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1.7.8Thermodynamics

Ideal gas law PV = nRT — derivation from kinetic theory

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🇮🇳 Yeh note Hinglish mein padho →

WHAT are we deriving?

The goal: show that for such a gas, PV=nRTPV = nRT where PP = pressure, VV = volume, nn = number of moles, RR = gas constant, TT = absolute temperature.

WHY start from collisions?

HOW — Derivation from scratch

Step 1 — One molecule hitting one wall

Put NN molecules in a cube of side LL (volume V=L3V=L^3). Take a molecule of mass mm moving with velocity components (vx,vy,vz)(v_x, v_y, v_z).

Consider the wall perpendicular to the xx-axis. The molecule hits it, bounces back elastically, so its xx-velocity flips sign: Δpx=(−mvx)−(mvx)=−2mvx\Delta p_x = (-mv_x) - (mv_x) = -2mv_x

Why this step? Elastic collision with a rigid wall reverses the perpendicular velocity but leaves speed unchanged; the magnitude of momentum given to the wall is 2mvx2mv_x.

Step 2 — How often does it hit?

After bouncing off one xx-wall, it travels to the opposite wall and back: distance 2L2L. Time between successive hits on the same wall: Δt=2Lvx\Delta t = \frac{2L}{v_x}

Why this step? Collisions on a given wall are periodic; rate of hitting = vx/2Lv_x/2L per second.

Step 3 — Force from one molecule

Average force on the wall from this one molecule (Newton's 2nd law, magnitude): F1=∣Δpx∣Δt=2mvx2L/vx=mvx2LF_1 = \frac{|\Delta p_x|}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}

Why this step? Momentum delivered per collision ÷ time per collision = average force.

Step 4 — Add up all NN molecules

F=mL∑i=1Nvx,i2=mLN⟨vx2⟩F = \frac{m}{L}\sum_{i=1}^{N} v_{x,i}^2 = \frac{m}{L} N \langle v_x^2\rangle where ⟨vx2⟩\langle v_x^2\rangle is the mean of vx2v_x^2.

Why this step? Total force is the sum of independent contributions; replacing the sum by N×N\timesaverage is just the definition of an average.

Step 5 — Use isotropy (randomness)

Motion is random, so no direction is special: ⟨vx2⟩=⟨vy2⟩=⟨vz2⟩\langle v_x^2\rangle = \langle v_y^2\rangle = \langle v_z^2\rangle Since v2=vx2+vy2+vz2v^2 = v_x^2+v_y^2+v_z^2: ⟨v2⟩=3⟨vx2⟩  ⇒  ⟨vx2⟩=13⟨v2⟩\langle v^2\rangle = 3\langle v_x^2\rangle \;\Rightarrow\; \langle v_x^2\rangle = \tfrac{1}{3}\langle v^2\rangle

Why this step? Isotropy is the physical statement that the gas has no preferred direction; it lets us swap ⟨vx2⟩\langle v_x^2\rangle for the full speed.

Step 6 — Pressure

P=Farea=FL2=mN⟨vx2⟩L3=mN⟨v2⟩3VP = \frac{F}{\text{area}} = \frac{F}{L^2} = \frac{m N \langle v_x^2\rangle}{L^3} = \frac{mN\langle v^2\rangle}{3V}

Figure — Ideal gas law PV = nRT — derivation from kinetic theory

Step 7 — Bring in temperature (the bridge)

Rewrite using kinetic energy. Multiply and divide by 2: PV=23N(12m⟨v2⟩)=23N ⟨Ek⟩PV = \frac{2}{3}N\left(\tfrac{1}{2}m\langle v^2\rangle\right) = \frac{2}{3}N\,\langle E_k\rangle

The definition of temperature for an ideal gas (equipartition, 32kBT\tfrac{3}{2}k_BT for 3 translational DOF): ⟨Ek⟩=12m⟨v2⟩=32kBT\langle E_k\rangle = \tfrac{1}{2}m\langle v^2\rangle = \frac{3}{2}k_B T

Why this step? Temperature is defined so that it measures average translational kinetic energy. This is the physical input that turns mechanics into thermodynamics.

Substitute: PV=23N⋅32kBT=NkBTPV = \frac{2}{3}N\cdot\frac{3}{2}k_B T = N k_B T

Step 8 — Switch to moles

N=nNAN = n N_A (moles × Avogadro), and define R=NAkBR = N_A k_B: PV=nRT\boxed{PV = n R T}

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Recall Quick self-test (hide answers)
  • Where does the factor 13\tfrac{1}{3} come from? → isotropy, ⟨vx2⟩=13⟨v2⟩\langle v_x^2\rangle=\tfrac13\langle v^2\rangle.
  • What is the physical definition of TT used in the bridge? → ⟨Ek⟩=32kBT\langle E_k\rangle=\tfrac32 k_B T.
  • What is RR in terms of kBk_B? → R=NAkBR=N_A k_B.
  • Why is ⟨vx⟩=0\langle v_x\rangle=0 but ⟨vx2⟩≠0\langle v_x^2\rangle\neq0? → directions cancel, squares don't.
Recall Feynman: explain to a 12-year-old

Imagine a box full of tiny bouncy balls flying around super fast. Every time a ball smacks a wall it gives the wall a little push. There are billions of balls smacking all the time, so the wall feels a steady push — that's pressure. If you heat the gas, the balls speed up, smack harder and more often, so the push gets stronger. Squeeze the box smaller and the balls hit the walls more often too. Add up all those tiny smacks with simple "push = how fast it bounces × how often" rules, and you discover the magic recipe: pressure × volume = (amount of gas) × (a constant) × (how hot it is). That's PV=nRTPV=nRT — it was hiding inside the bouncing all along.

Flashcards

What is pressure microscopically?
Average momentum delivered per second per unit area by molecules colliding with the wall.
Momentum change of a molecule hitting an x-wall elastically?
−2mvx-2mv_x (its xx-velocity reverses).
Time between hits on the same wall?
Δt=2L/vx\Delta t = 2L/v_x.
Average force from one molecule on a wall?
F1=mvx2/LF_1 = mv_x^2/L.
Why does ⟨vx2⟩=13⟨v2⟩\langle v_x^2\rangle=\frac13\langle v^2\rangle?
Isotropy: ⟨vx2⟩=⟨vy2⟩=⟨vz2⟩\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle and they sum to ⟨v2⟩\langle v^2\rangle.
Kinetic-theory pressure formula?
P=13Nm⟨v2⟩VP=\frac{1}{3}\frac{Nm\langle v^2\rangle}{V}.
Pressure in terms of average KE?
PV=23N⟨Ek⟩PV=\frac{2}{3}N\langle E_k\rangle.
Definition of temperature used in the bridge?
⟨Ek⟩=12m⟨v2⟩=32kBT\langle E_k\rangle=\frac{1}{2}m\langle v^2\rangle=\frac{3}{2}k_BT.
Relation between RR and kBk_B?
R=NAkB≈8.314R=N_Ak_B\approx 8.314 J/mol·K.
rms speed formula?
vrms=3RT/M=3kBT/mv_{rms}=\sqrt{3RT/M}=\sqrt{3k_BT/m}.
Why must TT be in Kelvin?
Because TT is proportional to average kinetic energy, which is zero only at absolute zero.
Why does ⟨vx⟩=0\langle v_x\rangle=0 but force isn't zero?
Directions cancel for vxv_x, but force depends on vx2v_x^2 (always positive).

Connections

Concept Map

models

assumes

assumes

defines

gives momentum kick 2 m vx

divided by hit time 2L/vx

sum over N molecules

vx2 equals one third v2

combined with

force over area, V equals L cubed

leads to

identify kinetic energy with T

Kinetic theory: bouncing molecules

Ideal gas point particles

Elastic collisions

Random isotropic motion

Newton 2nd law F equals dp/dt

Pressure as momentum flux

Momentum per collision

Force from one molecule

Total wall force

Mean square velocity

PV expression

PV equals nRT

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, gas ka matlab hai ek box mein crore-crore chhote balls (molecules) jo tezi se idhar-udhar uchhal rahe hain. Jab bhi ek ball wall se takraata hai, wo wall ko ek chhota sa dhakka deta hai. Pressure kuch aur nahi, bas in saare dhakkon ka average — force per unit area. Toh agar hum Newton ke simple laws lagaake ye dhakke count karein, to khud-ba-khud PV=nRTPV=nRT nikal aata hai. Yahi kinetic theory ka kamaal hai.

Derivation ka core: ek molecule wall se elastically takraata hai, to uska xx-velocity ulta ho jaata hai, momentum change =2mvx=2mv_x. Same wall pe wapas takraane mein time =2L/vx=2L/v_x lagta hai. Force = momentum change ÷ time =mvx2/L= mv_x^2/L. Saare NN molecules add karo, aur kyunki gas har direction mein equal random hai (isotropy), ⟨vx2⟩=13⟨v2⟩\langle v_x^2\rangle = \tfrac13\langle v^2\rangle. Isse milta hai P=13Nm⟨v2⟩VP=\tfrac13 \tfrac{Nm\langle v^2\rangle}{V} — ye sirf mechanics hai, abhi tak temperature aaya hi nahi.

Ab temperature ka bridge: physics mein temperature define hi aise hota hai ki 12m⟨v2⟩=32kBT\tfrac12 m\langle v^2\rangle = \tfrac32 k_B T (average kinetic energy). Ye daal do, to PV=NkBTPV = Nk_BT aata hai. Phir N=nNAN=nN_A aur R=NAkBR=N_Ak_B likho — bas, PV=nRTPV=nRT ready!

Do dhyaan dene wali baatein: (1) Temperature hamesha Kelvin mein, kyunki wo energy ke proportional hai. (2) Factor 13\tfrac13 isotropy se aata hai, vxv_x ki jagah vv mat use karna. Ye samajh lo to gas law ratta nahi, derive ho jaayega exam mein.

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Connections

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