Derive the relationship between ΔH and ΔU for an ideal gas. Explain each term involved in the equation.
From the first law of thermodynamics
∆U = q + W --- (1)
∆U= internal energy
q= heat absorbed
W= work done
We know W = -p∆V --- (2)
Therefore we can write equation-1 as
∆U = q - p∆V--- (3)
Now if the system absorb qpinternal energy changes from U1 to U2 and volume increase from V1 to V2
∆U = U2 - U1---(4)
∆V = V2 - V1 --- (5)
Substitute the value of ∆U and ∆V from equation-4 and equation-5 in equation-3 ie.
∆U = qp - p∆V
U2 - U1= qp - p (V2 - V1)
qp = U2 - U1+ p (V2 - V1)
U, P, V are state functions and U + PV is heat content or enthalpy of the system which is denoted by
H = U + PV
So qp = H2 – H1
qp = ∆H
Therefore,∆H = ∆U + p∆V--- (6)
At constant temperature when the heat is absorbed, we measure the changes in enthalpy
∆H = qp (when the heat is absorbed by the system at constant pressure)
∆H is positive for an endothermic reaction.
∆H is negative for an exothermic reaction.
At constant volume (∆V = 0)
∆U = qv
Therefore,
∆H = ∆U = qv
Now using ideal gas law if VA is the total volume of gas reactant and VB is the volume of gas product and nA and nB are moles of gas reactant and product respectively.
pVA = nART
pVB = nBRT
Thus, pVB- pVA = nBRT - nART
p(VB – VA) = RT (nB- nA)
p∆V = ∆ngRT --- (7)
Put the value of p∆V from equation-7 to equation-5
Therefore,
∆H = ∆U + ∆ngRT
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