Question

Derive the expression of Delta H = Delta u+ P Delta v

Answer 1

The relationship between delta H and delta U for an ideal gas is:

delta H = delta U + P delta V

delta H = change in enthalpy of the system (H2-H1)

delta U = change in internal energy of the system (U2-U1)

P = pressure

delta V = change in the volume of the system (V2-V1)

Derivation

We know that work done w = -P(V2-V1)

delta U = q+w (q is the qty of heat transferred from the surroundings to the system )

delta U = q - P(V2-V1)

U2-U1 = q - P(V2-V1)

q = (U2 +PV2)- (U1+PV1)

q = (U2-U1) + P(V2-V1)

The quantity U+PV is known as enthalpy of the system and denoted by H

i.e H = U+PV As U, Hand V are definite properties and depends upon the states(initial and final) of the system.

q = H2-H1=delta H

Therefore delta H = delta U + P delta V

Answer 2

Answer:

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