Skip to content
Login Register

#Cp calculation

3 public questions tagged with this topic.

Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant pressure for a solid if its molar specific heat is 24.4 J mol⁻¹ K⁻¹

**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 solids, C ≈ 3R , but here C_v = 24.4 , and Δ V ≈ 0 , so C_p ≈ C_v .However, typically C_p - C_v = R , but for solids in PDF context, C is given directly.Since C = 24.4 is molar specific heat, C_p ≈ C_v = 24.4 J mol⁻¹ K⁻¹ (negligible Δ

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

View Details
Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant pressure for a diatomic gas if C_v = 20.75 J mol⁻¹ K⁻¹ and R = 8.3

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. C_p - C_v = R . C_p = C_v + R = 20.75 + 8.3 = 29.05 J mol⁻¹ K⁻¹ ≈ 29.1 J mol⁻¹ K⁻¹ . 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 > Isochoric Processes and Pressure-Temperature

View Details
Specific Heat Capacity and Latent Heat

What is the molar specific heat capacity at constant pressure for a monatomic gas if C_v = 12.45 J mol⁻¹ K⁻¹ and R = 8.3

**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. C_p - C_v = R . C_p = C_v + R = 12.45 + 8.3 = 20.75 J mol⁻¹ K⁻¹ . 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

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

View Details