# Piston problem including force and work

• jessiemay1993
In summary, the problem given is to find the work done by a gas in a cylinder with a cross-sectional area of 0.02 m^2, when the gas exerts a constant pressure of 7.8x10^5 Pa on the piston, moving it a distance of 0.06 m. The equation Pressure*Area=Force was used to find the force, which was then plugged into the equation Force*distance=Work. The answer choices are A. 6.5*10^2 J, B. 9.4*10^2 J, C. 6.5*10^8 J, and D. 9.4*10^8 J. The correct answer

#### jessiemay1993

The problem given says: A cylinder has a cross-sectional area of 0.02 m^2. How much work is done by a gas in the cylinder if the gas exerts a constant pressure of 7.8x10^5 Pa on the piston, moving it a distance of 0.06 m?

I figured I needed to find the work, but I didn't have the force.

So I tried the equation Pressure*Area=Force

Then to get work, I used the equation Force*distance=Work

For the force, I got 15600. I then plugged that into the second formula as force and got 936. I have a really strong feeling it's not correct. My answer choices are as followed: A. 6.5*10^2 J B. 9.4*10^2 J C. 6.5*10^8 J and D. 9.4*10^8 J

I'm really confused on this and would really appreciate the help. My teacher doesn't really teach us what we need to know. Just throws it up on the board, briefly talks about it and expects us to learn from that. So would someone also help me understand Thermodynamics? It would be greatly appreciated :]

Thank you in advance,
Jessica

Change you answer to scientific notation or the 4 choices to standard notation.

A. 6.5×102 J B. 9.4×102 J C. 6.5×108 J and D. 9.4×108 J

Your answer is very close to one of them.

Wow! Thank you so much :]

Also: d(Work) = pressure*d(volume) ;)

gomunkul51 said:
Also: d(Work) = pressure*d(volume) ;)

What? Now I'm super confused

jessiemay1993 said:
What? Now I'm super confused
Don't be confused. It just arithmetic:

$$W = F \times \Delta d = (P \times A) \times \Delta d = P \times (A \times \Delta d) = P \times (\Delta V)$$

where $\Delta V$ is the change in volume of the gas.

AM

@jessiemay1993: if you want to be confused, take Engineering Thermodynamics ;)

Andrew Mason <--- what he said ;)

what answer you got was with the assumption that the pressure remains constant. but actually it does not as the pressure and volume are inter related. but since you have no initial value of he volume the assumption may be correct.

what equation gomunkul51 has written is the same as yours when keeping pressure constant. its just written in a different way. so no need to be confused