Enthalpy of a System

The quantity U + PV is known as the enthalpy of the system and is denoted by H. It represents the total energy stored in the system. Thus

      H = U + PV

It may be noted that like internal energy, enthalpy is also an extensive property as well as a state function. The absolute value of enthalpy can not be determined, however the change in enthalpy can be experimentally determined.

      ΔH = ΔU + Δ(PV)

Various kinds of processes:  

(i) Isothermal reversible expansion of an Ideal gas: Since internal energy of an Ideal gas is a function of temperature and it remains constant throughout the process hence  

ΔE = 0 and ΔH = ΔE + ΔPV  

            ΔE = 0  

            and P₁V₁ = P₂V₂ at constant temperature for a given amount of the gas  

            ΔH= 0  

Calculation of q and w:  

ΔE = q + w  

For an Isothermal process, w = -q  

This shows that in an Isothermal expansion, the work done by the gas is equal to amount of heat absorbed.  

            and w = - n RT ln(V₂/V₁) = - n RT ln(P₁/P₂).  

Solved examples.10 gm of Helium at 127°C is expanded isothermally from 100 atm to 1 atm Calculate the work done when the expansion is carried out (i) in single step (ii) in three steps the intermediate pressure being 60 and 30 atm respectively and (iii) reversibly.  

Solution:       (i)  Work done = V.ΔP  

                              V = (10/4) × 8.314×400 / 100 × 10⁵ = 83.14 × 10^–5 m³   

                              So W = 83×14/10⁵ (100-1) × 10⁵ =  8230.86 J.  

                   (ii) In three steps  

                              V_I = 83.14×10^-5 m³  

                              W_I = (83.14×10^-5)´(100-60)×10⁵  

                                  = 3325.6 Jules  

                              V_II = 2.5 × 8.314 × 400 / 60 × 10⁵ = 138.56 × 10⁵ m³  

                              W_II = V. ΔP  

                              W_II = 138.56×10^-5 (60-30)×10⁵  

                                   = 4156.99 »4157 J.  

                              V_III = 2.5 × 8.314 × 400 / 30 × 10⁵ = 277.13 × 10^–5 m³  

                              W_III = 277.13×10^-5 (30-1)´10⁵  

                              W_III =8036.86 J.  

                              W _total = W_I + W_II + W_III  

                              = 3325.6+4156.909+8036.86 = 15519.45 J.  

                     (iii) For reversible process  

                             W = 2.303 nRT log P₁/P₂  

                            = 2.303 × (10/4) × 8.314 × 400 × log (100/1)  

                              W = 38294.28 Jules  

Exercise:

Calculate the final volume of one mole of an ideal gas initially at 0°C and 1 atm pressure, if it absorbs 1000 cal of heat during a reversible isothermal expansion.  

Exercise:

Carbon monoxide is allowed to expand isothermally and reversibly from 10m³ to 20 m³at 300 K and work obtained is 4.754 KJ. Calculate the number of moles of carbon monoxide.    

(ii) Adiabatic Reversible Expansion of an Ideal gas:  

            q = 0  

            ΔE= -w.  

Total change in the internal energy is equal to external work done by the system.  

Work done by the system = ΔE= C_vΔT.  

            and C_p-C_v = R  

On dividing all the terms by C_v.  

            C_p/C_v – C_v/C_v – R/C_v  

            C_p/C_v = γ  

            (γ – 1) = R/C_v  

            and C_v = R / (γ – 1)  

                   

            and = R / (γ – 1) (T₂ – T₁) ΔH = ΔE + PΔV.  

Thus if T₂>T₁, w = +ve i.e. work is done on the system.  

Thus if T₂1, w = -ve i.e. work is done by the system.

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