#internal energy

Work Heat Distinction and Energy Transfer Modes

A gas expands adiabatically, doing 360 J of work. What is the change in its internal energy?

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

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Work Heat Distinction and Energy Transfer Modes

A system absorbs 730 J of heat and performs 190 J of work. What is the change in internal energy?

**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). First Law: Δ Q = Δ U + Δ W . Δ Q = 730 , Δ W = 190 (work by system). 730 = Δ U + 190 ⇒ Δ U = 730 - 190 = 540 J . Using first law ΔU = Q - W, W = ∫ P

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Heat Transfer Work Distinction and Internal Energy Change

In an isothermal process for an ideal gas, what happens to the internal energy?

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. For an ideal gas, internal energy ( U ) depends only on temperature. In an isothermal process, temperature remains constant ( Δ T = 0 ), so Δ U = 0 . Using first law ΔU = Q - W, W = ∫

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Heat Transfer Work Distinction and Internal Energy Change

A system absorbs 850 J of heat and has 300 J of work done on it. What is the change in internal energy?

**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. First Law: Δ Q = Δ U + Δ W . Δ Q = 850 J , Δ W = -300 J (work on system). 850 = Δ U - 300 ⇒ Δ U = 850 + 300 = 1150 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T

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Heat Transfer Work Distinction and Internal Energy Change

A gas is compressed adiabatically, doing 300 J of work on the system. What is the change in internal energy?

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. For adiabatic ( Δ Q = 0 ), First Law: Δ U = -Δ W . Work on system: Δ W = -300 J (negative by convention). Δ U = -(-300) = 300 J . Using first law ΔU = Q - W,

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Heat Transfer Work Distinction and Internal Energy Change

What is the change in internal energy for 0.8 moles of an ideal gas heated from 310 K to 370 K at constant volume? ( C_v

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. Δ U = μ C_v Δ T . μ = 0.8 , C_v = 20.8 , Δ T = 370 - 310 = 60 . Δ U = 0.8 × 20.8 × 60 = 998.4 J ≈ 998 J . Using first law ΔU = Q - W, W

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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

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

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Cyclic Processes and Reversibility Concepts

A system releases 700 J of heat and does 250 J of work. What is 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. First Law: Δ Q = Δ U + Δ W . Δ Q = -700 J (heat released), Δ W = 250 J (work by system). -700 = Δ U + 250 ⇒ Δ U = -700 - 250 = -950 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 =

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Cyclic Processes and Reversibility Concepts

What is the change in internal energy for 0.5 moles of an ideal gas heated from 250 K to 300 K at constant volume? ( C_v

**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. Δ U = μ C_v Δ T . μ = 0.5 , C_v = 20.8 , Δ T = 300 - 250 = 50 . Δ U = 0.5 × 20.8 × 50 = 520 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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Isochoric Processes and Pressure-Temperature

A system absorbs 600 J of heat and does 150 J of work. What is the change in internal energy?

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. First Law: Δ Q = Δ U + Δ W . Given Δ Q = 600 J , Δ W = 150 J (work by system). 600 = Δ U + 150 ⇒ Δ U = 600 - 150 = 450 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

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Isochoric Processes and Pressure-Temperature

A system releases 670 J of heat and has 230 J of work done on it. What is the change in internal energy?

**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. First Law: Δ Q = Δ U + Δ W . Δ Q = -670 (heat released), Δ W = -230 (work on system). -670 = Δ U - 230 ⇒ Δ U = -670 + 230 = -440 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

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