Im having trouble with a simple question (simple thermodynamics)

  • #1
HI! i was wondering if someone could please help me with this physics problem. it looks very simple, and im 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
 

Answers and Replies

  • #2
Andrew Mason
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nick727kcin said:
HI! i was wondering if someone could please help me with this physics problem. it looks very simple, and im 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
Andrew Mason
Science Advisor
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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
 

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