Homework Assignment #6
Name____________________________
Laboratory Section (circle one)
MW1.30
MW6.30
TTh7
1.
Chapter 5, Problem 45, 68, 112
5.45
Total moles = n
T
= PV/RT =
( )( )
( )
( )
3
626 mmHg
355 mL
1 atm
10
L
L • atm
760 mmHg
1 mL
0.0821
273
35 K
mol • K

+
= 0.011563657 mol (unrounded)
Moles Ne = n
Ne
= (0.146 g Ne) (1 mol Ne / 20.18 g Ne) = 0.007234886 mol Ne (unrounded)
Moles Ar = n
T
 n
Ne
= (0.011563657  0.007234886) mol = 0.004328771 =
0.0043 mol Ar
5.68
To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar
masses
(equation 5.14).
2
Rate O
Rate Kr
=
2
Molar Mass Kr
Molar Mass O
=
83.80 g/mol
32.00 g/mol
= 1.618255 =
1.618
5.112
Molar volume uses exactly one mole of gas along with the temperature and pressure given in the
problem.
V / n = RT / P =
( )
( )
L • atm
0.0821
730. K
mol • K
90 atm
= 0.66592 =
0.67 L/mol
2.
Consider two 1.0L samples of gas:
one is He and the other is N
2.
Both are at 1.00 atm and 25°C.
For each quantity given below, indicate which has:
a.
larger mass
b
for same number of moles nitrogen will have larger mass as the molar mass
of nitrogen is greater than helium
b.
greater density
b
nitrogen, as density is proportional to molar mass.
c.
faster average molecular speed
b
helium, as molecular speed is faster for gas with lower
molar mass
d.
slower rate of effusion
b
nitrogen, as effusion rate is inversely proportional to the square
root of the
molar mass
e.
3.
Chapter 6, Problem 9, 11, 18, 19, 25, 27, 30, 34, 38, 42, 45, 51, 59, 64, 68, 78, 85 (a, b only), 111
6.9
Δ
E
=
q
+
w
= 255 cal + (428 cal) =
683 cal
6.11
Δ
E
=
q
+
w
=
( )
3
10 J
0.615 kJ
1 kJ
+
( )
3
10 cal
4.184 J
0.247 kcal
1 kcal
1 cal
= 1648.4 =