Answer: 233mL
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
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.
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.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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