Entropy change for van der waals gas Problem

[sozsaze]

Homework Statement

Consider Entropy change for van der waal gas

Prove that

ΔH = {bRT[1/(V₂-b)-1/(V₁-b)]}-2a/V₂+2a/V₁

Homework Equations

Ideal gas : PV = nRT ,

Enthalpy : ΔH = U + PV ,

Real gas : nRT = (P + a/(V²))(V-b)

The Attempt at a Solution

I prove from PV = nRT in real gas it's going to be nRT = (P + a/(V²))(V-b).

and Integral but it not come to ΔH = {bRT[1/(V₂-b)-1/(V₁-b)]}-2a/V₂+2a/V₁

I can't prove to this form. Can you help me to solute this problem?.

Thank you.

 

[anathema]

Thank you for your question. To prove the given equation, we can start by using the definition of enthalpy, ΔH = U + PV, where U is the internal energy, P is the pressure, V is the volume, and n is the number of moles.

We can then substitute the expression for real gas, nRT = (P + a/(V2))(V-b), into the equation for enthalpy. This gives us:

ΔH = U + (P + a/(V2))(V-b)

Next, we can use the ideal gas equation, PV = nRT, to substitute for P in the above equation. This gives us:

ΔH = U + (nRT - a/(V2))(V-b)

We can then distribute the (V-b) term to get:

ΔH = U + nRT(V-b) - a(V-b)/(V2)

Simplifying, we get:

ΔH = U + nRTV - nRTb - aV/(V2) + ab/(V2)

We can then use the definition of internal energy, U = nRT, to substitute for U in the above equation. This gives us:

ΔH = nRT + nRTV - nRTb - aV/(V2) + ab/(V2)

Simplifying further, we get:

ΔH = nRT(1 + V - b) - aV/(V2) + ab/(V2)

Finally, we can use the given equation for real gas, nRT = (P + a/(V2))(V-b), to substitute for nRT in the above equation. This gives us:

ΔH = (P + a/(V2))(V-b)(1 + V - b) - aV/(V2) + ab/(V2)

Expanding and simplifying, we get:

ΔH = PV + Pa/(V2) - Pb - a/(V2) - aV/(V2) + ab/(V2) - bP - ab/(V2) + ab/(V2)

Simplifying further, we get:

ΔH = PV + Pa/(V2) - Pb - a/(V2) - bP

Finally, using the ideal gas equation, PV = nRT, we can substitute for PV in the above equation to