EQUILIBRIUM

IIT JEE Online › Physical Chemistry questions › EQUILIBRIUM

THE DENSITY OF GRAPHITE IS 1.5 AND DENSITY OF DIAMOND IS 3.5 CHANGE IN GIBBS ENERGY IS 1895 JOULE/MOLE.
WHAT IS THE PRESSURE WHEN GRAPHITE CHANGES TO DIAMOND

#Chemistry #Physical Chemistry

UP 0 DOWN 0 0 4
Post on FacebookPost anonymously?

4 Answers

Lokesh Verma ·2009-03-16 23:30:06

Î”S=nRln(Vfinal/Vinitial)

Î”G=-TÎ”Stotal

From these,

Î”G=-nRTln(Vfinal/Vinitial)

Î”G=-PVln(7/3)

we know V from density... since n=1 moles...

I am not 100% sure but this must work!

UP 0 DOWN 0
Post on FacebookPost anonymously?

deepanshu001 agarwal ·2009-03-16 23:45:20

@nishant sir

how is the delta H = 0

UP 0 DOWN 0
Post on FacebookPost anonymously?

deepanshu001 agarwal ·2009-03-17 03:46:00

sum1 plzzz.... reply

UP 0 DOWN 0
Post on FacebookPost anonymously?

rashi mathur ·2009-03-18 22:49:15

what will be the v in pv? why delta h is zero plz reply

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

Close [X]

EQUILIBRIUM

4 Answers

Your Answer

Add Image

Similar Questions