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#isochoric cooling

2 public questions tagged with this topic.

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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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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