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#pressure drop

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Isochoric Processes and Pressure-Temperature

A gas undergoes an adiabatic expansion from 25 L to 100 L , reducing its pressure from 16 atm to 1 atm . What is the val

**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 adiabatic: P₁ V₁^γ = P₂ V₂^γ . 16 × 25^γ = 1 × 100^γ . 16 = ((100)/(25))^γ ⇒ 16 = 4^γ . 4^γ = 2⁴ ⇒ 2²γ = 2⁴ ⇒ 2γ = 4 ⇒ γ = 2 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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

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Specific Heat Capacity and Latent Heat

In an adiabatic process, a gas expands from a volume of 1 L to 4 L , reducing its pressure from 16 atm to 1 atm . What i

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. For an adiabatic process, P₁ V₁^γ = P₂ V₂^γ .Substitute: 16 × 1^γ = 1 × 4^γ . 16 = 4^γ .Taking log: log(16) = γ log(4) . log(16) = log(2⁴) = 4 log(2) , log(4) = log(2²) = 2 log(2) . 4 log(2) = γ × 2 log(2) ⇒ γ = (4)/(2) = 2 . Using first law ΔU = Q -

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

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