Work done by air in piston with temperature change

In summary: In order to answer this, you must assume that the system comes to rest at state 2 and has achieved equilibrium with its surroundings; ie P2=200psi. Once you’ve got v2, then w=P(v2-v1) with P=200 psi.
  • #1
falsealarm715
2
0

Homework Statement


Two pounds of air contained in a cylinder expand without friction against a piston. The pressure on the back side of the piston is constant at 200 psia. The air initially occupies a volume of 0.5 ft3. What is the work done by the air in ft-lbf if the expansion continues until the temperature of the air reaches 100 F?

Homework Equations


Q - W = m(uf - ui), which translates to W = m*cv*(Tf - Ti) [Q is 0 because there's no heat transfer]
PV = mRT

The Attempt at a Solution


W = 2 lbm (0.171 Btu/lbm R)(559.67 R - Ti)

The text says the answer is 45300 ft-lbf

Basically I don't really know how to start this all off. I know I need to find the initial temperature which I think means I need to use PV = mRT but I can't find an agreeable R value in my book (also I have no idea if my cv value is correct). Could someone just point me in the right direction?
 
Last edited:
Physics news on Phys.org
  • #2
I don't know that Q-W formula, but I think you'll need the initial temperature and final volume, which you can get from the formula PV = mRT, if you think about which of those letters change and which are constant.
 
  • #3
vertigo said:
I don't know that Q-W formula, but I think you'll need the initial temperature and final volume, which you can get from the formula PV = mRT, if you think about which of those letters change and which are constant.

The thing is I can't really tell what is going to remain constant when I read the question. I think the pressure remains constant and I know mass does which leaves me with Ti/Vi = Tf/Vf. I've already got Vi and Tf but I don't really know what to do with that.
 
  • #4
In order to answer this, you must assume that the system comes to rest at state 2 and has achieved equilibrium with its surroundings; ie P2=200psi.
Now we can use the ideal gas equation to solve for v2:
V2=(mRT2)/P2.
Some notes here:
1) watch your units! P is in inches and R is in feet.
2) Use the right R value (85.67)
3) T is absolute.
Once you’ve got v2, then w=P(v2-v1) with P=200 psi.
I know pressure is not constant within the system, but it is outside the system. Think of it as the work done on the environment is the work done by the system.
 
  • #5
You need to make an assumption
 

1. What is work done by air in a piston?

The work done by air in a piston refers to the energy transferred by the air to the piston as it expands or contracts, resulting in a change in volume. This work can be calculated by multiplying the force exerted by the air on the piston by the distance the piston moves.

2. How does temperature change affect the work done by air in a piston?

As the temperature of the air in a piston changes, the pressure and volume of the air also change. This affects the work done by the air, as the force exerted by the air on the piston and the distance the piston moves will be different at different temperatures.

3. How is the work done by air in a piston with temperature change calculated?

The work done by air in a piston with temperature change can be calculated using the formula W = PΔV, where W is the work done, P is the pressure of the air, and ΔV is the change in volume of the air.

4. What factors can affect the work done by air in a piston with temperature change?

The work done by air in a piston with temperature change can be affected by factors such as the initial pressure and volume of the air, the type of gas used, and the type of piston. Additionally, external factors such as friction and air resistance can also impact the work done.

5. What is the significance of understanding the work done by air in a piston with temperature change?

Understanding the work done by air in a piston with temperature change is important in various fields of science and engineering, such as thermodynamics, fluid mechanics, and power generation. It allows for the efficient design and optimization of systems that use air pressure, such as engines and compressors.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
739
  • Introductory Physics Homework Help
Replies
16
Views
1K
Replies
5
Views
320
  • Introductory Physics Homework Help
Replies
1
Views
487
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Thermodynamics
Replies
8
Views
539
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
3K
Back
Top