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?

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")#