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#constant pressure

12 public questions tagged with this topic.

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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Adiabatic Processes and Gamma Determination

Why does an isobaric process involve both internal energy change and work?

**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. In an isobaric process ( P = constant ), heat added ( Δ Q ) increases internal energy ( Δ U ) and does work ( W = P Δ V ) due to volume expansion, as per the First Law: Δ Q = Δ U + P Δ V . Using first law ΔU = Q - W, W

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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Partial Pressures and Gas Mixtures

A gas at 2 atm and 300 K has a volume of 5 litres. If the temperature rises to 600 K at constant pressure, what is the n

**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 5 litres, T₁ = 300 K, T₂ = 600 K.V₂ = V₁ × (T₂)/(T₁) = 5 × (600)/(300) = 10 litres. Substituting values gives 10.0 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

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Molecular Mass Density and Ideal Gas Equation

A gas at 1 atm and 273 K has a volume of 15 litres. If the temperature increases to 819 K at constant pressure, what is

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 15 litres, T₁ = 273 K, T₂ = 819 K.V₂ = V₁ × (T₂)/(T₁) = 15 × (819)/(273) = 45 litres. Substituting values gives 45 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

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Degrees of Freedom and Molar Specific Heat

A gas at 2 atm and 400 K has a volume of 8 litres. If the temperature decreases to 200 K at constant pressure, what is t

**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 8 litres, T₁ = 400 K, T₂ = 200 K.V₂ = V₁ × (T₂)/(T₁) = 8 × (200)/(400) = 4 litres. Substituting values gives 4 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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Gas Laws and Volume-Temperature Relations

A gas has a volume of 22.4 litres at STP. How many moles are present if the temperature is raised to 546 K at constant p

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. At STP, 22.4 litres = 1 mole.Charles’ law: (V₁)/(T₁) = (V₂)/(T₂), but moles remain constant at constant P.Initial μ = 1 mol, remains 1 mole as V adjusts with T. Substituting values gives 1.0 mol, which matches expected kinetic theory result, confirming mean free path

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Mean Free Path and Molecular Diameter

A gas at 1 atm and 273 K has a volume of 11.2 litres. If the temperature increases to 546 K at constant pressure, what i

**Kinetic theory mean free path** λ = 1/(√2 π d² n) quantifies collision frequency. With n =1.0×10²⁵ m⁻³, λ=9×10⁻⁷ m, d² =1/(1.414×10²⁵×3.14×9×10⁻⁷)=2.5×10⁻²⁰ m², d≈1.58×10⁻¹⁰ m, typical molecular size ~10⁻¹⁰ m, consistent with gas kinetic theory. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 11.2 litres, T₁ = 273 K, T₂ = 546 K.V₂ = V₁ × (T₂)/(T₁) = 11.2 × (546)/(273) = 22.4 litres. Substituting values gives 22.4 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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Mean Free Path and Molecular Diameter

A gas at 1.5 atm and 300 K has a volume of 24 litres. If the temperature increases to 600 K at constant pressure, what i

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 24 litres, T₁ = 300 K, T₂ = 600 K.V₂ = V₁ × (T₂)/(T₁) = 24 × (600)/(300) = 48 litres. Substituting values gives 48 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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Thermodynamics

A system at constant pressure undergoes an enthalpy change of -200 kJ. This means:

The system releases heat is the scientifically accurate answer to this question. Within the study of Thermodynamics, this concept is well-established through extensive research and is documented in standard scientific literature. The specific properties, mechanisms, or characteristics of The system releases heat directly address what is being asked. Among the other options, The system absorbs heat, The system does no work, and The system remains unchanged do not correctly answer this question because they either refer to different concepts, describe properties of other molecules or processes, or represent common misconceptions about this topic.

Ref: Lehninger Principles of Biochemistry, Nelson & Cox, 8th Ed., Ch. 1

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