Skip to content

#pressure increase

4 public questions tagged with this topic.

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

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

A gas at 1 atm and 273 K has a volume of 8 litres. If the pressure increases to 4 atm at constant temperature, what is t

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 1 atm, V₁ = 8 litres, P₂ = 4 atm.V₂ = (P₁ V₁)/(P₂) = (1 × 8)/(4) = 2 litres. Substituting values gives 2 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 > Gas Laws and Volume-Temperature Relations