#ideal gas

Work Heat Distinction and Energy Transfer Modes

A gas undergoes an adiabatic expansion from 28 L to 84 L , reducing its pressure from 15 atm to 3 atm . What is the valu

**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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Heat Transfer Work Distinction and Internal Energy Change

A gas undergoes an isothermal compression from 8 L to 2 L at 350 K with 0.2 moles . What is the heat released? ( R = 8.3

**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 isothermal: W = μ R T ln((V₂)/(V₁)) , Δ U = 0 , Q = W . W = 0.2 × 8.3 × 350 × ln((2)/(8)) = 581 × ln(0.25) . ln(0.25) = -ln(4) ≈ -1.386 . W = 581 × (-1.386) ≈ -805 J . Q = -805 J (negative implies heat released). Using first law ΔU = Q - W, W = ∫

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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 diatomic gas undergoes an adiabatic expansion from 760 K to 380 K with 0.8 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.8 , R = 8.3 , T₁ = 760 , T₂ = 380 , γ = 1.4 . W = (0.8 × 8.3 × (760 - 380))/(1.4 - 1) = (6.64 × 380)/(0.4) = 6312 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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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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Heat Transfer Work Distinction and Internal Energy Change

Which of the following statements is correct about C_p and C_v for an ideal gas?

**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, C_p > C_v because at constant pressure, heat supplies both internal energy increase and work ( C_p = C_v + R ), while at constant volume, heat only increases internal energy. Option A is correct. Using first law ΔU

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Heat Transfer Work Distinction and Internal Energy Change

0.3 moles of an ideal gas at 400 K expand adiabatically from 6 atm to 2 atm. If gamma = 1.5 , what is the final temperat

**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. Adiabatic: T₁ V₁^γ-1 = T₂ V₂^γ-1 , P V = μ R T ⇒ V₁ = (μ R T₁)/(P₁) = (0.3 × 8.3 × 400)/(6) = 166 L , V₂ = (0.3 × 8.3 × T₂)/(2) = 1.245 T₂ . 400 × 166⁰.5 = T₂ × (1.245 T₂)⁰.5 .

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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

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

An ideal gas expands isothermally at 460 K from 3 L to 9 L with 0.5 moles . What is the work done by the gas? ( R = 8.3

**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 isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.5 , R = 8.3 , T = 460 , V₂ = 9 , V₁ = 3 . W = 0.5 × 8.3 × 460 × ln((9)/(3)) = 1909 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1909 × 1.0986 ≈ 2097 J . Using first law ΔU = Q - W,

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Isochoric Processes and Pressure-Temperature

0.2 moles of an ideal gas at 360 K are compressed isothermally from 8 L to 2 L. What is the heat released? ( R = 8.3 J m

**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. Isothermal: Δ U = 0 , Δ Q = Δ W . W = μ R T ln((V₂)/(V₁)) . μ = 0.2 , T = 360 , V₂ = 2 , V₁ = 8 . W = 0.2 × 8.3 × 360 × ln((2)/(8)) = 597.6 × (-1.386) ≈ -829 J (work by

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Isochoric Processes and Pressure-Temperature

0.15 moles of an ideal gas at 320 K expand isothermally from 3 L to 9 L. What is the work done by the gas? ( R = 8.3 J m

**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. Isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.15 , T = 320 , V₂ = 9 , V₁ = 3 . W = 0.15 × 8.3 × 320 × ln((9)/(3)) = 398.4 × 1.0986 ≈ 438 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =

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