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#gas laws

27 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 is compressed adiabatically from 16 L to 4 L , increasing its pressure from 2 atm to 8 atm . What is gamma ?

**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. P₁ V₁^γ = P₂ V₂^γ . 2 × 16^γ = 8 × 4^γ . (16^γ)/(4^γ) = (8)/(2) ⇒ ((16)/(4))^γ = 4 ⇒ 4^γ = 4¹ . γ = 1 , but check context—use γ = 1.33 as standard approximation. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

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

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Cyclic Processes and Reversibility Concepts

A gas is compressed adiabatically from 12 L to 3 L , increasing its pressure from 2 atm to 16 atm . What is gamma ?

**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. P₁ V₁^γ = P₂ V₂^γ . 2 × 12^γ = 16 × 3^γ . (12^γ)/(3^γ) = (16)/(2) ⇒ ((12)/(3))^γ = 8 ⇒ 4^γ = 8 . 4^γ = 2³ ⇒ 2²γ = 2³ ⇒ 2γ = 3 ⇒ γ = 1.5 . Using first law ΔU = Q - W, W = ∫ P

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

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Isochoric Processes and Pressure-Temperature

What is the physical significance of the equation of state in thermodynamics?

**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. The equation of state (e.g., P V = μ R T for an ideal gas) relates state variables (pressure, volume, temperature), defining the equilibrium state of a system and showing their interdependence. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

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

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Specific Heat Capacity and Latent Heat

A gas undergoes an adiabatic expansion from 10 L to 40 L , reducing its pressure from 8 atm to 1 atm . What is the value

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 8 × 10^γ = 1 × 40^γ . 8 = ((40)/(10))^γ ⇒ 8 = 4^γ . 4^γ = 2³ ⇒ 2²γ = 2³ ⇒ 2γ = 3 ⇒ γ = 1.5 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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 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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Isobaric and Isothermal Processes Work Calculation

A gas is compressed adiabatically, increasing its pressure from 1 atm to 4 atm in a 10 L container. What is the final vo

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. P₁ V₁^γ = P₂ V₂^γ . 1 × 10¹.6 = 4 × V₂¹.6 . V₂¹.6 = 10¹.64 . V₂ = (10¹.64)¹/1.6 = 10 × 4⁻¹/1.6 . 4⁻⁰.625 ≈ 0.315 , V₂ ≈ 10 × 0.315 ≈ 3.15 L . 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 > Isobaric and Isothermal Processes Work Calculation

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Isobaric and Isothermal Processes Work Calculation

An ideal gas expands isothermally at 480 K from 6 L to 18 L with 0.6 moles . What is the work done by the gas? ( R = 8.3

**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. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.6 , R = 8.3 , T = 480 , V₂ = 18 , V₁ = 6 . W = 0.6 × 8.3 × 480 × ln((18)/(6)) = 2390.4 × ln(3) .

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

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Mean Free Path and Molecular Diameter

The rms speed of a gas is 500 m/s at 250 K. At what temperature will the rms speed be 1000 m/s?

**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. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(1000)/(500) = √((T₂)/(250)), 2 = √((T₂)/(250)).Square both sides: 4 = (T₂)/(250), T₂ = 1000 K. Substituting values gives 1000 K, 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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