Ideal Gas Law Problems. 1) How many molecules are there in 985 mL of nitrogen at 0.0° ... Solutions. 1). P = 1.00 x 10. -6 mm Hg. T = 0.0° C + 273 = 2...
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This work is licensed under a Creative Commons Attribution - NonCommercial 4.0 International License. More chemistry tutorials and practice can be found at www.chemfiesta.com. Ideal Gas Law Practice Worksheet. Solve the following problems using the i
The ideal gas law is an equation that relates the volume, temperature, pressure and amount of gas particles to a constant. The ideal gas constant is abbreviated with the variable R and has the value of. 0.0821 atm·L/mol·K. The ideal gas law can be us
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• Jika n ≡ jumlah mol gas dan M ≡ berat molekul gas m =nM • misal: M O2 = 32 kg/kmol 1 kmol O2 ekivalen dengan 1 kg O2 • Dalam kmol: Pv = nMRT
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Avogadro's Law. The Ideal Gas Law. P V = n R T. P1 V1 n1 T1. = P2 V2 n2 T2. If pressure and moles do not change in the problem, omit those variables. Charles's Law. = V1. T1. V2. T2. P V n T. R = The Ideal Gas Law. P V = n R T d R T . P. M = A. Romer
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Ideal Gas Law Problems
1)
How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?
2)
Calculate the mass of 15.0 L of NH3 at 27° C and 900. mm Hg.
3)
An empty flask has a mass of 47.392 g and 47.816 g when filled with acetone vapor at 100.° C and 745 mm Hg. If the volume of the flask is 247.3 mL, what is the molar mass of the acetone?
4)
Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg.
5)
6.3 mg of a boron hydride is contained in a flask of 385 mL at 25.0° C and a pressure of 11 torr.
6)
(a)
Determine the molar mass of the hydride.
(b)
Which of the following hydrides is contained in the flask, BH3, B2H6, or B4H10?
A volume of 26.5 mL of nitrogen gas was collected in a tube at a temperature of 17° C and a pressure of 737 mm Hg. The next day the volume of the nitrogen was 27.1 mL with the barometer still reading 737 mm Hg. What was the temperature on the second day?
Solutions 1)
P = 1.00 x 10-6 mm Hg
T = 0.0° C + 273 = 273 K
V = 985 mL
R = 0.0821 L·atm/mol·K
PV = nRT n = PV/RT n = 1.00 x 10-6 mm x 1 atm/760 mm x 985 mL x 1 L/103 mL/ (0.0821 L·atm/mol·K x 273 K) = 5.78 x 10-11 moles N2 nmolecules = 5.78 x 10-11 moles N2 x 6.02 x 1023 N2 molecules/1 mol N2 = 3.48 x 1013 N2 molecules 2)
P = 900. mm Hg
T = 27° C + 273 = 300 K
V = 15.0 L
R = 0.0821 L·atm/mol·K
PV = nRT n = PV/RT n = 900. mm x 1 atm/760 mm x 15.0 L/(0.0821 L·atm/mol·K x 300 K) = n = 0.721 moles NH3 x 17.04 g NH3/1 mol NH3 = 12.3 g NH3
3)
P = 745 mm Hg
T = 100.° C + 273 = 373 K
V = 247.3 mL
R = 0.0821 L·atm/mol·K
mvapor = 47.392 g – 47.816 g = 0.424 g PV = nRT n = m/MM PV = mRT/MM MM = mRT/PV MM = 0.424 g x 0.0821 L·atm/mol·K x 373 K/(745 mm x 1 atm/760 mm x 247.3 mL x 1 L/103 mL) = 53.6 g/mol
4)
P = 671 mm Hg
T = 47° C + 273 = 320. K
V = 478 mL
R = 0.0821 L·atm/mol·K
PV = nRT n = m/MM D = m/V = P x MM/R x T D = 671 mm x 1 atm/760 mm x 83.80 g/mol/(0.0821 L·atm/mol·K x 320. K) = 2.82 g/L
5)
P = 11 torr
T = 25.0° C + 273 = 298 K
V = 385 mL
R = 0.0821 L·atm/mol·K
m = 6.3 mg
PV = nRT n = m/MM PV = mRT/MM MM = 6.3 mg x 1 g/103 mg x 0.0821 L·atm/mol·K x 298 K)/ (11 torr x 1 mm/1 torr x 1 atm/760 mm x 385 mL x 1 L/103 mL) = 27.7 g/mol B2H6 because its molar mass is 27.7 g. 6)
P1 = 737 mm Hg
P2 = 737 mm Hg
V1 = 26.5 mL
V2 = 27.1 mL
T1 = 17° C + 273 = 290. K
T2 = ?
P1V1 = nRT1 P2V2 = nRT2 P1V1/ P2V2 = nRT1/ nRT2 V1/V2 = T1/T2 (Charles’s Law) T2 = V2/V1 x T1 T2 = 27.1 mL/26.5 mL x 290. K = 297 K = 24° C
