Question

Pressure over `1000ml` of a liquid is gradually increased from `1` bar to `1001` bar under adiabatic conditions. If the final volume of the liquid is `990ml`, calculate `DeltaU` and `DeltaH` of the process?

Assuming a linear variation of volume with pressure.

Answer 1

I got about

`DeltaU ~~ "501 J"`
`DeltaH ~~ "99.5 kJ"`

It should be this big, seeing as how liquids are quite incompressible. Since the liquid was compressed, the work done is positive and `DeltaU > 0` and `DeltaH > 0`.

---

Well, since there was pressure-volume work done under adiabatic conditions, we know that

`dU = cancel(deltaq) + deltaw`

Assuming this variation of volume with pressure is linear, we simply take the area under the curve of a `PV` graph, which is a right triangle (vertical leg on the left) on top of a rectangle.

We are really traveling up the liquid curve here (right next to the dot), which is very steep:

The area of the right triangle is a significant portion of the amount of work done.

First, we note that its height is `"1000 bar"`. Then, the length is `"10 mL"`, or `"0.01 L"`.

Therefore,

`|w_"tri"| ~~ 1/2"0.01 L" cdot "1000 bar" = "5 L"cdot"bar"`

And converting the units,

`"5 L"cdot"bar" xx ("8.314472 J")/("0.083145 L"cdot"bar") = "499.998 J"`

The area of the rectangle is relative to `P_1 = "1 bar"` and the minimum `P = "0 bar"`:

`|w_"rect"| = "0.01 L" cdot "1 bar" = "0.01 L"cdot"bar" = "0.999997 J"`

So the total work is about:

`color(blue)(DeltaU_"adia") = w ~~ "499.998 J" + "0.999997 J"`

`~~` `"500.998 J"` `->` `color(blue)ul"501 J"`

For a liquid, since the pressure changed by so much, `DeltaH` should be significantly different. You know that `H = U + PV`, so:

`color(blue)(DeltaH_"adia") = DeltaU_"adia" + Delta(PV)`

`= DeltaU_"adia" + P_2V_2 - P_1V_1`

`= "500.998 J" + (1001 cancel"bar"cdot 0.990 cancel"L" - 1 cancel"bar" cdot 1 cancel"L") cdot (8.314472 cancel"J")/(0.083145 cancel("L"cdot"bar"))`

`~~` `color(blue)ul("99.5 kJ")`