Graham's Law of Effusion
Two gases at the same temperature
possess same kinetic energy:
KE1 =
KE2
½m1v1
= ½m2v2
m = mass, v = velocity
m1v1
= m2v2
m1u1=m2u2
u = rate of effusion
Rate of effusion
is inversely proportional to square root of mass:
u1/u2
= (m2/m1)½
Time of effusion is
directly proportional to square root of mass:
t1/t2
= (m1/m2)½
t = time of effusion
Maxwell's Root-Mean-Square Molecular Equation
u = [3RT/M]½
T = temperature in K
M = molar mass in kg/mol
R = 8.314 J/k-mol
The equation represents relationship
between molecular mass, average speed, and temperature. Because
two gases with
two different masses possess the same
average kinetic energy at the same temperature, the
gas with the greater molar mass
has lower average speed.
Practice Problems
1. By what factor does He effuse faster
than CO?
uHe/uCO
= (mCO/mHe)½
=
(28.0/4.00)½ = 2.65
2. Sample of N2O
effuses at 225 mL/s while under the same conditions X effuses at 375 mL/s.
Calculate
molar mass of X.
(uN2O/uX)2
= (mX/mN2O)
(225/375)2
= mX/44.0
mX
= 15.8 g/mol
3. If SO2
effuses in 125 s, how long does it take CO2
to effuse under the same conditions?
tCO2/tSO2
= (mCO2/mSO2)½
tCO2/125s
= (44.0/64.0)½ = 0.829
tCO2
= 104s
4. Calculate average velocity for SF6
at 100°C
u =
[3RT/M]½ = [(3)(8.31)(373)/0.146)]½
= 252 m/s
5. A gas has average velocity of 727
m/s at 200°C. What is molar mass?
u2
= 3RT/M
(727)2
= (3)(8.31)(473)/M
M = 22.3 g/mol
6. At what temperature (°C) does
He have average velocity of 1500 m/s?
u2
= 3RT/M
15002
= (3)(8.31)(T)/(0.00400)
T =
361K = 88°C
7. O2
at 400K versus CH4 at 300K: Which has the
higher average velocity and by what factor?
uO2
= [3RT/M]½
uCH4 = [3RT/M]½
uO2
= [(3)(8.31)(400)/(0.0320)]½
uCH4 = [(3)(8.31)(300)/(0.0160)]½
uO2
= 558 m/s
uCH4 = 684 m/s
CH4
faster by factor of 1.23 (684/558)