1.
a measure
of how much the volume of matter decreases under pressure compressibility
2.
the pressure
exerted by a gas in a mixture partial
pressure
3.
the escape
of gas through a small hole in a container effusion
4.
tendency of
molecules to move to regions of lower concentration diffusion
5.
Why is it
easier to compress a gas than a liquid or solid? It is easier to compress a gas than a
liquid or solid because unlike those two, gas particles are quite further away
from each other compared to liquid and solid particles, which are quite intact.
Thus, there’s more space between the gas particles causing them to be
compressed easily.
6.
Why does
the pressure inside a container of gas increase if more gas is added to the
container? The
pressure inside the container of a gas increases because less space is being
created by adding more gas particles into it. In addition, according to Boyle’s
Law, the volume of gas is inversely proportional to pressure, that is, when
adding more gas particles in the
container, less space is created, meaning less volume, thus pressure increases.
7.
What
happens to the temperature of a gas when it is compressed? The temperature of the gas increases as
the gas is compressed because the air particles are more brought closer to each
other, thus increasing the chance of colliding with each other causing the rise
in temperature.
8.
As the
temperature of the gas in a balloon decreases, which of the following occurs? When the temperature of the gas decreases,
the volume also decreases and eventually condenses.
9.
What
happens to the pressure of a gas inside a container if the temperature of the
gas decreases? As the gas
temperature decreases, the pressure of gas inside a container also decreases
because according to Gay – Lussac’s Law, pressure of gas is directly
proportional to its temperature.
10. If a balloon is heated, what happens to the
pressure of the air inside the balloon if the volume remains constant? When the balloon is heated, the
pressure inside it increases causing the balloon to expand.
11. If a balloon is heated, what happens to the
volume of the air in the balloon if the pressure is constant? If a balloon is heated, the
volume of the air in the balloon increases.
12. An ideal gas CANNOT be
a.
condensed c. heated
b.
cooled d. compressed
13. Under
what conditions of temperature and pressure is the behavior of real gases most
like that of ideal gases: Real
gases behave more like that of an ideal gases when they have high temperature and low pressure.
14. Which
of the following gases will effuse most rapidly? Hydrogen, nitrogen, oxygen, or helium. Why? Hydrogen will effuse most rapidly
because it has the least molar mass among the gases given. And according to
Graham’s Law of effusion, the rate of effusion is inversely proportional to the
square root the molar mass of the gas. Therefore, hydrogen, having the least
molar mass value, will effuse most rapidly.
For each of the following, show
your work, include correct units and box or highlight your answers. Hint: figure out which gas law you need to
use first.
15. The volume of a gas is 25L at
349.0 kPa pressure. What will the volume be when the pressure is reduced to
50.0 kPa, assuming the temperature remains constant?
Given: V1
= 25L, P1 = 349.0 kPa , P2 = 50. 0 kPa
Required: V2
Analysis: V1
x P1 = V2 x P2
Solve: 25L x 349.0 kPa = V2 x
50.0 kPa
V2
= 25L (349.0kPa) / 50.0 kPa
=174.5L
When the pressure is reduced to 50.0 kPa, the volume
increases to 174.5L .
16. A gas has a volume of 55 mL at
a temperature of –5.0
C. What volume will the gas occupy at 39.0C?
C. What volume will the gas occupy at 39.0C?
Given: V1 =
55 mL, T1 = -5.0oC or 268 K , T2 = 39oC or
312 K
Required : V2
Analysis: V1
/ T1 = V2 / T2
Solve: 55mL / 268 K = V2 / 312 K
V2
= 55mL (312 K) / 268 K
= 64.0 mL
The gas will occupy 64.0 mL at 39.0oC .
17. A rigid container of O
has
a pressure of 987 kPa at a temperature of 713 K. What is the pressure at 273 K?
has
a pressure of 987 kPa at a temperature of 713 K. What is the pressure at 273 K?
Given: P1
= 987 kPa, T1 = 713 K, T2 = 273 K
Required : P2
Analysis: P1/
T1 = P2 / T2
Solve: 987 kPa/
713 K = P2/ 273 K
P2
= 987 kPa (273 K) / 713 kPa
= 377.9 kPa
The pressure of O2 in the container at 273 K
is 377.9 kPa.
18. A gas occupies a volume of 567
mL at 35.0C and 99 kPa. What is the volume of the gas at 273 K and
101.3 kPa (STP)?
C and 99 kPa. What is the volume of the gas at 273 K and
101.3 kPa (STP)?
Given: V1
= 567 mL, T1 = 35.0oC or 308 K, P1 = 99 kPa, P2 = 101.3 kPa, T2
= 273 K
Required: V2
Analysis: P1 x V1/
T1 = P2 x V2 / T2
99kPa (567 mL) / 308 K = V2
101.3 kPa / 273 K
V2
= [(99 kPa)(567 mL)(273 K)] / 101.3 kPa (308 K)
= 491.2
mL
The gas will have a volume of 491.2 mL at temperature of
273 K and a pressure of 101.3 kPa.
19. How many moles of N are
in a flask with a volume of 0.870 L at a pressure of 300.0 kPa and a
temperature of 300.0 K?
are
in a flask with a volume of 0.870 L at a pressure of 300.0 kPa and a
temperature of 300.0 K?
Given: P = 300.0 kPa, V = .870 L, T = 300.0
K, R = 8.31 (L*kPa)/ (K*mol)
Required: n of N2
Analysis: n =
PV/ RT
Solve: n = 300.0 kPa
(0.870 L)/ [8.31 (L*kPa)/ (K*mol) x ( 300.0 K)]
= 0.1 mol N2
There are 0.1 mol of N2
in a flask with a volume of 0.870 L at a pressure of 300.0 kPa and a
temperature of 300.0 K
20. What is the pressure exerted by
87 g of O in
a 2.0-L container at 302.0C?
in
a 2.0-L container at 302.0C?
Given: mass = 87 g,
V= 2.0 L, T = 302.0C or 575 K
C or 575 K
Required: Pressure
(P)
Analysis: Convert
the mass of O2 to moles first, n O2 = 87 g x 1 mol/
2(16g)
= 2.7 mol O2
P = nRT/V
Solve: P = 2.7 mol x
[8.31 (L*kPa)/ (K*mol)] x 575 K / 2.0 L
=
6450.6 kPa
The pressure exerted by 87 g O2 in a 2.0 L
container at 302C is 6450.6 kPa.
C is 6450.6 kPa.
21. A mixture of gases at a total
pressure of 234 kPa contains N, CO, and O. The partial pressure of the CO is
124 kPa and the partial pressure of the N is
48 kPa. What is the partial pressure of the O?
, CO, and O. The partial pressure of the CO is
124 kPa and the partial pressure of the N is
48 kPa. What is the partial pressure of the O?
Given: Ptotal
= 234 kPa, Pnitrogen = 48 kPa, PCarbon dioxide = 124 kPa
Required: Poxygen
Analysis: Poxygen = : Ptotal - ( Pnitrogen + PCarbon dioxide )
Solve: Poxygen = 234 kPa – (48 kPa + 124 kPa)
= 62 kPa
The
partial pressure of oxygen is 62 kPa.
22. Use Graham’s law to calculate
how much slower bromine gas, Br2 , will effuse than chlorine gas, Cl, will.
, will.
Given:
M1 (bromine gas) = 2
(79.9g) , M2 (chlorine gas) = 2
(35.5 g)
= 159.8 g =
71g
Required: rate of effusion of Bromine gas and Chlorine gas
Analysis: R1/R2=√(M2/M1)
Solve: R1/R2= √ ( 71g / 159.8g)
= 0.45
Bromine gas will effuse 0.45 times slower than chlorine
gas.
23. How does the air pressure in a
balloon change when the balloon is squeezed? Explain why this change occurs. When the balloon is squeezed, it
creates a smaller room making the gas particles to come closer together, hence
resulting to more collision of the gas particles that increases the pressure.
24. How does the pressure of an
enclosed gas in a rigid container change when the gas is heated? Explain why
this change occurs. When the container
is heated, the gas particles became excited and moves rapidly colliding with
the container and other particles and as a result, the gas inside expand, thus
changing the pressure.
25.
Explain how pumping air into a
bicycle tire increases the pressure within the tire. As you pump air into the bike, more gas particles are added in
the tire. _With this, it creates less room for the gas particles that’s why
they are brought closer together increasing the tendency of collision among
particles resulting to higher pressure inside the tire.
#21-27 with the evaluation
#21-27 with the evaluation
- The volume of a gas is
25L at 349.0 kPa pressure. What will the volume be when the pressure is
reduced to 50.0 kPa, assuming the temperature remains constant?
Given: V1
= 25L, P1 = 349.0 kPa , P2 = 50. 0 kPa
Required:
V2
Analysis:
V1 x P1 = V2 x P2
Calculation:
25L x 349.0 kPa = V2 x 50.0 kPa
V2
= 25L (349.0kPa) / 50.0 kPa =174.5L
Paraphrase: When the pressure is reduced
to 50.0 kPa, the volume increases to 174.5L .
Evaluation: It makes sense because when
pressure decreases, volume will increase.
- A gas has a volume of
55 mL at a temperature of –5.0
C. What volume will the gas
occupy at 39.0
C?
Given: V1
= 55 mL, T1 = -5.0oC or 268 K , T2 = 39oC or 312 K
Required
: V2
Analysis:
V1 / T1 = V2 / T2
Calculation:
55mL / 268 K = V2 / 312 K
V2
= 55mL (312 K) / 268 K= 64.0 mL
Paraphrase:
The gas will occupy 64.0 mL at 39.0oC
Evaluation:
The answer makes sense because volume will increase if temperature increases.
- A rigid container of O
has a pressure of 987 kPa at a
temperature of 713 K. What is the pressure at 273 K? Given: P1 = 987 kPa, T1 = 713
K, T2 = 273 K
Required : P2
Analysis: P1/ T1 = P2 /
T2
Calculation: 987 kPa/ 713 K = P2/ 273 K
P2 = 987 kPa (273 K) / 713 kPa = 377.9 kPa
Paraphrase: pressure of O2 in the container at
273 K is 377.9 kPa.
Evaluation: The new pressure is lower than the last
pressure. This answer makes sense because pressure decreases as the temperature
decreases.
- A gas occupies a
volume of 567 mL at 35.0
C and 99 kPa. What is the
volume of the gas at 273 K and 101.3 kPa (STP)?
Given: V1 = 567 mL, T1
= 35.0oC or 308 K, P1 = 99 kPa, P2 = 101.3 kPa, T2
= 273 K
Required: V2
Analysis: P1 x V1/
T1 = P2 x V2 / T2
Calculation: 99kPa (567 mL) / 308 K = V2
101.3 kPa / 273 K
V2 = [(99 kPa)(567 mL)(273
K)] / 101.3 kPa (308 K) = 491.2 mL
Paraphrase: The gas will have a volume
of 491.2 mL at temperature of 273 K and a pressure of 101.3 kPa.
Evaluation: It’s reasonable that the new volume is smaller
than the last volume because the temperature drops and the pressure is bigger.
5.
How
many moles of N
are in a flask with a volume of 0.870 L at a
pressure of 300.0 kPa and a temperature of 300.0 K?
are in a flask with a volume of 0.870 L at a
pressure of 300.0 kPa and a temperature of 300.0 K?
Given: P = 300.0 kPa, V = 0.870 L, T =
300.0 K, R = 8.31 (L*kPa)/ (K*mol)
Required: n of N2
Analysis: n = PV/ RT
Calculation: n = 300.0 kPa (0.870 L)/
[8.31 (L*kPa)/ (K*mol) x ( 300.0 K)] = 0.1 mol N2
Paraphrase: There are 0.1 mol of N2
in a flask with a volume of 0.870 L at a pressure of 300.0 kPa and a
temperature of 300.0 K
Evaluation: It’s reasonable that the
number of moles is very little because the temperature is very high and the
volume is really small.
6.
What
is the pressure exerted by 87 g of O
in a 2.0-L container at 302.0
C?
in a 2.0-L container at 302.0C?
Given: mass = 87 g, V= 2.0 L, T = 302.0C
or 575 K
Required: Pressure (P)
Analysis: Convert the mass of O2
to moles first, n O2 = 87 g x 1 mol/ 2(16g) = 2.7 mol O2
P = nRT/V
Calculation: P = 2.7 mol x [8.31
(L*kPa)/ (K*mol)] x 575 K / 2.0 L = 6450.6 kPa
Paraphrase:
The pressure exerted by 87 g O2 in a 2.0 L container at 302C is
6450.6 kPa.
Evaluation:
It makes sense that the pressure is very high because the volume is really
small and the temperature is really high.
7.
A
mixture of gases at a total pressure of 234 kPa contains N
, CO
, and O
. The partial pressure of
the CO
is 124 kPa and the partial pressure of the N
is 48 kPa. What is the partial pressure of the
O
?
, CO, and O. The partial pressure of
the CO is 124 kPa and the partial pressure of the N is 48 kPa. What is the partial pressure of the
O?
Given: Ptotal = 234 kPa, Pnitrogen
= 48 kPa, PCarbon dioxide = 124 kPa
Required: Poxygen
Analysis: Poxygen = : Ptotal
- ( Pnitrogen + PCarbon dioxide )
Calculation: Poxygen =
234 kPa – (48 kPa + 124 kPa) = 62 kPa
Paraphrase: The partial pressure of
oxygen is 62 kPa.
Evaluation: The answer makes sense
because the partial pressure of oxygen is smaller than the total pressure.
I added evaluations
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