Thermodynamics

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Vishal Baghel

Contributor-Level 10

P = (RT/M? )ρ ⇒ P/ρT = constant
⇒ (12)/ (4 * 10? ³ * 10? ) * 280 = 6 * 10? * T ⇒ T = 1400K

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Vishal Baghel

Contributor-Level 10

Vibrational energy of a non-rigid gas molecule is K? T so, total energy = (5/2)K? T + K? T = (7/2)K? T
∴ C? = (7/2)R

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Vishal Baghel

Contributor-Level 10

ΔG° = – 2.303RT log Kp = – 2.303* 2 * 300log10? ³
= – 2.303* 2 * 300 * (– 3) = 4145.4cal mol? ¹

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Vishal Baghel

Contributor-Level 10

ΔH_rxn = ΔH_f (N? O, g) + 3ΔH_f (CO? , g) – (2ΔH_f NO? , g) – 3ΔH_f (CO, g)
= 81 + 3* (– 393) – 2 * 34 – 3 (–110)
= 81 – 1179 – 68 + 330 = – 836 kJ

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Vishal Baghel

Contributor-Level 10

One mole of water is converted to vapour at its boiling point which is 100°C and at 1 atm. For this process ΔG = 0. As phase transformation of water is an equilibrium process and at equilibrium, free energy change is always zero.

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Vishal Baghel

Contributor-Level 10

2CO (g) + O? (g) → 2CO? (g)
Δn_g = 2 – (2 + 1) = -1
ΔH = ΔE + Δn_g RT or ΔH = ΔE – 1RT
i.e. ΔH < E

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Vishal Baghel

Contributor-Level 10

(Q? - Q? )/Q? = (T? - T? )/T? ; Q = (300/600) * 1600 = 800 J
work done = Q? - Q? = 1600 – 800 = 800J

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Vishal Baghel

Contributor-Level 10

x = 8R/2 ; y = 5R/2 ; x/y = 8/5 = 1.6

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Vishal Baghel

Contributor-Level 10

Workdone by gas = area enclosed in P-V diagram
= π (2)² lt-atm

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Vishal Baghel

Contributor-Level 10

v = √ (γRT/M) ; V? /V? = √ (T? /T? ) ; V? ' = V√ (363/300)

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