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```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

VI = 83.14×10-5 m3

WI = (83.14×10-5)´(100-60)×105

= 3325.6 Jules

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

WII = V. ΔP

WII = 138.56×10-5 (60-30)×105

= 4156.99 »4157 J.

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

WIII = 277.13×10-5 (30-1)´105

WIII =8036.86 J.

W total = WI + WII + WIII

= 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= CvΔT.

and Cp-Cv = R

On dividing all the terms by Cv.

Cp/Cv – Cv/Cv – R/Cv

Cp/Cv = γ

(γ – 1) = R/Cv

and Cv = 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.

To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.
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