A gaseous reaction occurs at a constant pressure of 50.4 bar and releases 67.0 kJ of heat. Before the reaction, the volume of the system was 9.00 L . After the reaction, the volume of the system was 2.20 L .

Calculate the total internal energy change, ΔU, in kilojoules.

Question

A gaseous reaction occurs at a constant pressure of 50.4 bar and releases 67.0 kJ of heat. Before the reaction, the volume of the system was 9.00 L . After the reaction, the volume of the system was 2.20 L .

Calculate the total internal energy change, ΔU, in kilojoules.

owlet · Answer

So I know.. ΔU = q + W and W = - PΔV 

q=67.0 kJ
ΔV  = 2.20-9L= -6.8L
P= 50.4 bar
I also know 1 L•bar = 100 J

Substituting all my given values:
ΔU = q + W
ΔU = -67.0 kJ + (-50.4 bar)(-6.8 L )
ΔU = -67.0 kJ + 342.72 L•bar
ΔU =  -67.0 kJ + 34.272 kJ
ΔU =  -32.728 kJ


AM I RIGHT?

owlet · Answer

@aaronq

aaronq · Answer

yep! you got the signs and everything right, it looks right (as long as you didn't make a mistake multiplying or something)

owlet · Answer

okay thank you! I just wanted to make sure that it is right :)

aaronq · Answer

no problem! is this from a textbook?  they usually have answers at the back, or you can download the solution manual