Thermodynamics: Liquid System at High Pressure - Calculating ΔH and ΔU

[Kishlay]

Thermodynamics, Please Help...!

1. Homework Statement

100 ml of liquid contained in an insulated container at pressure of 1 bar. The pressure is steeply increased to 100 bars. The volume of the liquid decreased by 1 ml at this pressure. Find ΔH and ΔU

2. Homework Equations
How to deal with a system involving a liquid ??
3. The Attempt at a Solution
i am not getting how to attempt this question as i have faced only a system involving a gas. Please help me to clear this concept.

 

[Kishlay]

i know ΔH=ΔU + Δ(PV)
i have tried this equation for liquids, and hence doesn't work

 

[Chestermiller]

1. Homework Statement

100 ml of liquid contained in an insulated container at pressure of 1 bar. The pressure is steeply increased to 100 bars. The volume of the liquid decreased by 1 ml at this pressure. Find ΔH and ΔU

2. Homework Equations
How to deal with a system involving a liquid ??
3. The Attempt at a Solution
i am not getting how to attempt this question as i have faced only a system involving a gas. Please help me to clear this concept.

First law: ΔU=-∫PdV=-Δ(PV)+∫VdP

 

[Kishlay]

can you please elaborate it... so that i can understand more clearly...

 

[rude man]

@Chet, request permission to butt in! And please comment.

@Kishlay, what you need is some equation of state for your liquid. A simple model might be an inverse relationship between p and V, but T will rise so T should probably be involved also. In other words, you might assume a constant ∂V/∂p relating volume to pressure and another constant ∂V/∂T relating volume to temperature:

dV = ∂V/∂p dp + ∂V/∂T dT.

For a more rigorous approach the van der Waal equation of state is supposed to be useful in the liquid region.

 

[Chestermiller]

can you please elaborate it... so that i can understand more clearly...

You need to apply the first law of thermo to this problem. The container is insulated, so dQ = 0. So, the change in internal energy is going to be equal to minus the amount of work done on the surroundings. The amount of work done on the surroundings is ∫PdV. So,

ΔU=-∫PdV. If we integrate this by parts, we get:
\Delta U=-\Delta(PV)+∫VdP
Now, from the problem statement, we know the Δ(PV) exactly. In the integral term, the integrand (volume) varies monotonically with pressure from 100 ml to 99 ml over the course of the pressure change. This is only 1 percent, so we can get the value of the second integral to an accuracy within 1 percent simply by using the arithmetic average of the initial and final volumes, and writing:
∫VdP=\frac{(V_i+V_f)}{2}ΔP
Is this enough information to complete the solution to the problem?
Chet

 

[Kishlay]

hmmmmmmmmm ok thanks, i will try it once again. thanks!