Chemistry 201/211 Fall 2004 Oregon State University
Worksheet 4 October 28, 2004
1.) What volume of CH
4
at 0 o
C and 1.00 atm contains the same number of molecules as
0.50 L of N
2
measured at 27 o C and 1.50 atm? n
V


0.0821
1.50
atm
L
 atm
 mol
0.50L

K

0.0305
mol

0.0821
L

300 atm
1 .
00 atm
K
 mol

0 .
0305 mol
K

273 K

0 .
68 L
2.) Fluorine gas, which is dangerously reactive, is shipped in steel cylinders of 30.0 L capacity at a pressure of 160.0 lb/in
2
at 26 o
C. What mass of F
2
is contained in a cylinder? 14.70 lb/in
2
= 1atm
160 n
 lb in
2

1 atm
14 .
70 lb
10.88
atm in

0.0821
L
 atm mol
2

10
30.0L

K

.
88 atm
299K

13 .
3 mol
13.3
mol

38.00
g F
2
1 mol F
2

505 g F
2
3.) Starting at 5.0 atm, what would the final pressure be if the size of an airtight cylinder increased by 10.0%? Assume a constant temperature. a.) 4.5 atm b.) 5.5 atm c.) 50. atm d.) 0.50 atm e.) 2.3 atm
P i
V i
V
P f f

P f
V f


110 %V
5 atm

V i
1.1V
i i

1.1V
i

4 .
5 atm
4.) The gas pressure in an aerosol can is 1.5 atm at 25 o
C. What would the pressure be if the can were heated to 450 o
C?
The number of moles and the volume stays the same.
P
1
T
1

P
2
T
2
P
2

1.5
atm

7 23
2 98 o C o
C

3 .
6 atm

5.) Which of the graphs are not consistent with the ideal gas law for one mole of gas?
Graph A is consistent. The pressure-volume product does not change as the volume is increased (PV/V=P=constant). The other variables (n and T) remain constant since they are not on the graph.
Graph B is consistent. PV= const, so P

1/V.
Graph C is consistent. P and n are constant and V goes up as T increases.
Graph D is inconsistent. A graph of P vs. T should be a straight line.
6.) If 3.0 L of helium at 20.0 o
C is allowed to expand to 4.4 L, with the pressure remaining the same, what is the new temperature? a.) 702 o
C b.) 430 o C c.) 157 o
C d.) -30 o
C e.) -55 o
C
The amount of gas (n) and the pressure (P) remain the same.
V
1
T
1

V
2
T
2
3.0
L
293K

4 .
4 L
X
X

430 o C
7.) When 2.40 g of a certain volatile liquid is completely vaporized, the volume of the resulting vapor is 821 mL at a temperature of 127 o
C at standard pressure. The molecular weight of this substance is about a.) 32.0 g/mol b.) 64.0 g/mol c.) 96.0 g/mol d.) 128 g/mol e.) 192 g/mol

n

PV
RT

1 atm

0.0821
0.0821
L

L
 atm mol atm

K mol


400
K
K

0 .
025 mol
2.4
g
0.025
mol

96 g/mol
8.) Calculate the density of carbon dioxide gas in grams per liter at 2.00 atm and 0 o C. a.) 0.57 g/L b.) 0.88 g/L c.) 1.50 g/L d.) 1.96 g/L e.) 3.93 g/L d

MP
RT

44
0.0821
.
0
L g mol

2
 atm mol
.
00

K atm

273 K

3 .
93 g
L
9.) Page 123 of your book explains that hot air rises because a gas becomes less dense when its temperature is increased. A hot air balloon rises because the weight of the hot air inside of the balloon is lighter than the same volume of cool air that would normally occupy that volume. If a hot air balloon has a volume of 224,000 L and the average gas temperature is 100 o
C, what total mass could be lifted on a winter day when the air temperature is 0 o
C and the pressure is 1 atm? The molar mass of dry air is 29.0 g/mol d d c h


29.0
g/mol

1atm
.0821
L
 atm
29.0
mol g/mol


K

1atm
273 K
.0821
L
 atm mol

K

373 K


1
0
.
29387
.
94699 g g
L
L
The mass of the cold air is 1 .
29387 g/L

224000 L

289828g
The mass of the hot air is 0 .
94699 g/L

224000L

212126g
The difference
289828g in mass is equal to the weight th
212126g

77700g at the ballon can carry
10.) Five balloons are filled to the same volume. Each balloon contains a different gas,
CO
2
, O
2
, He, N
2
and CH
4
. Which of the following statements about the balloons is false. a.) the balloon with the He will have the fastest molecules b.) the average energy of the molecules in the He balloon will be the highest c.) the pressure of the balloons is the same d.) if the gasses effuse through the balloons, the one with the CO
2
will stay inflated the longest. e.) The helium, nitrogen, and methane (CH
4
) balloons should rise in the air. a.) True.
μ rms

3RT
The lighter the gas ( M is the molecular weight), the faster the gas.
M

b.) False. The energy of a gas is E t

3RT
. It is independent of the mass of the gas.
2N
A c.) True. V
 n. Since the gases have the same volume, they have the same number of moles. They are all at the same temperateure since it is not specified that it is different.
With V, n, T the same for all cases, P must also be the same. d.) Since effusion is inversely proportional to mass (see a.) then the one with the largest mass should move the slowest and thus stay inflated the longest. e.) True. A balloon will rise in the gas if the air inside is less dense than air. D

MP
RT
The gas with a molecular weight lower than that of air (29.0 g/mol- pg 122) will rise. He
(4 g/mol), N
2
(28 g/mol) and CH
4
(18g/mol) will all rise. O
2
(32 g/mol) and CO
2
(44g/mol) will not.
11.) ***(Difficult problem. Only for those who want a challenge)
Two flasks of equal volume are connected by a narrow tube of negligible volume.
Initially, both flasks are at 27.0 o
C and each flask contains 0.70 moles of hydrogen gas, the pressure being 0.50 atm. One of the flasks is then immersed in a hot oil bath at 127 o C, while the other is kept at 27 o C in a water bath. Calculate the pressure of the flasks.
We must first find the volume of each of the flasks
V

0.70
mol

.0821
L
 atm mol

K

300K
0.50atm
:

34 .
482 L
Both the pressure and the volume of each flask remains the same (the two volumes are connected through an open tube so the pressures must be equal).
Only the temperatu re and the amount of moles changes.
n c
T c
 n h
T h
In the equation above, there are two unknowns, the moles on the hot side and the moles on the cold side.
To solve for two unknowns, we must have at least two equations.
The second equation can be found by realizing n c n h
that the
 n h

1.4

1.4
n c
total number of moles must add up to 1.4
(0.70
on each side).
combining both equations : n n c c
T c

( 1.4
(300K)
(300K)n
(700K)n n c

0.80
c c
n c
)T h


(1.4
n c
)(400K)
(560K)

400n c

560K

300K

0 .
0821
L
 atm
P c

0.80
mol
34 .
5 L mol

K 
0 .
57 atm