Internal energy and the ideal gas flashcards

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Internal energy
The sum of the randomly distributed kinetic and potential energies of all the molecules of a body.
Absolute zero
The lowest possible temperature, 0 K = −273 °C, at which the molecules have their minimum possible energy and the pressure of an ideal gas would be zero.
Ideal gas
A gas that obeys pV = NkT exactly; in the kinetic model its molecules have negligible volume and no forces between them except during collisions.
Root-mean-square (r.m.s.) speed
The square root of the mean of the squares of the speeds of the molecules: c_rms = √⟨c²⟩.
Boltzmann constant k
k = 1.38 × 10⁻²³ J K⁻¹: links the mean kinetic energy of a molecule to the absolute temperature.
Ideal gas equation
pV = NkT (p in Pa, V in m³, N = number of molecules, T in K). (Yes)
Mean kinetic energy of a gas molecule
½m⟨c²⟩ = 3/2 kT (m = mass of one molecule in kg, T in K). (Yes)
Celsius to kelvin
T / K = θ / °C + 273. (No — learn it)
Fixed mass of gas changing state
p₁V₁/T₁ = p₂V₂/T₂ (from pV = NkT with N constant). (No — derive it)
Kinetic theory pressure equation
pV = ⅓Nm⟨c²⟩, or p = ⅓ρ⟨c²⟩. (No — derive it)
r.m.s. speed from temperature
c_rms = √(3kT/m). (No — derive it)
Sketch: p against V at constant T (Boyle's law)
Curve (hyperbola) approaching both axes; a hotter isotherm lies further from the origin.
Sketch: p against 1/V at constant T
Straight line through the origin; gradient = NkT.
Sketch: p against θ (°C) at constant V
Straight line with positive gradient that extrapolates to p = 0 at −273 °C.
Sketch: V against T (K) at constant p (Charles's law)
Straight line through the origin (dashed extrapolation to 0 K).
Distribution of molecular speeds at a higher temperature
Peak moves to higher speed and becomes lower and broader; area (number of molecules) unchanged.
Internal energy of an ideal gas
All random kinetic energy: no intermolecular forces, so no potential energy. It is proportional to T in kelvin.
What happens to molecular energies during melting at constant temperature?
Mean kinetic energy unchanged; potential energy increases, so internal energy increases.
How does temperature relate to molecular kinetic energy?
The mean kinetic energy of the molecules is proportional to the absolute temperature.
Kinetic model assumptions
Large number of molecules in random motion; negligible molecular volume; no forces except in collisions; elastic collisions; collision time negligible.
How does a gas exert pressure?
Molecules rebound from the walls; each collision changes their momentum, so the wall exerts a force on them and they exert an equal force on the wall. Pressure = total force per unit area.
Why does pressure rise when a gas is heated at constant volume?
Molecules move faster: larger momentum change per collision and more frequent collisions with the walls.
Why does pressure rise when a gas is compressed at constant temperature?
Same molecular speeds, but molecules hit each unit area of wall more often.
Core Practical 14: key precautions
Fixed mass of gas (no leaks); wait after each change so the gas returns to room temperature; read the scale at eye level; plot p against 1/V.
24 litres of gas at 1.0 × 10⁵ Pa and 17 °C. Number of molecules?
N = pV/kT = (1.0 × 10⁵ × 0.024) ÷ (1.38 × 10⁻²³ × 290) = 6.0 × 10²³.
Gas heated from 27 °C to 127 °C. Factor by which the mean molecular kinetic energy increases?
400 K ÷ 300 K = 1.33.
Fixed mass at 2.0 × 10⁵ Pa, 300 cm³ is expanded to 450 cm³ at constant T. New pressure?
p₂ = 2.0 × 10⁵ × 300 ÷ 450 = 1.3 × 10⁵ Pa (1.33 × 10⁵ Pa).
Mean kinetic energy of a gas molecule at 400 K?
3/2 kT = 1.5 × 1.38 × 10⁻²³ × 400 = 8.28 × 10⁻²¹ J.
r.m.s. speed of helium atoms (4.0 u) at 300 K?
c_rms = √(3 × 1.38 × 10⁻²³ × 300 ÷ 6.64 × 10⁻²⁷) = 1.37 × 10³ m s⁻¹.
Three molecules: 300, 500 and 700 m s⁻¹. Mean speed and r.m.s. speed?
Mean = 500 m s⁻¹; r.m.s. = √((300² + 500² + 700²) ÷ 3) = 526 m s⁻¹.
r.m.s. speed of a gas at 250 K is tripled. New temperature?
T ∝ c_rms², so T = 250 × 9 = 2250 K.
Using °C in gas calculations
Always convert to kelvin first: ratios of Celsius temperatures are meaningless.
N in pV = NkT
N is the number of molecules, not moles. If given moles and N_A, use N = n × N_A.
Same temperature, different gases
Same mean kinetic energy; c_rms ∝ 1/√m, so lighter molecules are faster.
Mass in ½m⟨c²⟩ = 3/2 kT
Mass of ONE molecule in kg (e.g. 28 u = 28 × 1.66 × 10⁻²⁷ kg), never the molar mass in grams.
Showing Boyle's law from a table
Calculate pV for several readings and show it is constant, or plot p against 1/V for a straight line through the origin.
Uncertainty in V = πd²l/4
Add percentage uncertainties: 2 × %d + %l (the diameter is squared).

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