Qatar University Department of Chemistry: CHEM 241 201011 Fall Term
Assignment 1: Gases
Hand in: Tuesday October 5 in class.
ANSWERS
1.
(4 points)
Benzene is a common industrial petrochemical. It is however a known carcinogen, and there are
very strict workplace safety standards for benzene exposure. The generally accepted TLV (Threshold Limit
Value) for benzene in air in the workplace is 1 ppm for an 8 hr/day 40 hr/wk exposure. Calculate the amount of
benzene (i.e., its total mass in the laboratory air) in a laboratory of 10x20x3m at 1 atm pressure and at 20
o
C
when the benzene vapour concentration is 1 ppm (parts per million mass
concentration).
Assume air to be 80%
N
2
and 20% O
2
and behaving as an ideal gas.
Answer
:
ppm benzene = (mass benzene/mass air)x10
6
Therefore we first have to calculate the mass of air in the space of 10x20x3=600 m
3
.
We could say we have 0.80 atm N
2
and 0.20 atm O
2
n
N2
= pV/RT = 0.80x101,325x600/(8.314x293) = 2.00x10
4
moles N
2
mass N
2
= 2.00x10
4
x28
= 5.6x10
5
g (2 sign. figs)
n
O2
= pV/RT = 0.20x101,325x600/(8.314x293) = 4.99x10
3
moles O
2
mass O
2
= 4.99x10
3
x32 = 1.6x10
5
g (s sign. Figs)
mass
air
total
=
7.19x10
5
g
Or we could say the average MM of air is 0.8x28 + 0.2x22 = 28.8 g/mole, with n
total
= 2.50x0
4
. The answer for
the mass of air of course will be the same.
TLV = 1 ppm, so then
1 ppm benzene = (mass benzene/mass air)10
6
Mass benzene = 1(ppm)xmass air(g)/10
6
= 1x7.19x10
5
/10
6
= 0.719 g benzene
Answer:
mass of benzene =
0.719 g
Notes:
1) Always report units in your answer.
2) Pay attention to the significant figures
3) Imagine what the benzene concentration would be if by accident you left a bottle with just 100 g of
benzene open to evaporate in the air, it would be 140x the TLV!
2.
Calculate the density of CO
2
gas at 40
o
C and
p = 1 atm. (This temperature is just above the
critical temperature of carbon dioxide, see problem 3 on the following page),
a).
(3 points)
Assuming that
CO
2
is an ideal gas under these conditions.
Hint: to calculate the density, use the ideal gas equation to calculate the number of moles e.g. in 1 m
3
.
Note that the answer requires units!
Answer:
With V = 1.00 m
3
, T = 313K, p = 1.01x10
5
Pa, we calculate n = pV/RT = 1.01x10
5
x1.00/(8.314x313)
= 38.8 moles N
2
(in 1 m
3
)
38.8 moles = 38.8x44.0 = 1.69x10
3
g
d = m/V = 1.69x10
3
(g)
/1.00(m
3
) =
1.69x10
3
g.m
–3
or 1.69 kg.m
–3
Again, units must be shown, and correct (or reasonable) significant figures
b).
(3 points)
Assuming that CO
2
is a nonideal gas following the van der Waals equation, with the
constants a and b as in the Lecture Notes Table 1.1 (p.6)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 huda
 Chemistry, Scientific Notation, Van der Waals

Click to edit the document details