Need help with Ideal Gas Question

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  • #1
jojosg
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Homework Statement
On a day when the barometer reads 75.23 cm, a reaction vessel holds 250 mL of ideal gas at 20 celsius. An oil manometer ( density= 810 kg/m^3) reads the pressure in the vessel to be 41 cm of oil and below atmospheric pressure. What volume will the gas occupy under S.T.P.?

Answer: 233mL
Relevant Equations
PV/T = constant
Need help solving this question. Can't seem to get the right answer using PV/T=constant

P1V1/T1 = P2V2/T2

Patm = 75.23cmHg T1+20+273=293K

STP: P=1.01 x 10^5 N/m^2 Pabs=41cmOil

P1 = density x g x h = (810 kg/m^3)(9.8 m/s^2)(75.23-41)x10^-2 mOil=2717.18 N/m^2

V2=(P1V1T2)/(T1P2)=(2717.18N/m^2 x 250mL x 273K)/(293K x 1.01 x 10^5 N/m^2)= 6.2666 m^3
 
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  • #2
Show your work. Also, I suggest reading the forum rules.
 
  • #3
jojosg said:
(75.23-41)
So you are subtracting heights of columns of different liquids? Won't work, if anything, you should multiply them by their densities, not just both by the oil density.

That's just the first thing that caught my attention, doesn't mean there are no other problems, what you wrote is a bit chaotic and difficult to follow.
 
  • #4
Ok so I got Pressure of the barometer to be 10231.3 N/m^2 if I multiply 75.23cm with 13.6 x 10^3 kg/m^3, & Pressure of the oil manometer to be 389.5 N/m^2 if multiplied by 950 kg/m^3. Is that the Pressure for P1 and P2 respectively? (For the P1V1/T1 = P2V2/T2)
 
  • #5
Welcome, @jojosg ! 😎

Please, consider the following facts, and compare them with your equations:

1) The internal volume of the reaction vessel must be variable; otherwise, the volume that the ideal gas that it contains will occupy under S.T.P. would be 250 ml (same as under initial atmospheric conditions).

2) The 75.23 cm of Hg column that our barometer reads are equivalent to the absolute pressure acting on the surface of the oil column that is exposed to the atmosphere in our oil manometer.

3) The statement "the pressure in the vessel to be 41 cm of oil and below atmospheric pressure" indicates that the absolute initial pressure inside out vessel is equivalent 75.23 cm of Hg column minus 41.00 cm of oil column, which is acting on the surface of the oil column that is exposed to the vessel in our oil manometer.

4) It is safe to assume that the temperature of 20° Celsius measured inside our ideal gas equals the atmospheric temperature at the initial time.

5) It is safe to assume that the temperature of 0° Celsius corresponding to the new S.T.P. atmospheric conditions equals the new temperature inside our ideal gas at the final time.
 
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  • #6
Lnewqban said:
Welcome, @jojosg ! 😎

Please, consider the following facts, and compare them with your equations:

1) The internal volume of the reaction vessel must be variable; otherwise, the volume that the ideal gas that it contains will occupy under S.T.P. would be 250 ml (same as under initial atmospheric conditions).

2) The 75.23 cm of Hg column that our barometer reads are equivalent to the absolute pressure acting on the surface of the oil column that is exposed to the atmosphere in our oil manometer.

3) The statement "the pressure in the vessel to be 41 cm of oil and below atmospheric pressure" indicates that the absolute initial pressure inside out vessel is equivalent 75.23 cm of Hg column minus 41.00 cm of oil column, which is acting on the surface of the oil column that is exposed to the vessel in our oil manometer.

4) It is safe to assume that the temperature of 20° Celsius measured inside our ideal gas equals the atmospheric temperature at the initial time.

5) It is safe to assume that the temperature of 0° Celsius corresponding to the new S.T.P. atmospheric conditions equals the new temperature inside our ideal gas at the final time.
Thank You so much! Managed to solve it this time.
 
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  • #7
Ideal gas expansion.jpg


Ideal gas graph.jpg
 
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