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CHAPTER 5: GASES 136 
 Dividing r1 by r2 gives: 
 
 11
2
2
3
 = 
3
RT
r
r RT
M
M
 
 
 Canceling 3RT from the equation gives: 
 
 1 1 2
2 1
2
1
 = 
1
=
r
r
M M
M
M
 
 
 Does the derived equation make sense? Assume that gas1 is a lighter gas (has a smaller molar mass) than 
gas2. Dividing a larger molar mass (M2) by a smaller molar mass (M1) will give a number larger than 1. This 
indicates that the lighter gas will diffuse at a faster rate compared to the heavier gas. 
 
5.82 The separation factor is given by: 
 
 1 2
2 1
= =
r
s
r
M
M
 
 
 This equation is the same as Graham’s Law, Equation (5.17) of the text. For 235UF6 and 238UF6, we have: 
 
 238 (6)(19.00)
235 (6)(19.00)
+
= =
+
1.0043s 
 
 This is a very small separation factor, which is why many (thousands) stages of effusion are needed to enrich 
235U. 
 
5.83 The rate of effusion is the number of molecules passing through a porous barrier in a given time. The longer 
it takes, the slower the rate of effusion. Therefore, Equation (5.17) of the text can be written as 
 
 1 2 2
2 1 1
r t
r t
= =
M
M
 
 
 where t1 and t2 are the times of effusion for gases 1 and 2, respectively. 
 
 The molar mass of N2 is 28.02 g/mol. We write 
 
 15.0 min
12.0 min 28.02 g/mol
=
M 
 
 where M is the molar mass of the unknown gas. Solving for M, we obtain 
 
 
2
15.0 min 28.02 g/mol
12.0 min
⎛ ⎞
= × =⎜ ⎟
⎝ ⎠
43.8 g/molM 
 
 The gas is carbon dioxide, CO2 (molar mass = 44.01 g/mol). During the fermentation of glucose, ethanol 
and carbon dioxide are produced.