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#constant volume

29 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

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

**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. For diatomic gas: C_v = (5)/(2) R . C_v = (5)/(2) × 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

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

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Heat Transfer Work Distinction and Internal Energy Change

0.5 kg of a substance at 15°C absorbs 1800 J of heat at constant volume, reaching 45°C. What is its specific heat capaci

**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. Specific heat: s = (Δ Q)/(m Δ T) . Δ Q = 1800 J , m = 0.5 kg , Δ T = 45 - 15 = 30 K . s = (1800)/(0.5 × 30) = 120 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W,

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

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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

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

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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

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

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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

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

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Specific Heat Capacity and Latent Heat

Why does the specific heat capacity of a gas differ at constant pressure and constant volume?

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. At constant pressure ( C_p ), heat supplies energy for both internal energy increase and work done due to expansion ( Δ Q = Δ U + P Δ V ). At constant volume ( C_v ), no work is done ( Δ V = 0 ), so heat

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

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Specific Heat Capacity and Latent Heat

A gas at 3 atm in a 4 L container is heated from 25°C to 75°C at constant volume. What is the final pressure?

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. Isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 3 atm , T₁ = 25 + 273 = 298 K , T₂ = 75 + 273 = 348 K . (3)/(298) = (P₂)/(348) ⇒ P₂ = (3 × 348)/(298) ≈ 3.5 atm . Using first law ΔU = Q - W,

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

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the change in internal energy when 1 mole of an ideal gas is heated from 300 K to 350 K at constant volume? ( C_

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. Δ U = μ C_v Δ T . μ = 1 , C_v = 20.8 , Δ T = 350 - 300 = 50 . Δ U = 1 × 20.8 × 50 = 1040 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

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

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the molar specific heat capacity at constant volume for a monatomic ideal gas? (Take R = 8.3 J mol⁻¹ K⁻¹ )

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. For a monatomic gas, C_v = (3)/(2) R . C_v = (3)/(2) × 8.3 = 12.45 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 12.45 J

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

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First Law of Thermodynamics Applications

What is the change in internal energy for 0.5 moles of an ideal gas heated from 260 K to 310 K at constant volume? ( C_v

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. Δ U = μ C_v Δ T . μ = 0.5 , C_v = 20.8 , Δ T = 310 - 260 = 50 . Δ U = 0.5 × 20.8 × 50 = 520 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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First Law of Thermodynamics Applications

What is the change in internal energy for 0.4 moles of an ideal gas heated from 290 K to 330 K at constant volume? ( C_v

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Δ U = μ C_v Δ T . μ = 0.4 , C_v = 20.8 , Δ T = 330 - 290 = 40 . Δ U = 0.4 × 20.8 × 40 = 332.8 J ≈ 333 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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