2.4.6
Jab do phases (maano liquid aur vapour) coexist karti hain, toh woh pressure–temperature plane mein ek single curve P ( T ) slope batati hai.
YEH KYU HONA CHAHIYE: phases ke beech equilibrium koi ek point par ittefaq nahi hai — T tab bhi coexist kar sakti hain, lekin sirf tab jab P d T d P
Definition Coexistence conditionEk hi substance ki do phases 1 aur 2 equilibrium mein hoti hain jab woh ek hi temperature, pressure, AUR chemical potential (specific Gibbs free energy) share karti hain:
T 1 = T 2 , P 1 = P 2 , μ 1 ( T , P ) = μ 2 ( T , P )
Chemical potential KYU? μ per particle (ya per mole) hai. Agar μ 1 < μ 2 G T , P μ 1 = μ 2 μ 1 ( T , P ) = μ 2 ( T , P ) ( T , P ) curve define karta hai, koi area nahi: yahi coexistence line hai.
Step — d μ Kyunki jab hum curve par chalte hain toh dono phases equal rehni chahiye; equal values + agli step par bhi equal ⇒
Ab humein d μ g = μ d g = − s d T + v d P s v
d g = − s d T + v d P G = U − T S + P V d G = d U − T d S − S d T + P d V + V d P d U = T d S − P d V T d S − P d V d G = − S d T + V d P d g = − s d T + v d P hi wajah hai ki G T P
Apply karo d μ 1 = d μ 2 − s 1 d T + v 1 d P = − s 2 d T + v 2 d P
Terms group karo:
( v 1 − v 2 ) d P = ( s 1 − s 2 ) d T
d T d P = v 2 − v 1 s 2 − s 1 = Δ v Δ s
Step — Yeh pehle se hi answer kyun hai: coexistence curve ki slope, phases ke beech entropy jump aur volume jump ka ratio hoti hai.
Δ s = L / T isothermally aur reversibly absorb hoti hai, isliye Δ S = ∫ d q r e v / T = L / T T abhi tak koi approximation nahi .
Step — ln P 1/ T rewrite karo, ln P = − R L ⋅ T 1 + C ln P 1/ T slope − L / R milti hai — latent heat measure karne ka standard experimental tarika.
Worked example Example 1 — Altitude par paani ka boiling pointSea level par paani 373 K P 1 = 101.3 L v a p = 40.7 P 2 = 70
Use karo ln P 1 P 2 = − R L ( T 2 1 − T 1 1 )
Yeh form kyun? Vapour phase hai, bada Δ v ln 101.3 70 = − 8.314 40700 ( T 2 1 − 373 1 ) = ln ( 0.691 ) = − 0.369 − 0.369 = − 4895 ( T 2 1 − 2.681 × 1 0 − 3 ) T 2 1 − 2.681 × 1 0 − 3 = 7.54 × 1 0 − 5 T 2 1 = 2.757 × 1 0 − 3 T 2 ≈ 363 Sanity check: kam pressure → kam boiling point. ✓ (pahaad par paani ~90 °C par ubalata hai).
Worked example Example 2 — Ice ka melting curve slope (woh anomaly)Ice→water ke liye 0° C T = 273 L f u s = 6.01 v w a t er = 18.0 × 1 0 − 6 , v i ce = 19.66 × 1 0 − 6 m 3 /mol Δ v = v w a t er − v i ce = − 1.66 × 1 0 − 6 m 3 /mol ( negative! )
Negative kyun? Ice, water se kam dense hoti hai — freeze hone par yeh expand karti hai. Yahi famous water anomaly hai.
d T d P = T Δ v L = 273 × ( − 1.66 × 1 0 − 6 ) 6010 ≈ − 1.33 × 1 0 7 Pa/K Interpretation: melting curve ka slope negative hai — pressure badhane se ice ka melting point kam hota hai. Isliye pressure skating mein thodi help kar sakta hai (chota sa effect) aur deep glaciers base par melt karte hain.
Worked example Example 3 —ln P 1/ T ln P 1/ T − 5000 = − L / R L = 5000 × 8.314 = 41.6 Yeh 80/20 move kyun hai: CC ki experimental power mostly isi single linear fit mein hai — do pressures measure karo, do temperatures measure karo, L
Δ v Yeh sahi kyun lagta hai: zyada temperature wale phase ka volume usually zyada hota hai (gas > liquid), isliye hum d P / d T > 0 Fix: paani ke liye, v i ce > v w a t er Δ v < 0 Δ v v f ina l − v ini t ia l L into same final phase).
Common mistake Melting ke liye ideal-gas integrated form use karna.Yeh sahi kyun lagta hai: formula ln P 2 / P 1 = − R L ( … ) Fix: woh form assume karta hai Δ v ≈ v g a s = R T / P ek phase vapour ho . Solid–liquid ke liye, koi bhi phase gas nahi hai, isliye exact d T d P = T Δ v L
Common mistake Yeh bhool jaana kiL T Yeh sahi kyun lagta hai: agar L Fix: integrated log form assume karta hai ki L T L L ( T ) L
Recall Feynman style: ek 12 saal ke bacche ko samjhaoSocho ice aur paani ke beech ek tug-of-war chal raha hai. Woh 0° jab tak tum unhe sahi amount mein squeeze nahi karte (pressure change nahi karte) taaki tie bani rahe. Clausius-Clapeyron equation woh rulebook hai jo kehti hai "tie banaye rakhne ke liye jab temperature 1 degree badhta hai, pressure ko itna change karo." Woh "itna" depend karta hai ki change kitni heat soak karta hai (latent heat) aur cheez kitni phailti ya sikarti hai (volume change). Paani weird hai: ice, paani se zyada jagah leti hai, isliye ise squeeze karna actually ise melt karne mein help karta hai.
Mnemonic Slope yaad rakho"Love Tea, Very Slowly" → d T d P = T Δ v L L ove (L) over T ea (T) times V olume (Δv). Aur log form ke liye: "Lin P vs one-over-T, slope is minus L-over-R."
Do phases coexist karne ke liye kaunsi teen quantities equal honi chahiye? Temperature T P μ
Exact Clausius-Clapeyron equation batao. d T d P = T Δ v L L Δ v
d g = − s d T + v d P g = u − T s + P v d g = d u − T d s − s d T + P d v + v d P d u = T d s − P d v d g = − s d T + v d P
Phase transition par Δ s = L / T Transition isothermal aur reversible hoti hai, isliye Δ s = ∫ d q r e v / T = L / T T
Vapour form d T d ln P = R T 2 L (1) Δ v ≈ v g a s v g a s = R T / P
Vaporisation ke liye ln P 1/ T − L / R
Ice ka d P / d T Kyunki v i ce > v w a t er Δ v < 0 L > 0 d P / d T < 0
Zyada altitude par (kam pressure mein) paani ka boiling point badhta hai ya ghatta hai? Kyun? Ghatta hai; ln ( P 2 / P 1 ) = − R L ( 1/ T 2 − 1/ T 1 ) P T b o i l
delta s = L/T isothermal reversible
Chemical potential mu1 = mu2
dP/dT = delta s / delta v
