Change in Pressure: Find Gas P (Pa) Experiment

  • Thread starter Thread starter Devin Longo
  • Start date Start date
  • Tags Tags
    Change Pressure
AI Thread Summary
The experiment aims to determine the change in pressure of a gas using a water manometer, with initial atmospheric pressure set at 1.5 x 10^5 Pa. The user attempted to calculate the pressure change using the formula ΔP = ρgΔh, where they mistakenly included atmospheric pressure in their calculations. The correct approach requires using the density of water and the change in height, which was noted as 4 cm. The user was advised to convert units properly and focus on the density of the fluid rather than the atmospheric pressure. Accurate unit conversion and understanding of the variables involved are crucial for obtaining the correct pressure change.
Devin Longo
Messages
11
Reaction score
0

Homework Statement



An experiment to determine the specific heat of a gas makes use of a water manometer attached to a flask (the figure below ). Initially the two columns of water are even. Atmospheric pressure is 1.5 105 Pa. After heating the gas, the water levels change to those shown. Find the change in pressure of the gas in Pa.



2. Homework Equations
\DeltaP = \rhog\Deltah



The Attempt at a Solution



I basically did 1.5 x 10 ^5 * 9.8 m/s2 * .01m (what I assume given the picture to be the change in height)
 

Attachments

  • fig-035.gif
    fig-035.gif
    5.4 KB · Views: 468
Physics news on Phys.org
Difference in water level is 4 cm.
So ΔP = ρgΔh.
 
If I change 4 cm to meters I get 58800 as the calculated total from the equation, which is incorrect according to the website. Is there some factor I'm neglecting?
 
Devin Longo said:
If I change 4 cm to meters I get 58800 as the calculated total from the equation, which is incorrect according to the website. Is there some factor I'm neglecting?

Rho is the density of the fluid, not the pressure, so I'm not sure why you are using 1.5e5 in the equation. Also, pay attention to units when you are doing your computation (doing so would have revealed this error).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top