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#pressure calculation

47 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A gas at 11 atm and 90^circ C in a 10 L container is cooled isochorically to 30^circ C . What is the final pressure?

**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). For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 11 atm , T₁ = 90 + 273 = 363 K , T₂ = 30 + 273 = 303 K . (11)/(363) = (P₂)/(303) ⇒ P₂ = (11 × 303)/(363) ≈ 9.18 atm . Using first law ΔU = Q - W, W =

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

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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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Specific Heat Capacity and Latent Heat

A gas at 9 atm and 70^circ C in a 6 L container is cooled isochorically to 10^circ C . 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. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 9 atm , T₁ = 70 + 273 = 343 K , T₂ = 10 + 273 = 283 K . (9)/(343) = (P₂)/(283) ⇒ P₂ = (9 × 283)/(343) ≈ 7.42 atm . Using first law ΔU = Q -

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

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Zeroth Law Thermal Equilibrium and Quasi-static

A gas at 4 atm and 27^circ C in a 3 L container is cooled isochorically to -73^circ C . What is the final pressure?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 4 atm , T₁ = 27 + 273 = 300 K , T₂ = -73 + 273 = 200 K . (4)/(300) = (P₂)/(200) ⇒ P₂ = (4 × 200)/(300) = (8)/(3) ≈ 2.67 atm . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic

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

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Isobaric and Isothermal Processes Work Calculation

A gas is compressed adiabatically from 8 L to 2 L . If the initial pressure is 1 atm and gamma = 1.4 , what is the final

**Isobaric process** constant pressure, work W = P ΔV = P(V₂ - V₁) = n R ΔT, for expansion ΔV positive W positive, for compression negative, heat Q = n C_p ΔT, ΔU = n C_v ΔT. Isothermal process constant temperature ΔU=0, work W = n R T ln(V₂/V₁) = n R T ln(P₁/P₂), Q = W, heat absorbed equals work done. P₁ V₁^γ = P₂ V₂^γ . P₁ = 1 atm , V₁ = 8 L , V₂ = 2 L , γ = 1.4 . 1 × 8¹.4 = P₂ × 2¹.4 . P₂ = 8¹.42¹.4 = ((8)/(2))¹.4 = 4¹.4 . 4¹.4 =

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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Isobaric and Isothermal Processes Work Calculation

A gas at 3 atm and 400 K is cooled isochorically to 200 K . What is the final pressure?

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 3 , T₁ = 400 , T₂ = 200 . (3)/(400) = (P₂)/(200) ⇒ P₂ = (3 × 200)/(400) = 1.5 atm . 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 =

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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Partial Pressures and Gas Mixtures

What is the pressure exerted by 0.1 mole of an ideal gas in a 2-litre container at 127°C? (R = 8.31 J mol⁻¹ K⁻¹)

**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. PV = μ R T, P = (μ R T)/(V).T = 127 + 273 = 400 K, V = 2 × 10⁻³ m³.P = (0.1 × 8.31 × 400)/(2 × 10⁻³) = 1.663 × 10⁵ Pa ≈ 1.66 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 1.66 atm, which matches expected kinetic theory result, confirming mean free path λ =

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 mixture of 0.5 moles of helium and 1.5 moles of oxygen is at 350 K in a 25-litre container. What is the total pressure

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. PV = μ R T, P = (μ R T)/(V).Total moles = 0.5 + 1.5 = 2, V = 25 × 10⁻³ m³.P = (2 × 8.31 × 350)/(25 × 10⁻³) = 2.326 × 10⁵ Pa ≈ 2.33 atm. Substituting values gives 2.33 atm, 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

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

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Molecular Mass Density and Ideal Gas Equation

What is the pressure of 0.8 moles of an ideal gas in a 16-litre container at 427°C? (R = 8.31 J mol⁻¹ K⁻¹)

**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. PV = μ R T, P = (μ R T)/(V).T = 427 + 273 = 700 K, V = 16 × 10⁻³ m³.P = (0.8 × 8.31 × 700)/(16 × 10⁻³) = 2.90625 × 10⁵ Pa ≈ 2.91 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 2.91 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

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

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Molecular Mass Density and Ideal Gas Equation

A mixture of 1 mole of argon and 0.5 moles of nitrogen is at 450 K in a 15-litre container. What is the total pressure?

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. PV = μ R T, P = (μ R T)/(V).Total moles = 1 + 0.5 = 1.5, V = 15 × 10⁻³ m³.P = (1.5 × 8.31 × 450)/(15 × 10⁻³) = 3.74 × 10⁵ Pa ≈ 3.74 atm. Substituting values gives 3.74 atm, 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

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

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RMS Speed and Temperature Dependence

What is the pressure of 0.25 moles of an ideal gas in a 5-litre container at 227°C? (R = 8.31 J mol⁻¹ K⁻¹)

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. PV = μ R T, P = (μ R T)/(V).T = 227 + 273 = 500 K, V = 5 × 10⁻³ m³.P = (0.25 × 8.31 × 500)/(5 × 10⁻³) = 2.0775 × 10⁵ Pa ≈ 2.08 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 2.08 atm, which matches expected kinetic theory result,

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

A mixture of 1 mole of neon and 2 moles of argon is at 500 K in a 30-litre container. What is the total pressure? (R = 8

**Temperature dependence of RMS speed** v_rms ∝ √T, doubling T increases v_rms by √2≈1.414, e.g., at 300 K v_rms for O₂ ≈483 m/s, at 600 K ≈683 m/s, illustrating kinetic theory relation between temperature and molecular motion, average kinetic energy ½ m v_rms² =3/2 k_B T. PV = μ R T, P = (μ R T)/(V).Total moles = 1 + 2 = 3, V = 30 × 10⁻³ m³.P = (3 × 8.31 × 500)/(30 × 10⁻³) = 4.155 × 10⁵ Pa ≈ 4.16 atm. Substituting values gives 4.16 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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