Thermodynamics

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Vishal Baghel

Contributor-Level 10

PV? = C ⇒ V? (dP/dV) + P (γV^ (γ-1) = 0 ⇒ dP/dV = -γ (P/V) ⇒ dP/P = -γ (dV/V)

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alok kumar singh

Contributor-Level 10

Pressure decreases with an increase in volume in both isothermal and adiabatic processes. In an adiabatic process, as volume decreases, pressure increases with the increase in temperature.

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alok kumar singh

Contributor-Level 10

In an LCR series AC circuit, the phase difference φ between current and voltage is: φ = tan? ¹ (X? - X? ) / R)

If each vibrational mode contributes two degrees of freedom and the total degrees of freedom f = 3 + 3 + 4 = 10, then:
β = 1 + 2/f = 1 + 2/10 = 1.2

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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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