Temperature — thermal equilibrium, thermometers, scales: Visual walkthrough

1.7.1 · D2 · Physics › Thermodynamics › Temperature — thermal equilibrium, thermometers, scales

Hum sirf yeh ideas use karte hain, jo jab bhi aate hain wahi define ho jaate hain: ek box mein gas, woh push jo woh wall par karta hai (pressure), thermometer se padhne wala temperature, aur ek graph par seedhi line. Kuch aur assume nahi kiya gaya.

Step 1 — Ek box of gas jo ek lid par push karta hai

KYA. Kuch gas — hawa, helium, kuch bhi thin — ko ek aisi lid ke neeche band karo jo slide kar sakti hai lekin is tarah rokhi hui hai ki volume kabhi nahi badalti (constant volume). Gas andar se lid par drumming karta rahta hai. Woh steady drumming, lid ke area mein spread hokar, pressure $P$ hai: gas kitna hard push karta hai surface ke har unit par.

KYUN. Hume ek thermometric property chahiye — ek aisi physical quantity jo reliably change hoti ho jab hotness change ho. Hum fixed volume par pressure choose karte hain kyunki (parent ne bataya) sabhi dilute gases yahan same tarah behave karti hain, mercury ya alcohol ke unlike jo ek slightly alag story bataati hain. Pressure haara imandaar witness hai.

PICTURE. Orange arrows woh gas particles hain jo walls par hammering kar rahe hain; gauge par teal number woh pressure hai jo woh produce karte hain. Zyada warmth daalo aur hammering tez ho jaati hai.

Figure — Temperature — thermal equilibrium, thermometers, scales

Step 2 — Ise thanda karo, push ko kamzor hote dekho

KYA. Temperature $t_{C}$ (Celsius mein) kam karo aur pressure record karo. Warm gas → harder push; cool gas → gentler push. Har measurement ko ek dot ki tarah plot karo: temperature bottom mein, pressure side mein.

KYUN. Koi bhi formula likhne se pehle, hum dekhte hain. Hum raw experimental sawaal pooch rahe hain: kya koi pattern hai? Agar dots randomly scatter hote hain, toh koi law dhundhne ko nahi hai. Nature is se zyada kind hai.

PICTURE. Jab hum warm (right) se cold (left) ki taraf sweep karte hain, pressure gauge girta hai aur dots ek surprisingly seedhi line mein neeche aur baayein ki taraf march karte hain.

Recall Dekhne se pehle kyun sochna chahiye?

Question ::: Formula assume karne se pehle data plot kyun karte hain? Answer ::: Kyunki data ki shape (ek seedhi line) hi linear law ko justify karti hai. Hum linearity guess nahi kar rahe — hum ise graph se padh rahe hain.

Step 3 — Dots ek seedhi line par hain

KYA. Dots ke through best straight line draw karo. Ek seedhi line ek equation se capture hoti hai: $P = P_{0} (1 + α t_{C})$

Har symbol wahan padhte hain jahan woh hai:

$P$ — pressure jo hum abhi measure kar rahe hain (ek dot ki height).
$t_{C}$ — Celsius mein temperature jo humne set kiya (us dot ki horizontal position).
$P_{0}$ — ice point par pressure, $t_{C} = 0 ° C$ . Yeh woh jagah hai jahan line vertical axis cross karti hai. Ise anchor height kaho.
$α$ — fractional slope: warming ke har degree par $P_{0}$ ka kitna hissa pressure gain karta hai. Experiment se har dilute gas ke liye $α \approx \frac{1}{273.15} ° C^{- 1}$ milta hai.

Seedhi line kyun aur curve kyun nahi? Kyunki dots yahi dikhate hain, aur yeh sabse simple law hai jo fit hota hai — ek intercept ( $P_{0}$ ) aur ek slope ( $α P_{0}$ ). Curve ko extra parameters ki zaroorat hogi jo data demand nahi karta.

PICTURE. Teal line axis par $P_{0}$ par hit karti hai jab $t_{C} = 0$ hota hai; chhota step-triangle $1 ° C$ par rise $α P_{0}$ dikhata hai.

Step 4 — Line ko neeche follow karo: push kahan gayab hoga?

KYA. Line thanda karne par slope karte huye neeche jaati rehti hai. Ek sharp sawaal poochho: kis temperature par pressure exactly zero ho jaayega? $P = 0$ set karo aur $t_{C}$ ke liye solve karo.

$0 = P_{0} (1 + α t_{C})$

$P = 0$ kyun set karein? Kyunki Step 1 ne bataya ki pressure negative nahi ho sakta — yeh ek push hai, aur push sirf kuch nahi tak kamzor ho sakta hai. Toh $P = 0$ natural floor hai. Woh $t_{C}$ dhundhna jo ise produce kare, sabse thanda conceivable temperature dhundhna hai.

Ab solve karo, ek honest move at a time:

$P_{0} \neq = 0$ (box mein real gas hai), toh hum ise divide kar sakte hain: $0 = 1 + α t_{C}$ .
$1$ subtract karo: $α t_{C} = - 1$ .
$α$ se divide karo: $t_{C} = - \frac{1}{α}$ .

Measured slope $α = \frac{1}{273.15}$ daalo: $t_{C} = - \frac{1}{1/273.15} = - 273.15 ° C$

PICTURE. Seedhi line extend ki gayi hai (dashed) data se aage jaake jab tak woh horizontal axis ko pierce nahi karti. Woh piercing point — jahan push khatam hoti hai — $- 273.15 ° C$ par hai. Yahi absolute zero hai.

Step 5 — Edge case: kya answer depend karta hai kis gas par?

KYA. Experiment helium ke saath chalao, phir nitrogen ke saath, phir argon ke saath. Har ek ka apna anchor height $P_{0}$ hota hai (gas ki alag amount, alag starting push). Toh hume teen alag lines milti hain. Kya woh axis ko teen alag jagahon par cross karti hain?

KYUN yeh matter karta hai. Agar absolute zero substance par depend karta, toh yeh universal constant nahi ho sakta — yeh sirf helium ki ek quirk hogi. Hume yeh degenerate-looking possibility check karni hai ki crossing wander kare.

Crossing ki algebra. Har line, chahe $P_{0}$ kuch bhi ho, $0 = P_{0} (1 + α t_{C})$ follow karti hai. Kyunki $P_{0}$ divide out ho jaata hai, crossing sirf $α$ par depend karti hai — aur $α$ saari dilute gases ke liye same hai. Toh teen saari lines, chahe kitni bhi steep hon, axis ko ek hi same $- 273.15 ° C$ par pierce karti hain.

PICTURE. Teen lines jo alag steepness ki hain (teen gases) teen alag intercepts se fan out karti hain — phir bhi woh saari temperature axis par ek shared point par converge karti hain.

Step 6 — Origin slide karo: Kelvin scale banao

KYA. Humne temperature ka ek true zero dhundh liya hai. Ise ugly number $- 273.15$ par rakhna wasteful hai. Toh ek naya scale $T$ invent karo jiska zero exactly absolute zero par baithe, har degree ka size Celsius degree jaisa rakhte huye: $T = t_{C} + 273.15$

Symbols padhte hain:

$T$ — naya Kelvin temperature (unit: K, koi degree sign nahi).
$t_{C}$ — wahi Celsius reading jaise pehle.
$+ 273.15$ — woh shift jo origin ko ice point se absolute zero tak move karta hai.

Rescale kyun nahi, sirf shift kyun? Hum Celsius degree ka size pasand karte hain (ice aur steam ke beech 100 steps), toh hum slope rakhte hain. Hum sirf origin slide karte hain taaki physically special point — jahan pressure vanish hota hai — numerically zero ban jaaye.

Payoff. $t_{C} = - 273.15$ (absolute zero) shift mein daalo: $T = - 273.15 + 273.15 = 0$ K. Aur pressure law ko naye scale par rewrite karo: $P = P_{0} (1 + α t_{C}) = P_{0} α (t_{C} + 1/ α) = P_{0} α (t_{C} + 273.15) = (P_{0} α) T$ Toh Kelvin scale par, pressure simply temperature ke proportional hai: $P \propto T$ . Clean, koi offset nahi.

PICTURE. Do rulers stack kiye: Celsius ruler apne awkward $- 273.15$ ke saath, aur Kelvin ruler slide kiya hua taaki uska $0$ absolute zero par aa jaaye. Same tick spacing, alag starting number.

Step 7 — Scale ko ek fixed point se anchor karo

KYA. Origin se jaane waali line ko poori tarah pin karne ke liye sirf ek known point chahiye. Hum triple point of water use karte hain — woh exact temperature jahan ice, liquid, aur vapour ek saath exist karte hain — jo $T_{tr} = 273.16$ K define kiya gaya hai. Wahan gas pressure $P_{tr}$ measure karo, aur koi bhi doosra temperature proportion se aata hai: $T = T_{tr} \frac{P}{P _{tr}}$

$T$ — woh temperature jo hum chahte hain.
$P$ — unknown bath mein gas ka pressure.
$P_{tr}$ — usi gas ka triple point par pressure.
$T_{tr} = 273.16$ K — fixed anchor.

Ab sirf ek point kyun? Kyunki Step 6 ne origin se jaane waali line di (zero physics se fixed hai, choice se nahi). Origin se jaane waali line mein ek free number hota hai — uska slope — toh ek measurement sab kuch lock kar deta hai.

Degenerate check. Agar $P \to 0$ , toh $T \to 0$ — Step 4 se consistent. Agar $P = P_{tr}$ , toh $T = T_{tr}$ — anchor khud ko reproduce karta hai. Dono sanity checks pass hote hain.

PICTURE. Origin se ek line, $(P_{tr}, T_{tr})$ par ek marked dot; koi bhi pressure $P$ ray par ride karke temperature read karta hai.

Ek-picture summary

Upar sab kuch, ek graph par: data dots, fitted line, uska dashed extension neeche axis tak $- 273.15 ° C$ par, aur doosra Kelvin axis slide kiya hua taaki woh piercing point $0$ K padhta ho. Gas ko thanda karna tumhe line ke saath left le jaata hai; push exactly absolute zero par khatam hoti hai, aur Kelvin sirf Celsius hai jiska zero wahan shift kar diya gaya ho.

Recall Feynman retelling — plain words mein poora walk

Question ::: Absolute zero ki derivation ek curious bachche ko retell karo. Answer ::: Kuch hawa ek sealed jar mein daalo taaki woh grow ya shrink na kar sake, aur us par ek pressure gauge lagao. Hawa walls par bang karti hai — woh banging pressure hai, uska "push." Ab jar ko thanda karo. Jitna thanda hoga, utni lazy banging hogi, toh push girti hai. Agar tum har (temperature, push) pair ko dot karo, dots ek perfectly seedhi line par giraage jo neeche baayein ki taraf slope karti hai. Ek ruler lo aur woh line wahan se aage draw karte raho jahan tumne measure karna band kiya: woh "zero push" mark par $- 273.15 ° C$ par cross karti hai. Lekin gas inward-less-than-nothing push nahi kar sakti — zero push road ka end hai, toh $- 273.15 ° C$ sabse thanda temperature hai jo ho sakta hai. Helium, nitrogen, argon ke saath try karo: har ek alag push par start karta hai, toh har line alag steepness ki hai — phir bhi woh saari us same crossing point par aim karti hain, kyunki ek kamzor gas zyada gently bhi fade hoti hai. Kyunki woh point itna special hai, hum ek naya ruler (Kelvin) banate hain jiska zero bilkul wahan baithe, same degree size rakhte huye. Us ruler par push Kelvin temperature ke plain proportional hoti hai — Kelvin temperature double karo, push double ho jaati hai. Water ke triple point ( $273.16$ K) par ek accha measurement poora scale set kar deta hai.

Prerequisites & neighbours: parent topic · Ideal Gas Law · Kinetic Theory of Gases · Thermal Expansion · Zeroth Law of Thermodynamics · Heat and Internal Energy