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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

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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