Thermodynamics
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New answer posted
a month agoContributor-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
New answer posted
a month agoContributor-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? )
New answer posted
a month agoContributor-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
New answer posted
a month agoContributor-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
New answer posted
a month agoContributor-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
New answer posted
a month agoContributor-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
New answer posted
a month agoContributor-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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