Practice
converting pressure units and Kinetic Theory
1. What
are the 3 assumptions of the kinetic theory? The particles in a gas are considered to be small, hard
spheres with an insignificant volume, the motion of the particles in a gas is
rapid, constant, and random, all collisions between particles in a gas are
perfectly elastic
2. An
empty space with no particles and no pressure is called a vacuum
3. What
is the source of pressure? Source of
pressure is from the force exerted by a gas per unit surface area of an object.
With collisions of billions of rapidly moving particles in a gas with an
object, as the result there is gas pressure
4. Describe
the particle motion at absolute zero: At
absolute zero, the temperature at which the motion of particles theoretically
ceases. The particles would have no kinetic energy at absolute zero, this means
the object doesn’t have energy because there is no kinetic energy.
5. If
a gas heats up, do particles move faster or slower? Faster Assuming
the container is rigid how does an increase in temperature impact pressure? Increasing
in temperature will cause the gas particle to move faster, so they will collide
with the walls of the container more often and more forceful. Due to more
collisions with the wall of container, the pressure is increasing. If a gas
is compressed, do particles hit the walls of the container more or less often? More often
How does this impact pressure (assuming
the container is rigid? If
the particles hit the walls of the container more often, so the pressure is
increasing.
6. If
more gas particles are added to a container, what happens to the pressure of
the container? Why? The pressure will increase. Because as the gas
particles is added to the container, the number of particles is increased so
they will be more collisions with the container. Therefore, pressure is
increasing.
7.
For each of the
following, convert from the units given to the desired units using dimensional
analysis. SHOW YOUR WORK if you wish to
receive credit. Include all units and box
or highlight your answers
a.
130.5 kPa = ? atm
Given: pressure=130.5 kPa; 1
atm= 101.3 kPa
Required: pressure expressed in
atm
Solve: 130.5 kPa x 1 atm /
101.3 kPa = 1.3 atm
Paraphrase: The pressure is 1.3
atm
Evaluate: The result makes
sense because the conversion factor is much less than 1, so the values
expressed in atm is smaller than the value expressed in kPa
b.
45 kPa = ? mmHg
Given: pressure= 45 kPa; 1 atm=
760 mmHg; 1 atm = 101.3 kPa
Required: Pressure expressed in
mmHg
Solve: 45 kPa x 760 mmHg /
101.3 kPa = 337.6 mmHg
Paraphrase: The pressure is
337.6 mmHg
Evaluate: The result makes
sense because the conversion factor is much greater than 1, so the values
expressed in mmHg would be bigger than the value expressed in kPa
c.
1.2 atm = ? kPa
Given: pressure = 1.2 atm; 1
atm= 101.3 kPa
Required: pressure expressed in
kpa
Solve: 1.2 atm x 101.3 kPa / 1
atm = 121.6 kPa
Paraphrase: The pressure is
121.6 kPa
Evaluate: The result makes
sense because the conversion factor is much greater than 1, so the values
expressed in kPa would be bigger than the value expressed in atm
d.
0.76 atm = ? mmHg
Given: pressure = 0.76 atm ; 1
atm = 760 mmHg
Required: pressure expressed in
mmHg
Solve: 0.76 atm x 760 mmHg / 1
atm = 577.6 mmHg
Paraphrase: The pressure is
577.6 mmHg
Evaluate: The result makes
sense because the conversion factor is much greater than 1, so the values
expressed in mmHg would be bigger than the value expressed in atm
e.
770 mmHg = ? kPa
Given: pressure= 770 mmHg ; 1
atm = 760 mmHg; 1 atm = 101.3 kPa
Required: pressure expressed in
kPa
Solve: 770 mmHg x 101.3 kPa /
760 mmHg = 102.6 kPa
Paraphrase: The pressure is
102.6 kPa
Evaluate: The result makes
sense because the conversion factor is much less than 1, so the values
expressed in kPa would be smaller than the value expressed in mmHg
f.
25 mmHg = ? atm
Given: pressure = 25 mmHg ; 1
atm = 760 mmHg
Solve: 25 mmHg x 1 atm / 760
mmHg = 0.03 atm
Paraphrase: The pressure is
0.03 atm
Evaluate: The result makes
sense because the conversion factor is much less than 1, so the values
expressed in atm would be smaller than the value expressed in mmHg
Gas Laws (there are 9 questions)
1. For
each of the following, indicate which gas law is described:
a. As
volume increases, pressure decreases: Boyle’s
Law
b. As
temperature increases, volume increases: Charles’s
law
c. As
temperature increases, pressure increases: Guy- Lussac’s law
2. If
a balloon is squeezed, what happens to the pressure within the balloon? The pressure within the balloon will increase
3. What
happens to the pressure inside your tires if the temperature decreases? The pressure also decreases
4. When
a balloon is taken from a room at 25oC to the outside at 125oC,
what happens to the volume of the balloon? volume increases
Solve the following problems.
Show your work for each problem. Include correct units. Enclose your
answers in a box. Make sure you determine which gas law you should use before
you attempt the problem.
5.
What is the new volume
of a 87 L sample of gas at 89oC and 107 kPa when it is heated to 95 oC
and the pressure increases to 112 kPa? Gas law you will use: Combined gas law Work and solution:
Given: V1 = 87 L; T1=
89 C+ 273 = 362 K; P1= 107 kPa; T2= 95 C+273= 368K; P2=
112 kPa
Required: V2
Solve: V2 = P1 x
V1 x T2 / P2 x T1
V2 = 107 kPa x 87 L x 368K
/ 112 kPa x 362K = 85 L
Paraphrase: The new volume is approximately 84.5 L
Evaluation: An increase in temperature and pressure has
opposite effects on the volume of a gas. To evaluate the increase in volume, multiply
the original volume(87L) by the ratio of P1 to P2 (0.95)
and the ratio of T2 to T1 (1.017). The result is 84.5 L
6.
What is the new volume of a 1.9 mL container at 76o C
when the temperature is adjusted to 283 oC? (Assume pressure is
constant) Gas law you will use: Charles’s
law Work and solution:Given:
V1: 1.9 mL; T1: 76C + 273 = 349K; T2= 283C +
273= 556K Required: V2 Solve: V2=
V1 x T2 / T1 V2= 1.9 mL x 556K / 349 K =
3.03 mL Paraphrase: The new volume is approximately
3.03 mL
evaluation: The volume decrease as the temperature
decrease. This result agrees with both kinetic theory and Charles’s law
7.
A container of gas at 75 oC
and 101 kPa increases in pressure to 200 kPa.
What is the new temperature of the gas? (assume that volume is constant)
Gas law you will use: Gay-Lussac’s Law Work and solution: Given:
T1= 75C+273 = 348K;P1= 101 kPa; P2= 200 kPa
Required: T2
Solve: T2= P2 x T1 / P1 T2= 200 kPa x 348K / 101 kPa= 689.1 K
Paraphrase: The new temperature of the gas is 689.1 K
Evaluation: From the kinetic theory, the
expected outcome if the pressure increase, the temperature is also increasing
if the volume remains constant. The calculated value does show such an
increase.
8.
What
is the new volume when a 65.0 mL container at standard pressure (101.3 kPa) is
expanded until the new pressure is 204.3 kPa? (assume the temperature is
constant) Gas law you will use: Boyle’s law. Work and solution: Given: V1= 55.0 mL; P1=
101.3 kPa; P2= 204.3m kPa Required: V2
Solve: V2= P1 x V1 / P2
V2= 101.3 kPa x 55.0 mL / 204.3 kPa = 27.3 mL
Paraphrase: The new volume is 27.3 mL
Evaluation: An increase in pressure at constant temperature must correspond to
a proportional decrease in volume. The calculated result agrees with both
kinetic theory and the pressure-volume relationship. Also, the units have
canceled correctly and the answer is expressed to the proper number of
significant figures
9.
What is the new pressure of a 15 L sample of gas at 85oC
and 101 kPa when it is heated to 125oC and the volume increases to
17.5L? Gas law you will use: Combined
gas law Work and
solution: Given: V1= 15 L; T1= 85C+273= 358K; P1=101
kPa; T2= 125C + 273 = 398K; V2= 17.5 L Required: P2
Solve: P2 = P1 x V1 x T2 / V2
x T1
P2= 101 kPa x 15 L x 398 K / 17.5 L x 358 K= 96.3 kPa
Paraphrase: the new pressure is 96.3 kPa
Evaluation: Increase in volume has an opposite effects on the pressure. To
evaluate the decrease in pressure, multiply the original pressure (101 kPa) by
the ratio of V1 to V2 (0.857) and the ratio of T2
to T1 (1.112). The result is 96.3 kPa
you missed what 's required in #7f
ReplyDeleteRequired: pressure expressed in atm