chrisfris Posted December 23, 2016 Posted December 23, 2016 Hello all, I have been stuck in this question for at least 3 hours, and i would like to get some tips on how to solve it. First, please check the drawing in the files to understand the water-barometer. The question is as followed: "The water-barometer is very sensitive for temperature changes in the flask. For the increase Δh in water level per degree of increase in temperature, the following formula can roughly be applied: Δh / ΔT = p / (T * ρwater * g) T is the absolute temperature and p the pressure inside the flask. The atmospherical pressure (outside the flask) is equal to 10^5 Pa. Now roughly calculate the increase of the water level in cm when the temperature of the gas increases from 20° C to 23° C. Useful data: ΔT = 3 K T = 293 K ρwater = 10^3 kg/m^3 g = 9,81 m/s^2 patmosferical = 10^5 Pa pflask = unknown Δh = unknown h = unknown Δp = unknown V = 1,1 dm^3 = (almost) unchanged n = amount of gas in mole = unknown How can i calculate p or Δh? Thanks a lot and merry Christmas! PS: I am not a native speaker, so excuse me if my language was incorrect.
swansont Posted December 23, 2016 Posted December 23, 2016 How does the pressure inside relate to the pressure outside? 1
chrisfris Posted December 23, 2016 Author Posted December 23, 2016 I does not relate in the way that you can calculate the pressure inside with the pressure outside. You can, however, calculate the height in the pipe with the difference between both pressures in the formula: Δp = ρwater * g * h
swansont Posted December 23, 2016 Posted December 23, 2016 I does not relate in the way that you can calculate the pressure inside with the pressure outside. You can, however, calculate the height in the pipe with the difference between both pressures in the formula: Δp = ρwater * g * h But you have the outside pressure. So you can estimate the inside oressure. The height is most likely of order a meter, at most. So the pressure difference is ~10^4 Pa, or 0.1 of atmosphere, at biggest. IOW, if you are estimating the answer, you can ignore this difference. Now you have an equation where you know everything but the result you want. Calculate away. 1
chrisfris Posted December 23, 2016 Author Posted December 23, 2016 Okay thanks I think this can really help me!
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now