Thermodynamics

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Vishal Baghel

Contributor-Level 10

3CaO + 2Al → 3Ca + Al? O?
ΔH? = ΣΔH? (Products) - ΣΔH? (Reactants)
ΔH? = [ (0) + (-1675) ] - [ (3 * -635) + (0) ]
ΔH? = -1675 - (-1905) = -1675 + 1905 = 230 kJ
Ans = 230

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Vishal Baghel

Contributor-Level 10

U = U? + U? = (n? /N_A) (F? R/2)T? + (n? /N_A) (F? R/2)T?

For the mixture: U = (n? +n? )/N_A * (FR/2)T

F = (n? F? + n? F? ) / (n? + n? )

Equating the expressions for U and solving for T gives:
T = (n? F? T? + n? F? T? ) / (n? F? + n? F? )

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Vishal Baghel

Contributor-Level 10

η = 1 - T_C / T_H

η = 1 - 400/800 = 1 - ½ = ½

η = W/Q_H ⇒ ½ = W/Q_H ⇒ Q_H = 2W = 2 * 1200 = 2400 J

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Vishal Baghel

Contributor-Level 10

f = f_Translational + f_Rotational + f_Vibrational

f = 3 + 3 + 48 = 54 (Since each vibrational mode has two degrees of freedom)

γ = 1 + 2/f = 1 + 2/54 = 1 + 1/27 = 28/27 ≈ 1.03

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alok kumar singh

Contributor-Level 10

1/4 m (210)² = m (0.03) x (4.2) x 1000 x ΔT ; Q = mSΔt
ΔT = (210) (210)/ (4) (4.2) (0.03) (1000) = 87.5°C

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Vishal Baghel

Contributor-Level 10

Pext is zero so W = zero

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Vishal Baghel

Contributor-Level 10

H can easily gain electron to form its anion.

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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Vishal Baghel

Contributor-Level 10

T = constant
P = constant

PV = nRT

PdV = nRdT

PdV + VdP = 0

ΔV = nRΔT/P

dV = (-)VdP/P
|ΔV| = V (ΔP/P)
V/P ΔP = nRΔT/P
ΔT = V/nR ΔP
C = V/nR T/P = 300/2 = 150

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Vishal Baghel

Contributor-Level 10

ΔQ = heat supplied
ΔW = work done
ΔU = change in internal energy
(i) adiabatic (B) Δθ = 0
(ii) isothermal (D) ΔU = 0
(iii) isochoric (A) ΔW = 0
(iv) isobaric (C) ΔU ≠ 0, ΔW ≠ 0, ΔQ ≠ 0

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