Gas Laws Questions! Help

[moonhoplite]

Hi everybody,

Q1 In a Boyle's Law apparatus the surface of the mercury in the open tube is 6 cm higher than that in the closed tube, the length of the enclosed gas column being 12cm. Mercury is then added until the length of the gas column is 7cm. Calculate the new difference in levels of mercury surfaces, the barometric pressure being 101.3kPa or 760 mm of mercury.

Q2 A faulty barometer has some air trapped above the mercury. The height of the mercury column is 73.2cm when the air space is 10cm long. When the tube is pushed down into the bowl so that the air space is 5cm long, the height of the mercury column is 71.6cm. What is the true barometer reading?

Q3 The closed limb of a U-tube contains a quantity of gas of volume 6 x 10^-5 m^3 at a temperature of 35 degrees Celsius. The level of mercury in the closed limb is 9cm below the level in the open limb. A barometer reads 97.57 kPa. What is the volume of the gas at standard conditions of 0 degrees Celsius and 101.3 kPa?

Could you please help me with these questions? I'm fairly competent with gas laws (I know PV = k, V/T = k, PV/T = k....) I just need help on how to do these, or give me a direction on how to do this, methods, equations, etc... I'm clueless on these ones...

The book shows answers, but no solving, the answers are:
7: 0.646 m
8: 99.7 kPa
9: 5.75 x 10^-5 m^3

thanks,

moon

[TheMasterMind]

Hi everybody,

Q2 A faulty barometer has some air trapped above the mercury. The height of the mercury column is 73.2cm when the air space is 10cm long. When the tube is pushed down into the bowl so that the air space is 5cm long, the height of the mercury column is 71.6cm. What is the true barometer reading?



technically, there should be no air in the barometer. with air in a barometer, you can never get a precise reading. the space above the mercury is a vacuum with a little vapour in it. if the mercury column is broken so that there is a space, shaking will remove it. these spaces also exist in thermometers but the diameter and weight of the mercury works so any space created by transporting the barometer would dissapear when it returns motionless.

[moonhoplite]

Hi,

I know, the question is dodgy, it came from an old text book. I was able to do the 1st question recently...
i had:
(760 + 60) x 12 = 7 x (760 + X)
this is all in mm...
eventually i got X = 645.7 mm
= 0.646 m





BUT~!!!!!!!!!!!

The rest of the questions are still hard, different methods? Any help?

[SuBNoIze]

Hi,

The height of the mercury is showing the DIFFERENCE in pressure between the atmosphere and the air in the tube. So these lengths are M1 and M2.

The length of the air part is showing the change in pressure in the air. PV=constant, assuming the change in temperature and diameter of the tube is to be ignored, and its an ideal gas.

Since V is related to length, you can get the ratio between the before and after pressures.

Change in Air pressure is = AP2 / AP1 = (1/AL2) / (1 / AL1) = L1/L2.

Turn the mm of mercury into pascal's , pascals per mm hg is a constant in your text book. These are P1 and P2.



Atmospheric Pressure .. room pressure RP ..

= ( P1 + AP1) , because the RP is the pressure caused by that height of mercury plus the pressure of the air in the tube.

Also (P2 + AP2) = Rp2.. theres two different measurements with different pressure in the tube.

We know the ratio of the difference in the air pressure in the tube.


If AP1= RP - P1
and AP2= RP - P2

then
AP2/AP1=(Rp-P1)/(RP- P2)=L1/L2

Now we know the ratio is related to the room pressure and the mercury heights.

So plug in knowns - P1, P2, L1 and L2- simplify (or simplify then plug in unknowns) and you get RP.