 4. Thermal energy needed to vaporise 0.150 kg of water at 100.0°C: Q₂ = m_s l_v = 0.150 kg × 2.26 × 10⁶ J kg⁻¹ = 3.39 × 10⁵ J [1 mark]. 5. Additional time to boil off 0.150 kg: t₂ = Q₂ ÷ P_useful = (3.39 × 10⁵ J) ÷ 2208 W = 154 s [1 mark].
- Q2
A scuba diving cylinder of internal volume 0.0150 m³ contains 0.360 kg of helium gas (molar mass = 4.00 × 10⁻³ kg mol⁻¹) at a temperature of 22.0°C. Calculate: (a) the number of moles and number of helium atoms in the cylinder, (b) the pressure of the helium gas in MPa, and (c) the root-mean-square speed (c_rms) of the helium atoms. (R = 8.31 J K⁻¹ mol⁻¹, N_A = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ J K⁻¹). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Number of moles n = m ÷ M_r = 0.360 kg ÷ (4.00 × 10⁻³ kg mol⁻¹) = 90.0 mol; Number of atoms N = n × N_A = 90.0 × 6.02 × 10²³ = 5.42 × 10²⁵ atoms [1 mark]. 2. (b) Convert temperature to Kelvin: T = 22.0 + 273.15 = 295.15 K (295 K) [1 mark]. 3. Use pV = nRT to find pressure p: p = (90.0 × 8.31 × 295.15) ÷ 0.0150 = 1.47 × 10⁷ Pa = 14.7 MPa [1 mark]. 4. (c) Mass of a single helium atom m_atom = (4.00 × 10⁻³ kg mol⁻¹) ÷ (6.02 × 10²³ mol⁻¹) = 6.645 × 10⁻²⁷ kg [1 mark]. 5. Use ½m(c_rms)² = (3/2)kT (or pV = ⅓Nm(c_rms)²): c_rms = √(3kT ÷ m_atom) = √(3 × 1.38 × 10⁻²³ × 295.15 ÷ 6.645 × 10⁻²⁷) = √(1.839 × 10⁶) = 1.36 × 10³ m s⁻¹ (1356 m s⁻¹) [1 mark].
- Q3
Starting from the consideration of a single gas molecule of mass m moving with velocity component u_x inside a rectangular box of length L and end-wall area A, derive the expression pV = ⅓Nm(c_rms)² for the pressure exerted by N ideal gas molecules. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. When a single molecule of mass m and x-velocity u_x collides elastically with the end wall of area A, its x-velocity reverses to -u_x, so the magnitude of its change in momentum is Δp = 2m u_x [1 mark]. 2. The molecule travels a distance of 2L (to the opposite wall and back) between successive collisions with the same wall, so the time interval between collisions is Δt = 2L ÷ u_x [1 mark]. 3. By Newton's Second and Third Laws, the average force exerted by this one molecule on the wall is F = Δp ÷ Δt = (2m u_x) ÷ (2L ÷ u_x) = m u_x² ÷ L [1 mark]. 4. For N molecules with mean-square x-speed <u_x²>, total pressure on the wall of area A (where container volume V = A × L) is p = F_total ÷ A = N m <u_x²> ÷ (A L) = N m <u_x²> ÷ V, so pV = N m <u_x²> [1 mark]. 5. By 3D Pythagoras, <c²> = <u_x²> + <u_y²> + <u_z²>, and because molecular motion is completely random in all three directions (<u_x²> = <u_y²> = <u_z²>), we have <u_x²> = ⅓<c²> = ⅓(c_rms)², giving pV = ⅓Nm(c_rms)² [1 mark].
- Q4
Calculate the root-mean-square speed (c_rms) of the following four gas molecules having speeds of 200 m s⁻¹, 300 m s⁻¹, 400 m s⁻¹, and 500 m s⁻¹, and explain why c_rms is always slightly larger than the mean speed. (3 marks)
Click “Check Mark Scheme” to compare your answer, 1. Mean speed = (200 + 300 + 400 + 500) ÷ 4 = 350 m s⁻¹ [1 mark]. 2. Square each speed and find the mean of the squares (<c²>): <c²> = (200² + 300² + 400² + 500²) ÷ 4 = (40,000 + 90,000 + 160,000 + 250,000) ÷ 4 = 540,000 ÷ 4 = 135,000 m² s⁻² [1 mark]. 3. Take the square root: c_rms = √135,000 = 367 m s⁻¹ (which is greater than 350 m s⁻¹ because squaring gives disproportionately greater mathematical weight to the faster speeds) [1 mark].
Common Exam Questions (4)
In the continuous-flow electrical calorimeter experiment (V₁I₁ = m₁cΔT + H), why do we repeat the experiment with a different water flow rate m₂ and different power V₂I₂ that gives the exact same temperature rise ΔT?
If the inlet and outlet temperatures (and therefore ΔT) are kept identical in both runs, the rate of heat loss to the surroundings (H) is identical in both runs. Subtracting the two equations (V₂I₂ - V₁I₁ = (m₂ - m₁)cΔT) eliminates the unknown heat loss H completely.
Why is specific latent heat of vaporisation (l_v) of water (~2.26 MJ kg⁻¹) over six times larger than the specific latent heat of fusion (l_f) of ice (~0.334 MJ kg⁻¹)?
When ice melts at 0°C, hydrogen bonds are only partially loosened so molecules can slide past each other while remaining closely packed in liquid water. When liquid water boils at 100°C, all intermolecular hydrogen bonds must be completely broken to separate the molecules far apart into a gas and work (W = pΔV) must be done pushing back the atmosphere.
What happens to the pressure of a gas in a rigid container if a valve leaks so that 20% of the gas molecules escape while the absolute temperature increases by 10%?
Use the molecular ideal gas equation pV = NkT at constant V (so p ∝ N × T): N becomes 0.80 N and T becomes 1.10 T, so the new pressure is p_new = 0.80 × 1.10 × p_initial = 0.88 p_initial (pressure decreases by 12%).
Why does the First Law of Thermodynamics state ΔU = Q + W (or Q = ΔU + W), and what happens in an isothermal expansion of an ideal gas?
For an ideal gas, internal energy U depends only on temperature T (U = 3/2 nRT). Therefore, in an isothermal (constant temperature, ΔT = 0) expansion, the change in internal energy is zero (ΔU = 0), meaning the thermal energy Q supplied to the gas from the surroundings must exactly equal the work done W = pΔV by the expanding gas.
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