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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Derive the relation between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.

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प्रश्न

Derive the relation between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.

संख्यात्मक
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उत्तर

When the system at constant pressure undergoes changes from an initial state with H1, U1, V1 and P parameters to a final state with H2, U2, V2 and P parameters, the change in enthalpy ∆H, is given by

H = U + PV

At initial state H1 = U1 + PV1    ...(1)

At final state H2 = U2 + PV2    ...(2)

change in enthalpy is (2) – (1)

⇒ (H2 – H1) = (U2 – U1) + P(V2 – V1)

∆H = ∆U + P∆V     ...(3)

As per first law of thermodynamics,

∆U = q + w; w = - P∆V 

Equation (3) becomes

∆H = q + w + P∆V

∆H = qp – P∆V + P∆V

∆H = qp    ...(4)

qp is the heat absorbed at constant pressure and is considered as heat content.

Consider a closed system of gases which are chemically reacting to form gaseous products at constant temperature and pressure with Vi and Vf, as the total volumes of the reactant and product gases respectively, and ni and nf as the number of moles of gaseous reactants and products, then,

For reactants: P Vi = ni RT   ...(5)

For products: P Vf = nf RT   ...(6)

(6) - (5)

P (Vf – Vi) = (nf – ni) RT

P∆V = ∆ng RT   ...(7)

Substituting in (7) in (3)

∆H = ∆U + P∆V

∆H = ∆U + ∆ng RT 

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पाठ 7: Thermodynamics - Evaluation [पृष्ठ २२५]

APPEARS IN

सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 11 TN Board
पाठ 7 Thermodynamics
Evaluation | Q II. 25. | पृष्ठ २२५

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