How Do You Calculate Work Done by an Expanding Gas?

[brilyn]

A fas is in a cylinder is at pressure of 6000 Pa and a piston has an area of .1m^2. As heat is slowly added the piston is pushed up a distance of 5cm. Calculate the work done on the surroundings by the expanding gas. (Assume pressure stays constant).

Thus far i found the force using
P=f/a 6000Pa or 6000N/m^2=F/.1m^2 therefore F=600N Correct? i think I'm mixing up units, but then i used
W=1/2F(s^2) > 1/2(600N)(5cm)^2 > i get 7500N/cm^2 but these units don't seem right for work?
Can you help me with my units and this is the work done by the gas right?

To find the internal energy if 52J of heat is added do i just use Q=IE+W?

 

[Andrew Mason]

A fas is in a cylinder is at pressure of 6000 Pa and a piston has an area of .1m^2. As heat is slowly added the piston is pushed up a distance of 5cm. Calculate the work done on the surroundings by the expanding gas. (Assume pressure stays constant).

Thus far i found the force using
P=f/a 6000Pa or 6000N/m^2=F/.1m^2 therefore F=600N Correct? i think I'm mixing up units, but then i used
W=1/2F(s^2) > 1/2(600N)(5cm)^2 > i get 7500N/cm^2 but these units don't seem right for work?
Can you help me with my units and this is the work done by the gas right?

To find the internal energy if 52J of heat is added do i just use Q=IE+W?

$$W = P\Delta V$$

You are trying to use W = force x distance but Fds = PAdS = PdV. It is much easier to use PdV.

What is the change in volume? Multiply that by the pressure to get the work done. On a PV diagram, Work is the area under the graph.

AM

 

[kyle8921]

Yes, your calculation for the force is correct. However, when calculating work, you need to make sure that the units are consistent. In this case, the force is in Newtons (N) and the displacement is in meters (m), so the work done will be in Joules (J). Therefore, the work done by the expanding gas would be 1/2(600N)(0.05m)^2 = 7.5J.

To find the internal energy, you can use the equation Q = ΔU + W, where Q is the heat added, ΔU is the change in internal energy, and W is the work done. So in this case, ΔU = Q - W = 52J - 7.5J = 44.5J. This is the change in internal energy of the gas.