Assumptions about gases have negligible volume exert no forces on one another

# Assumptions about gases have negligible volume exert

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10.8 Molecular Effusion and Diffusion Root mean square speed = μ rms 3 R T Mol. Wt. ½ μ rms =
T = =
Effusion: Escape of a gas through a hole into evacuated space Diffusion Spread of gas molecules through space. Move quick, but collisions slow them down.
=
Graham found: MW 2 MW 1 ½ rate 1 rate 2 = NH 3 (g) + HCl (g) à NH 4 Cl (s) Experiment Where along the tube will NH 4 Cl be detected? 36.5 g / mol 17.0 g / mol ½ rate NH3 rate HCl = rate NH3 = 1.46 ( rate HCl ) Demo
10.9 Real Gases: Deviations from Ideal Behavior Real Gasses deviate at (1) High Pressure (2) Low Temperature P > 10 atm T < 200 - 1000 K
Pressure Open Area
Gases do attract one another Gases have finite volume ( P + n 2 a / V 2 ) ( V – n b ) = nRT P V = n R T Correction Factors: a --- units of L 2 atm / mol 2 b --- units of L / mol
( P + n 2 a / V 2 ) ( V – n b ) = nRT a --- corrects for the attractive forces b --- corrects for the volume differences
Calculate the pressure exerted by 1.00 mole of methane, CH 4 , in a 500.0 mL vessel at 25.0 °C assuming (a) Ideal behavior (b) Nonideal behavior (a = 2.25 L 2 atm/mol; b = 0.0428 L/mol) Ideal behavior ------ P V = n R T P = ( n R T ) / V ( 1.00 mole ) (0.0821 L atm / mol K) (298 K) 0.5000 L P = 48.93 atm P = 48.9 atm
Calculate the pressure exerted by 1.00 mole of methane, CH 4 , in a 500.0 mL vessel at 25.0 °C assuming (a) Ideal behavior ---- 48.9 atm (b) Nonideal behavior (a = 2.25 L 2 atm/mol; b = 0.0428 L/mol) ( P + n 2 a / V 2 ) ( V – n b ) = nRT n 2 a / V 2 = ( 1.00 mol ) 2 (2.25 L 2 atm/mol) / (0.5000L) 2 ( V – n b ) = 0.5000 L - 0.0428 L/mol (1.000 mole) n 2 a / V 2 = 9.00 atm ( V – n b ) = 0.4572 L
Calculate the pressure exerted by 1.00 mole of methane, CH 4 , in a 500.0 mL vessel at 25.0 °C assuming (a) Ideal behavior ---- 48.9 atm (b) Nonideal behavior (a = 2.25 L 2 atm/mol; b = 0.0428 L/mol) ( P + n 2 a / V 2 ) ( V – n b ) = nRT ( 1.00 mole ) (0.0821 L atm / mol K) (298 K) 0.4572 L ( P + 9) = ( P + 9) = 53.512 P = 44.5 atm
Cyanogen, a highly toxic gas, is composed of 46.2% C and 53.8% N by mass. At 25°C and 751 torr, 1.05 g of cyanogen occupies 0.500 L. (a) What is the molecular formula of cyanogen? (b) Predict its molecular structure. (c) Predict the polarity of the compound. SAMPLE INTEGRATIVE EXERCISE --- Page 424
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together Cyanogen, a highly toxic gas, is composed of 46.2% C and 53.8% N by mass. At 25°C and 751 torr, 1.05 g of cyanogen occupies 0.500 L. (a) What is the molecular formula of cyanogen? (b) Predict its molecular structure. (c) Predict the polarity of the compound. Because the ratio of the moles of the two elements is essentially 1:1, the empirical formula is CN.