Thermodynamics
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Which one is the correct option for the two different thermodynamic processes?
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Which one is the correct option for the two different thermodynamic processes?
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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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If one mole of the polyatomic gas is having two vibrational modes and β is the ratio of molar specific heats for polyatomic gas (β = C_p / C_v), then the value of β is:
If one mole of the polyatomic gas is having two vibrational modes and β is the ratio of molar specific heats for polyatomic gas (β = C_p / C_v), then the value of β is:
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
Upvote The standard enthalpies of formation of Al₂O₃ and CaO are –1675 kJ mol⁻¹ and -635 kJ mol⁻¹ respectively.
For the reaction 3CaO + 2Al → 3Ca + Al₂O₃ the standard reaction enthalpy ΔᵣH° = _______ kJ.
(Round off to the nearest integer).
The standard enthalpies of formation of Al₂O₃ and CaO are –1675 kJ mol⁻¹ and -635 kJ mol⁻¹ respectively.
For the reaction 3CaO + 2Al → 3Ca + Al₂O₃ the standard reaction enthalpy ΔᵣH° = _______ kJ.
(Round off to the nearest integer).
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
Upvote Two ideal polyatomic gases at temperatures T₁ and T₂ are mixed so that there is no loss of energy. If F₁ and F₂, m₁ and m₂, n₁ and n₂ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is:
Two ideal polyatomic gases at temperatures T₁ and T₂ are mixed so that there is no loss of energy. If F₁ and F₂, m₁ and m₂, n₁ and n₂ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is:
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? )
Upvote A Carnot's engine working between 400 K and 800 K has a work output of 1200J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is:
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A Carnot's engine working between 400 K and 800 K has a work output of 1200J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is:
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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
Upvote A polyatomic ideal gas has 24 vibrational modes. What is the value of γ ?
A polyatomic ideal gas has 24 vibrational modes. What is the value of γ ?
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
Upvote A bullet of mass 5 g, traveling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030cal/(g- °C)(1cal = 4.2 × 10⁷ergs) close to:
A bullet of mass 5 g, traveling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030cal/(g- °C)(1cal = 4.2 × 10⁷ergs) close to:
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
Upvote H can easily gain electron to form its anion.
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