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#cyclic process

12 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A system in a cyclic process absorbs 940 J of heat and rejects 360 J . What is the net work done?

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 940 - 360 = 580 J . W = 580 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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Heat Transfer Work Distinction and Internal Energy Change

A system in a cyclic process performs 300 J of work and rejects 200 J of heat. What is the heat absorbed?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . Q_absorb - 200 = 300 ⇒ Q_absorb = 500 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

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

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Cyclic Processes and Reversibility Concepts

In a cyclic process, what is true about the change in internal energy?

**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 a cyclic process, the system returns to its initial state. Since internal energy ( U ) is a state variable, its change ( Δ U ) is zero over a complete cycle, regardless of the path taken. 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

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

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Isochoric Processes and Pressure-Temperature

A system in a cyclic process absorbs 1020 J of heat and rejects 380 J . What is the net work done?

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 1020 - 380 = 640 J . W = 640 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Isochoric Processes and Pressure-Temperature

A gas undergoes a cyclic process where 600 J of heat is absorbed. What is the net work done by the gas?

**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. For a cyclic process, Δ U = 0 . Δ Q = Δ U + Δ W ⇒ 600 = 0 + Δ W . Δ W = 600 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 =

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Isochoric Processes and Pressure-Temperature

A system in a cyclic process absorbs 900 J of heat and rejects 400 J . What is the net work done?

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. For cyclic process: Δ U = 0 , W = Q_net . Q_net = Q_absorb - Q_reject = 900 - 400 = 500 J . W = 500 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Specific Heat Capacity and Latent Heat

Which of the following statements is incorrect about a cyclic process?

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. In a cyclic process, Δ U = 0 since U is a state function, and net heat equals net work. Option A is incorrect; internal energy does not change over a complete cycle. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Specific Heat Capacity and Latent Heat

A system in a cyclic process absorbs 650 J of heat and performs 200 J of work. What is the heat rejected?

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . 650 - Q_reject = 200 ⇒ Q_reject = 650 - 200 = 450 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

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Zeroth Law Thermal Equilibrium and Quasi-static

A system in a cyclic process absorbs 980 J of heat and performs 420 J of work. What is the heat rejected?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . 980 - Q_reject = 420 ⇒ Q_reject = 980 - 420 = 560 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

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

A system in a cyclic process absorbs 860 J of heat and performs 340 J of work. What is the heat rejected?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . 860 - Q_reject = 340 ⇒ Q_reject = 860 - 340 = 520 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

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Second Law Heat Engines and Kelvin-Planck

What thermodynamic property ensures that Delta U is zero in a cyclic process?

**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. Internal energy ( U ) is a state function, meaning its value depends only on the state, not the path. In a cyclic process, returning to the initial state means Δ U = 0 , regardless of intermediate changes. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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