Isochoric Processes and Pressure-Temperature

Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant pressure for a solid if its molar specific heat is 24.4 J mol⁻¹ K⁻¹

**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. For solids, C ≈ 3R , but here C_v = 24.4 , and Δ V ≈ 0 , so C_p ≈ C_v .However, typically C_p - C_v = R , but for solids in PDF context, C is given directly.Since C = 24.4 is molar specific heat, C_p ≈ C_v = 24.4 J mol⁻¹ K⁻¹ (negligible Δ

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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

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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 =

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Isochoric Processes and Pressure-Temperature

How much heat is required to raise the temperature of 1 kg of copper from 20^circ C to 50^circ C ? (Specific heat of cop

**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. Heat capacity: Δ Q = m s Δ T . m = 1 kg , s = 386.4 J kg⁻¹ K⁻¹ , Δ T = 50 - 20 = 30 K . Δ Q = 1 × 386.4 × 30 = 11592 J . Using first law ΔU = Q - W, W = ∫ P dV,

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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

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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

A gas undergoes an adiabatic expansion from 25 L to 100 L , reducing its pressure from 16 atm to 1 atm . What is the val

**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. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 16 × 25^γ = 1 × 100^γ . 16 = ((100)/(25))^γ ⇒ 16 = 4^γ . 4^γ = 2⁴ ⇒ 2²γ = 2⁴ ⇒ 2γ = 4 ⇒ γ = 2 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant pressure for a diatomic gas if C_v = 20.75 J mol⁻¹ K⁻¹ and R = 8.3

**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. C_p - C_v = R . C_p = C_v + R = 20.75 + 8.3 = 29.05 J mol⁻¹ K⁻¹ ≈ 29.1 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

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Isochoric Processes and Pressure-Temperature

A monatomic gas undergoes an adiabatic expansion from 820 K to 410 K with 0.9 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.9 , R = 8.3 , T₁ = 820 , T₂ = 410 , γ = 1.67 . W = (0.9 × 8.3 × (820 - 410))/(1.67 - 1) = (7.47 × 410)/(0.67) ≈ 4570.15 J ≈ 4570 J . Using first law ΔU = Q - W, 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

An ideal gas expands isothermally at 510 K from 8 L to 24 L with 0.2 moles . What is the work done by the gas? ( R = 8.3

**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 isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.2 , R = 8.3 , T = 510 , V₂ = 24 , V₁ = 8 . W = 0.2 × 8.3 × 510 × ln((24)/(8)) = 846.6 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 846.6 × 1.0986 ≈

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Isochoric Processes and Pressure-Temperature

A system absorbs 850 J of heat and has 400 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 = 850 , Δ W = -400 (work done on system). 850 = Δ U - 400 ⇒ Δ U = 850 + 400 = 1250 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

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