GAS LAWS Chemistry Questions and Answers
QUESTION 1.
Explain why the volume of a gas increases when its temperature is increased at a constant pressure.
ANSWER
The volume of a gas increases when its temperature is increased at a constant pressure due to the effect of temperature on the kinetic energy of gas molecules.
Gases are composed of a large number of molecules that are in constant motion. The kinetic theory of gases states that the kinetic energy of gas molecules is directly proportional to their temperature. When the temperature of a gas is increased, the average kinetic energy of the gas molecules also increases.
As the gas molecules gain kinetic energy with an increase in temperature, they move faster and collide more frequently with the walls of the container. These collisions exert pressure on the walls of the container, resulting in an increase in pressure. However, since the pressure is kept constant in this scenario, the gas molecules cannot escape or expand beyond the confines of the container.
As the gas molecules gain kinetic energy and move faster, they also push against each other with greater force. This increases the distance between the gas molecules, which results in an increase in the average distance between gas molecules, or the intermolecular spacing. The increased intermolecular spacing leads to an increase in the volume of the gas.
In summary, when the temperature of a gas is increased at constant pressure, the gas molecules gain kinetic energy, move faster, and collide more frequently with the walls of the container. This results in an increase in pressure.
Additionally, the increased kinetic energy also causes the gas molecules to push against each other more forcefully, leading to an increase in intermolecular spacing and hence an increase in the volume of the gas.
This relationship between temperature and volume at constant pressure is described by Charles's Law, which states that the volume of a gas is directly proportional to its temperature at constant pressure.
QUESTION 2.
Cotton wool pads were socked with concentrated solutions of gas “p” and gas “Q” the
pads were then placed on the opposite ends of a long horizontal glass tube at the same
time.
The tube was then immediately corked at both ends as shown the diagram
below.
After sometimes the gases were observed to meet at point “R” which of the two gases
is dense? Explain your answer.
ANSWER
“Q” it diffuses more slowly i.e, it covered a short distance.
QUESTION 3.
A mixture containing equal volumes of hydrogen and carbon (IV) oxide was
introduced as shown below.
ANSWER
Hydrogen; it is less dense than carbon (iv) oxide and diffuses faster
QUESTION 4
In an experiment to study the diffusion of gases, a student set up the apparatus shown
in diagram I.
After some time the student noticed a change in the water level as
shown in diagram II.
Give an explanation for the change in water level.
ANSWER
Air is less dense than carbon (IV) oxide and so it enters the porous pot faster than
carbon (IV) oxide and so it enters the porous pot faster than carbon (IV) oxide
leaves out of it.
This creates a high pressure in the pot and the level of water rises up as shown.
QUESTION 5.
A fixed mass of gas has a volume of 250 cm3
at a temperature of 2700
and 750 mm
Hg pressure. Calculate the volume the gas would occupy at 420
c and 750 mm
pressure.
ANSWER
V1/T1 = V2/T2
∴V2 = (V1* T2) / T1
∴T2 =(250 x 315) / 300
= 262.5 cm 3
QUESTION 6.
A gas occupies a volume of 400 cm3
at 500k and atmospheric pressure. What will be
the temperature of the gas when the volume and pressure of the gas is 100 cm3
and 0.5 atmospheric pressure respectively?
ANSWER
P1 V1 / T1 = P2 V2 /T2
∴T2= (T1,P2V2) / P1V1
∴T2 = (500x 0.5 x 100) / 1x 400
= 62.5K
QUESTION 7.
A sealed glass tube containing air at S.T.P was immersed in water at 1000C.
Assuming there was no increase in the volume of the glass tube due to the expansion of
the glass.
Calculate the pressure of the air inside the tube.
Standard pressure = 760mmHg: Standard temperature = 273 K.
ANSWER
P1/T1= P2/T2
∴P2 = P1 T2 / T1
∴P2 = (760 x 373) / 273
= 1038.39 mmHg
QUESTION 8.
A given volume of Ozone (03) diffused from a certain apparatus in 96 seconds.
Calculate the time taken by an equal volume of carbon (IV) oxide (Co2) to diffuse
under the same condition (O= 16) (C=12)
ANSWER
Rmm of O3 = 16x 3 = 48
Rmm of CO2 = 12+ 36 =44
TCO2/96 = √44 / √48
Tcoz = (96x 6.63) /6.92
= 91.9 seconds
QUESTION 9
A few crystals of potassium manganate VII were carefully placed in a beaker at
one spot.
The beaker was left undisturbed for two hours. State and explain the
observation that was made.
ANSWER
The entire solution turns pink/purple.
- potassium permanganate/potassium manganate (VII) particles diffused into the
water molecules.
QUESTION 10
The graph below shows the behavior of a fixed mass of a gas at a constant
temperature.
(a) What is the relationship between the volume and the pressure of the gas.
(b) 3 liters of oxygen gas at one atmospheric pressure were compressed to two
atmospheres at a constant temperature.
Calculate the volume occupied by
oxygen gas.
ANSWER
a) The volume of a fixed mass of gas is inversely proportional to the
pressure at a constant temperature.
b) P1V1 = P2V2
∴V2 = P1V1 /P2
V2 = (3 x 1) /2
= 1.5lts
QUESTION 11.
A small crystal of potassium (VII) was placed in a beaker containing water.
The
beaker was left standing for two days without shaking. State and explain the
observations that were made.
ANSWER
This is because potassium (VII), also known as potassium permanganate (KMnO4), is a highly soluble compound in water. It dissolves readily, forming purple or pinkish-colored ions in the solution.
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