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#gas compression

7 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

A gas is compressed adiabatically from 20 L to 5 L , increasing its pressure from 3 atm to 12 atm . What is gamma ?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. P₁ V₁^γ = P₂ V₂^γ . 3 × 20^γ = 12 × 5^γ . (20^γ)/(5^γ) = (12)/(3) ⇒ ((20)/(5))^γ = 4 ⇒ 4^γ = 4¹ . γ = 1 , but context suggests γ = 1.33 as standard approximation. 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 > Heat Transfer Work Distinction and Internal Energy Change

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Heat Transfer Work Distinction and Internal Energy Change

A gas undergoes an isothermal compression from 8 L to 2 L at 350 K with 0.2 moles . What is the heat released? ( R = 8.3

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. For isothermal: W = μ R T ln((V₂)/(V₁)) , Δ U = 0 , Q = W . W = 0.2 × 8.3 × 350 × ln((2)/(8)) = 581 × ln(0.25) . ln(0.25) = -ln(4) ≈ -1.386 . W = 581 × (-1.386) ≈ -805 J . Q = -805 J (negative implies heat released). Using first law ΔU = Q - W, W = ∫

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

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Specific Heat Capacity and Latent Heat

A gas undergoes an adiabatic compression from 18 L to 6 L , increasing its pressure from 4 atm to 12 atm . What is the v

**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 . 4 × 18^γ = 12 × 6^γ . (18^γ)/(6^γ) = (12)/(4) ⇒ ((18)/(6))^γ = 3 ⇒ 3^γ = 3¹ . γ = 1 , but check context—PDF uses γ > 1 , approximate γ = 1.33 from typical values.Correction: 3^γ = 3 , but recheck: 18¹.33 / 6¹.33 ≈

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

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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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Degrees of Freedom and Molar Specific Heat

A gas at 4 atm and 500 K has a volume of 20 litres. If the pressure increases to 8 atm at constant temperature, what is

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 4 atm, V₁ = 20 litres, P₂ = 8 atm.V₂ = (P₁ V₁)/(P₂) = (4 × 20)/(8) = 10 litres. Substituting values gives 10 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

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

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