Pressure and Mm Hg Essay

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AP Practice Gas Law Problems 2

A mixture of H2 (g), O2(g), and 2 millilitres of H2O(l) i s present in a 0.500 litre rigid container at 25 C. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1,146 millimetres mercury. (The equilibrium vapor pressure of pure water at 25 C is 24 millimetres mercury.)
The mixture is sparked, and H2 and O2 react until one reactant is completely consumed.

(a) Identify the reactant remaining and calculate the number of moles of the reactant remaining.
(b) Calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90 C. (The equilibrium vapor pressure of water at 90 C is 526 millimetres mercury.) (c) Calculate the number of m oles of water present a s vapor in the container at 90 C.
Answer:
(a) 2 H2 + O2 2 H2O mol H2 = mol O2 initially, but 2 mole H2 r eact for every 1 mol of O2, therefore, O2 is left.
PT = PH2 + PO2 + PH2O
1146 mm Hg = PH2 + PO2 + 24 mm Hg
PH2 + PO2 = 1122 mm Hg
1122 mm Hg / 4 = P O2 left (1/2 of initial PO2 which is 1 /2 total)
PO2 = 280.5 mm Hg
280.5
atm (0.500 L)
PV
760 n 7.55 10 3 mol O2
L atm
RT
0.0821 mol K (298K)
(b)

PO

2 8 0. 5m mHg
2 9 8K

2

3 6 3K

; PO
2

3 4 2m mHg

PT = PO2 + PH2O = (342 + 526)mm H g = 868 mm H g
(c)

n

PV
RT

5 26

( 7 60 a tm )( 0 . 5 0 0 L )
L atm
K

0 . 0 8 2 1 mo l

( 3 63 K )

0 . 0 11 6 m o l

A student collected a sample of hydrogen gas by the displacement of water as shown by the diagram above. The relevant data are given in the following table.
GAS SAMPLE DATA
Volume of sample

90.0 mL

Temperature

25 C

Atmospheric Pressure

745 mm Hg

Equilibrium Vapor Pressure of H2O (25 C)

23.8 mm Hg

(a) Calculate the number of moles of hydrogen gas collected.
(b) Calculate the number of molecules of water vapor in the sample of gas.
(c) Calculate the ratio of the average speed of the hydrogen molecules to the average speed of the water vapor molecules in the sample.
(d) Which of the two gases, H2 or H2O, deviates more from ideal behavior? Explain your answer. Answer:
(a) PH2 = Patm - PH2O = (745 - 23.8) mm Hg
= 721.2 mm Hg n = (PV)/(RT) = (721.2 mm Hg 90.0 mL)/(62400 mm Hg.mL/mol.K 298.15K)
= 3.49 10-3 mol
(b) nH O = (23.8 mm Hg 90.0 mL)/(62400 mm Hg.mL/mol.K 298.15K) 6.022 1023 molecules/mol
= 6.93 1019 molecules
(c) (massH2)(velocityH2)2 = (massH2O)(velocityH2O)2
2(vH )2 = 18(vH O)2 v 2 H / v 2H O = 9 ; v H / v H O = 3
(d) H2O deviates more from ideal behavior:
(i) greater number of electrons = greater van der Waal attraction
(ii) it is a polar molecule with strong polar attraction
(iii) it hydrogen bonds to other water molecules
(iv) larger molecule and is slower at a given temp. and occupies more space.
2

2

2

2

2

2

2

A rigid 5.00 L cylinder contains 24.5 g of N 2(g) and 28.0 g of O2(g)
(a) Calculate the total pressure, in atm, of the gas
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