How Does Reversible Expansion Calculate Work in Thermodynamics?

[Puchinita5]

Homework Statement

Okay i feel like I'm going crazy. Here is a sample problem my professor gave me. But I cannot figure out for my life how he got his answer and it hsould be very simple.


A 10.0 mol sample of a perfect gas having Cv = 2.0R undergoes a reversible expanion form 5.00 L to 20.0 L at 5.00 atm of pressure. Calculate q, w, delta U, and delta H for this process..

Solving for work, he has...


w = -PdV = -5(20-5) = -22.5 L atm = -22.8 J

HOW ON EARTH does he get those numbers?

Homework Equations

The Attempt at a Solution

 

[ptolema]

dV is the change in volume, which is (20.0-5.00 L)=15.0 L
-P is the given pressure x(-1), -P = -5.00 atm
w = -PdV= -5.00*15.0 L atm = -75.0 L atm
Since L atm is a unit of work, it can be converted to J:
1 L atm = 101.325 J, so -75.0(101.325) J = -7.60x10^3 J = w

I'm not sure what else your professor did to get to his answer, but this is what I got based on what was given.

 

[ptolema]

dV is the change in volume, which is (20.0-5.00 L)=15.0 L
-P is the given pressure x(-1), -P = -5.00 atm
w = -PdV= -5.00*15.0 L atm = -75.0 L atm
Since L atm is a unit of work, it can be converted to J:
1 L atm = 101.325 J, so -75.0(101.325) J = -7.60x10^3 J = w

I'm not sure what else your professor did to get to his answer, but this is what I got based on what was given.

 

[Puchinita5]

thank you! I just needed confirmation that I wasn't missing something. That was exactly what I thought the answer should be