How Is Work Calculated When Compressing Gas?

[Negan57]

Homework Statement

Pt. A) A gas is compressed at a constant pressure of 0.632 atm from 7.11 L to 5 L. In the process,
440 J of energy leaves the gas by heat.
What is the work done on the gas?
Answer in units of J

Pt. B)

What is the change in its internal energy?
Answer in units of J

Homework Equations

PV = nRT
U = Q + W
1 atm = 101325 pa
1L = .001m^3

The Attempt at a Solution

A)

.632atm = 64037.4 pa
7.22L = .00711 m^3
5L = .005 m^3

W = P(Vf -Vi)
=
64037.4(.005-.00711)
=
-135.189

U = Q+W

U = -135.189 -440
=
-575.1189

OR (done other ways)
: 304.881
: -304.881
: 575.118914

All of these were wrong. Please help?

B) ...

 

[Andrew Mason]

The Attempt at a Solution

A)

.632atm = 64037.4 pa
7.22L = .00711 m^3
5L = .005 m^3

W = P(Vf -Vi)

Is this the work done ON the gas or the work done BY the gas? What does the question ask for?

B) ...

What is W in U = Q + W? (ie is it work done ON or BY the gas?).

AM

 

[Negan57]

It asks for the work done ON the gas. The reasons its W = P(Vf-Vi) and not W = -P(Vf-Vi) is because its work ON gas not BY the gas. In U = Q + W, W is work on gas presumably, though I'm not 100% sure.

 

[Andrew Mason]

It asks for the work done ON the gas. The reasons its W = P(Vf-Vi) and not W = -P(Vf-Vi) is because its work ON gas not BY the gas. In U = Q + W, W is work on gas presumably, though I'm not 100% sure.

This is important. Positive work is done BY the gas when it expands. The work done BY the gas is PΔV where ΔV>0. Positive work is done ON the gas when it is compressed. So the work done ON the gas is -PΔV.

So, in the expression U = Q+W, the W is work done ___ the gas.(hint: if W is positive is there an expansion or compression?).

AM