Work done on compressed 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

 

[Nickie]

There is no information given about the initial or final internal energy, so it cannot be determined. In order to calculate the change in internal energy, the initial and final states of the gas must be known.