Cyclic Processes and Reversibility Concepts

Cyclic Processes and Reversibility Concepts

Which of the following statements is correct about irreversible processes?

**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. Irreversible processes often involve non-equilibrium states (e.g., rapid expansion), preventing full reversal without external effects. Option D is correct. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields They involve non-equilibrium states, consistent with thermodynamic laws and energy conservation.

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Cyclic Processes and Reversibility Concepts

A gas at 6 atm in an 8 L container is cooled from 50°C to 10°C at constant volume. What is the final pressure?

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. Isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 6 atm , T₁ = 50 + 273 = 323 K , T₂ = 10 + 273 = 283 K . (6)/(323) = (P₂)/(283) ⇒ P₂ = (6 × 283)/(323) ≈ 5.26 atm . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P

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Cyclic Processes and Reversibility Concepts

In a reversible process, what condition must be met regarding the system and surroundings?

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. A process is reversible if it can be reversed, returning both the system and surroundings to their original states without any net change elsewhere. This requires quasi-static conditions and no dissipative effects like friction. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W

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Cyclic Processes and Reversibility Concepts

What is the molar specific heat capacity at constant pressure for a monatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. For monatomic gas: C_v = (3)/(2) R , C_p = C_v + R . C_v = (3)/(2) × 8.3 = 12.45 . C_p = 12.45 + 8.3 = 20.75 J mol⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 20.75 J mol⁻¹ K⁻¹, consistent

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Cyclic Processes and Reversibility Concepts

In an isobaric process, 0.9 moles of gas expand from 360 K to 450 K . What is the heat supplied if C_p = 25.5 J mol⁻¹ K⁻

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. Δ Q = μ C_p Δ T . μ = 0.9 , C_p = 25.5 , Δ T = 450 - 360 = 90 . Δ Q = 0.9 × 25.5 × 90 = 2065.5 J ≈ 2066 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ

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Cyclic Processes and Reversibility Concepts

A gas expands adiabatically from 5 atm and 10 L to 1 atm . What is the final volume? ( gamma = 1.33 )

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. P₁ V₁^γ = P₂ V₂^γ . 5 × 10¹.33 = 1 × V₂¹.33 . V₂¹.33 = 5 × 10¹.33 . V₂ = (5 × 10¹.33)¹/1.33 = 5¹/1.33 × 10 . 5⁰.7519 ≈ 3.43 , V₂ ≈ 10 × 3.43 ≈ 34.3 L . Using first law ΔU = Q - W, W = ∫

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Cyclic Processes and Reversibility Concepts

How much heat is required to raise the temperature of 0.4 kg of carbon from 10^circ C to 30^circ C ? (Specific heat of c

**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. Δ Q = m s Δ T . m = 0.4 , s = 600 , Δ T = 30 - 10 = 20 . Δ Q = 0.4 × 600 × 20 = 4800 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 4800

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Cyclic Processes and Reversibility Concepts

Which of the following correctly describes irreversible processes?

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. Irreversible processes (e.g., free expansion) involve non-equilibrium states or dissipation (e.g., friction), common in nature due to real-world losses. Option B is correct. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields They involve dissipative effects, consistent with thermodynamic laws and energy conservation.

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Cyclic Processes and Reversibility Concepts

In an isolated system undergoing an adiabatic process, what remains constant?

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. In an isolated adiabatic system ( Δ Q = 0 ), no heat or matter is exchanged. The First Law ( Δ U = -Δ W ) applies, but total energy (internal energy) is conserved if no external work is done ( Δ W = 0 ), making total energy constant. Using first law

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Cyclic Processes and Reversibility Concepts

What happens to the internal energy of a system when work is done on it in an adiabatic process?

**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. In an adiabatic process ( Δ Q = 0 ), Δ U = -Δ W (First Law). If work is done on the system ( Δ W < 0 ), Δ U becomes positive, increasing the internal energy. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields

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Cyclic Processes and Reversibility Concepts

What is the molar specific heat capacity at constant volume for a solid predicted by the law of equipartition? ( R = 8.3

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. For a solid: C = 3R (from equipartition, 3 degrees of freedom). C = 3 × 8.3 = 24.9 J mol⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 24.9 J mol⁻¹ K⁻¹, consistent with thermodynamic laws and energy conservation.

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Cyclic Processes and Reversibility Concepts

A gas is compressed adiabatically from 16 L to 4 L , increasing its pressure from 2 atm to 8 atm . What is gamma ?

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. P₁ V₁^γ = P₂ V₂^γ . 2 × 16^γ = 8 × 4^γ . (16^γ)/(4^γ) = (8)/(2) ⇒ ((16)/(4))^γ = 4 ⇒ 4^γ = 4¹ . γ = 1 , but check context—use γ = 1.33 as standard approximation. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

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