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dU = dq + dw

dU = dq + dw. Now 1st Law becomes :. dU = C v dT + PdV. Can also define constant – pressure heat capacity:. C P ≡ (∂q/∂T) P. Finally:. C P – C V = R. At constant volume: dU = dq If system can change volume, dU ≠ dq Some heat into the system is converted to work ∴ dU < dq

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dU = dq + dw

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  1. dU = dq + dw Now 1st Law becomes: dU = CvdT + PdV Can also define constant – pressure heat capacity: CP≡ (∂q/∂T)P Finally: CP– CV = R

  2. At constant volume: dU = dq • If system can change volume, • dU ≠ dq • Some heat into the system • is converted to work • ∴ dU < dq • Constant pressure processes • much more common than constant volume processes

  3. dU = dq + dw • = dq + PdV • Enthalpy, H defined as: H ≡ U + PV • At constant pressure: dH = dqP • For a measurable change: ΔH = qP • Properties of H: • absolute H cannot be determined • state function • H is an extensive physical property

  4. Plot of enthalpy as a function of temperature H = U + PV Slope = (∂H/∂T)P The heat capacity at constant pressure:

  5. Variation of Enthalpy with Temperature At constant pressure: dH = CP dT ∫dH = CP ∫dT (accurate only for a monatomic gas and/or small changes in T) ΔH = CPΔT For molecular gases and large changes in T:

  6. Use an approximate empirical expression: CP = a + bT + cT-2 e.g. For N2: a = 1021, b = 134.6, c = -17.9 (all in J K-1 kg-1) Determine ΔH when air is heated from -25°C to 20°C. (Assume air is pure N2.)

  7. Consider dry air in a layer heating and expanding: • The air expands and rises doing work vs gravity • U increases according to dU = CvdT • Work is done according to w = PdV • Assume air is N2 and O2 alone, then heat (dq) added • is partitioned in a ratio of 5:2 as: • dq = dU + PdV • U receives 2.5R units of heat, w receives R units

  8. Consider dry air in a layer heating and expanding: • A more general expression for a moving dry air • parcel which varies in P as it rises or sinks: • dΦ = - V dP • dq = dH - V dP • dH = Cp dT • dq = d(H + Φ) = d(Cp dT + Φ) • d(H + Φ) ≡ dry static energy

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