Im having trouble with a simple question (simple thermodynamics)

In summary, the conversation discusses a physics problem involving a cylinder of nitrogen gas with a movable copper piston and an evacuated air space. The missing value is the mass of the piston, which can be calculated using the density and volume of the copper. The correct pressure calculation is \rho hg, and the final result is 3.5 kpa.
  • #1
pureouchies4717
99
0
HI! i was wondering if someone could please help me with this physics problem. it looks very simple, and I am only missing one value. thanks

A 6.0-cm-diameter cylinder of nitrogen gas has a 4.0-cm-thick movable copper piston. The cylinder is oriented vertically, as shown in the figure, and the air above the piston is evacuated. When the gas temperature is 20 degree C, the piston floats 20 cm above the bottom of the cylinder.What is the gas pressure?
knight_Figure_17_80.jpg


gas pressure is: p= F/A = (mg)/A(of the cylinder)

my problem is: how can i get the mass of the piston? its really annoying me. please help. thanks
 
Physics news on Phys.org
  • #2
nick727kcin said:
HI! i was wondering if someone could please help me with this physics problem. it looks very simple, and I am only missing one value. thanks


knight_Figure_17_80.jpg


gas pressure is: p= F/A = (mg)/A(of the cylinder)

my problem is: how can i get the mass of the piston? its really annoying me. please help. thanks
What is the volume and density of the piston? (ie. what is the density of copper?).

AM
 
  • #3
thanks for responding!

the density is: 8.96 g/cm^3

the volume of the piston is: [(3.14) x (.03)^2].04 = .000113 m^3 = 113.0973 cm^3

so then, i need to multiply : M= 113.0973 x 8.96 = 1013.35212g

The area of the container is: A= [(3.14 x .03^2)] = .002827

So then: p = Mg/A = 3512858.979

this is wrong though... can you plesae help me find what i did wrong :grumpy:
 
Last edited:
  • #4
nick727kcin said:
thanks for responding!

the density is: 8.96 g/cm^3

the volume of the piston is: [(3.14) x (.03)^2].04 = .000113 m^3 = 113.0973 cm^3

so then, i need to multiply : M= 113.0973 x 8.96 = 1013.35212g

The area of the container is: A= [(3.14 x .03^2)] = .002827

So then: p = Mg/A = 3512858.979

this is wrong though... can you plesae help me find what i did wrong :grumpy:
First of all, work out the solution algebraically and then plug in numbers. Second, you should stick to significant figures.

Mass of piston = [itex] \rho V = \rho\pi r^2h[/itex]
Pressure = Mg/A = [itex]\rho\pi r^2hg/\pi r^2 = \rho hg[/itex]

g = 9.80 m/sec^2
[itex]\rho = 8.96 g/cm^3 = 8.96 x 10^-3 kg/10^{-6} m^3=8.96 x 10^3 kg/m^3[/itex]

Pressure = 8960 x .04 (9.80) = 3512 N/m^2 = 3512 Pa. = 3.512 kpa

In significant figures, this would be 3.5 kpa

AM
 

1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the study of energy and its transformations, particularly in relation to heat and work.

2. What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern the behavior of energy in a system. The four laws are: 1) conservation of energy, 2) entropy always increases, 3) the entropy of a perfect crystal at absolute zero temperature is zero, and 4) absolute zero is unattainable.

3. How does thermodynamics apply to everyday life?

Thermodynamics plays a crucial role in many aspects of our daily lives, from the functioning of our bodies to the production of energy and the design of technologies. It helps us understand and improve processes such as energy conversion, refrigeration, and combustion.

4. What are some common misconceptions about thermodynamics?

One common misconception is that thermodynamics only applies to large-scale systems, when in fact it also applies to small-scale systems, such as at the molecular level. Another misconception is that energy can be created or destroyed, when in fact it can only be transformed from one form to another.

5. What are some practical applications of thermodynamics?

Thermodynamics has numerous practical applications, including the design of engines, refrigerators, and air conditioning systems. It is also used in the production of electricity, chemical reactions, and the study of climate change and global warming.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
735
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Replies
5
Views
518
Replies
22
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Thermodynamics
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top