Rawalpindi Board Bise Physical States of Matter Class 9 Physics Notes Chapter 5 short long questions, Problems and Mcqs.
Short Questions
Q.1) What is diffusion? Explain with an example?
Answer:
When two gases are placed in contact, they mix spontaneously. This is due to the movement of one gas molecule into the other. This process of mixing gases by the random motion of molecules is called diffusion.
For example, when bromine is added to a gas jar filled with air, the red-brown gas is seen to fill the gas jar slowly. This is due to the diffusion of bromine. The rate of diffusion of a lighter gas is more compared to a heavier gas. H2 diffuses four times faster than O2.
Q.2) Define standard atmospheric pressure. What are its units? How is it related to Pascal?
Answer:
Standard atmospheric pressure (atm) is the pressure exerted by the earth’s atmosphere at sea level. It is the pressure exerted by a mercury column of 760 mm height at sea level.
Atmospheric pressure is expressed in several systems of units: millimeters of mercury (mm of Hg), standard atmospheres, torr, or Pascal (Pa).
Normal sea-level pressure, by definition, equals 760 mm of Hg, 1 atm, 760 torrs, 101325 Pa.
Q.3) Why are the densities of gases lower than that of liquids?
Answer:
Gases are light in mass, and their molecules occupy more volume than liquids. Moreover, molecules of gases are farther apart. As the liquid molecules have strong intermolecular forces, they are closely packed, and the spaces between them are negligible. Therefore, the densities of gases are lower than that of liquids.
Q.4) What do you mean hv evaporation, and how it is affected by surface area.
Answer:
This process of converting a liquid to a gas or vapor at all temperatures is called evaporation. It is a continuous process.
The greater surface area exposed increases the rate of evaporation. For example, a wet cloth spread out dries faster than when folded.
Q.5) Define the term allotropy with examples.
Answer:
Allotropy is the property of some chemical elements to exist in two or more different forms in the same physical state. They are known as the allotropes of these elements.
Examples: Allotropy elements include tin, carbon, sulfur, phosphorus, and oxygen.
| Element | Allotropic forms |
| Tin | Gray tin, white tin |
| Carbon | diamond, graphite |
| Sulphur | Rhombic crystals, monoclinic crystals |
| Phosphorus | White phosphorus, red phosphorus |
| Oxygen | Diatomic oxygen, triatomic oxygen |
Q.6) In which form sulfur exist at 100 °C.
Answer:
Sulfur forms two allotropic forms, rhombic crystals, and monoclinic crystals. Rhombic crystals are stable below 96°C, and monoclinic crystals are stable above 96°C.
Therefore, sulfur exists as monoclinic crystals at 100oC.
Q.7) What is the relationship between evaporation and the boiling point of a liquid?
Answer:
The evaporation and boiling point of a liquid is inversely proportional to each other. If the boiling point of a liquid is high, its evaporation is slow and vice versa.
The boiling point of a liquid is high due to strong intermolecular forces between the molecules of the liquid. Because of this, molecules overcome the attractive forces among the molecules slowly, and thus evaporation is slow.
Related Post: (Structure of Molecules)
Long Questions physical states of matter g9 bise rawalpindi notes
Q.1) Define Boyle’s law and verify it with an example.
Answer:
Boyle’s Law
Boyle’s law states that the pressure and volume of a gas have an inverse relationship when the temperature is held constant. It means as pressure increases, the volume of the gas decreases in proportion. Similarly, as pressure decreases, the volume of the gas increases.
Mathematically it can be written as:
or
PV = k
where ‘P’ is the pressure of the gas, ‘V’ is the volume of the gas, and ‘k’ is a constant.
The equation above states that the product of pressure and volume is constant for a given mass of confined gas at a constant temperature.
If P1V2 = k then P2V2 =k
Where P1 = initial pressure V1 = initial volume
P2 = final pressure V2 = final volume
As both equations have the same constant, their variable has the same value.
Thus P1V1 = P2V2
Experimental verification of Boyle’s law
Suppose there’s a gas confined in a jar with a piston at the top, as shown in the figure below. The initial state of the gas has a volume equal to 8.0 dm3, and the pressure is 1.0 atm. The temperature and number of moles are constant. Weights are slowly added to the top of the piston to increase the pressure. When the pressure is 2 atm, the volume decreases to 4.0 dm3. The product of pressure and volume remains constant.
P1V1 = P2V2
1 x 8 = 2 x 4
Q.2) Define and explain Charles’ law of gases.
Answer:
Charles’s Law
Charles’s law states that the volume is directly proportional to the kelvin temperature for a fixed mass of gas at constant pressure. As the temperature of the gas increases, the volume increases. This is because the gas molecules begin to move around more quickly and hit the walls of their container with more force. Similarly, as the temperature decreases, the volume also decreases, as shown in the figure below:
Mathematically it can be written as:
V ∝ T
Or
V / T = k
where ‘V’ is the volume of the gas, ‘T’ is the temperature of the gas, and ‘k’ is a constant.
The equation above states that the ratio of volume to the temperature of a given mass of a gas is constant at constant pressure.
Where V1 = initial volume T1 = initial temperature
V2 = final volume T2 = final temperature
As both equations have the same constant, their variable has the same value.
Thus
Q.3) What is vapor pressure, and how it is affected by intermolecular forces.
Answer:
Vapour pressure
Vapour pressure is the pressure of a vapor in equilibrium with its condensed phases in a closed container.
Liquids tend to evaporate into vapor form, and gases tend to condense back to their liquid form. In an open container, liquid molecules evaporate and escape to mix up with the air. However, evaporated molecules cannot escape and gather over the liquid surface when a liquid is in a closed container. Some of the molecules condense to return to liquid.
After some time, the gas condensation rate becomes equal to the rate of evaporation of the liquid. This state is called dynamic equilibrium.
Effect of intermolecular forces on vapor pressure
Vapour pressure is inversely proportional to the strength of intermolecular forces. The vapor pressure is low if the intermolecular forces between molecules are strong. And if the intermolecular forces between molecules are weak, the vapor pressure is high.
For example, water has strong hydrogen bonding and has less vapor pressure than that alcohol at room temperature.
Q.4) Define boiling point and explain how it is affected by different factors.
Answer:
Boiling point
The boiling point of a liquid is the temperature at which its vapor pressure is equal to the pressure of the gas above it.
When a liquid is heated, its molecules gain energy and overcome the intermolecular forces—the rate of evaporation increases which increases vapor pressure. A stage is reached when the vapor pressure of the liquid becomes equal to atmospheric pressure. At this point, the liquid starts boiling.
Factors affecting the boiling point
i) Pressure
Boiling points depend upon the external pressure over the surface of the liquid. The boiling point is high at high external pressure and vice versa.
ii) Nature of liquid
If the intermolecular forces between molecules are strong, the boiling point is high and vice versa. Polar liquids have high boiling points than non-polar liquids because there are strong intermolecular forces in polar liquids.
iii) Intermolecular forces
If the intermolecular forces between molecules are relatively strong, the boiling point will be relatively high and vice versa. Liquids with strong intermolecular forces attain a vapor pressure equal to external pressure at high temperatures.
iv) Impurities
The presence of impurities in a liquid increases its boiling point.
Q.5) Describe the diffusion phenomenon in liquids and the factors that influence it.
Answer:
Diffusion is the movement of a fluid from an area of higher concentration to an area of lower concentration. Diffusion is a result of the kinetic properties of particles of matter.
It is the intermixing of liquids by the random motion of their molecules and collisions to form a homogeneous mixture.
For example, a drop of ink added to a glass of water diffuses slowly to form a uniformly colored solution.
Factors affecting diffusion
i) Temperature
Generally, the rate of diffusion increases with an increase in temperature; as temperature increases, the average kinetic energy of particles increases. Thus the intermolecular forces become weak, and the rate of diffusion increases.
ii) Size of molecules
A smaller particle at a given temperature moves faster than a larger particle. The rate of diffusion is inversely proportional to the size of molecules. For example, honey diffuses more slowly in water than that alcohol.
iii) Concentration difference
The rate of diffusion is directly proportional to the concentration gradient. The greater the concentration difference is between the two areas, the faster the diffusion.
iv) Intermolecular forces
Liquids with strong intermolecular forces diffuse slower than liquids with weak intermolecular forces.
Q.6) Differentiate between crystalline and amorphous solids.
Answer:
| Crystalline solids | Amorphous solids |
| A crystalline solid is a solid that is composed of orderly, repeating three-dimensional arrangement of particles. | Amorphous solid does not have a well-defined arrangement of its particles. |
| Crystalline solids have sharp melting points. | Amorphous solids do not melt at a definite temperature but gradually soften when heated. |
| Crystalline solids give clean cleavage. | Amorphous solids give irregular cut. |
| Examples: Diamond, sodium chloride etc | Examples: Glass, plastic, rubber etc |
Numericals
Q.1) Convert the following units:
a) 850 mm Hg to atm (b) 205000 Pa to atm
a) 560 torr to cm Hg (d) 1.25 atm to Pa
Answer:
a) 850 mm Hg to atm
760 mm of Hg = 1 atm
b) 205000 Pa to atm
101325 Pa = 1 atm
c) 560 torr to cm Hg
560 torr = 560 mm Hg
d) 1.25 atm to Pa
1 atm = 101325 Pa
1.25 atm = 101325 x 1.25 = 126656.25 Pa
Q.2) Convert the following units:
a) 750 °C to K (b) 150 °C to K
c) 100 K to °C (d) 172 K to °C
Answer:
a) 750oC to K
TC = 750oC
TK =?
TK = TC + 273
= 750 + 273 = 1023 K
b) 150oC to K
TC = 150oC
TK =?
TK = TC + 273
= 150 + 273 = 423 K
c) 100 K to oC
TK = 100 K
TC =?
TC = TK – 273
= 100 – 273 = -73oC
d) 172 K to oC
TK = 172 K
TC =?
TC = TK – 273
= 172 – 273 = -101oC
Q.3) A gas at pressure 912 mm of Hg has volume 450cm3. What will be its volume at 0.4 atm?
Answer:
Given that
P1 = 912 mm of Hg V1 = 450 cm3
= 912 / 760 = 1.2 atm
P2 = 0.4 atm V2 =?
Formula applied
P1V1 = P2V2
Putting values
1.2 x 450 = 0.4 x V2
Q.4) A gas occupies a volume of 800 cm3 at 1 atm, when it is allowed to expand up to 1200 cm3 what will be its pressure in mm of Hg.
Answer:
Given that
P1 = 1 atm V1 = 800 cm3
P2 (in mm of Hg) =? V2 = 1200 cm3
Formula applied
P1V1 = P2V2
Putting values
1 x 800 = P2 x 1200
We know that
1 atm = 760 mm of Hg
0.667 atm = 760 x 0.667 = 506.7 mm of Hg
Q.5) It is desired to increase the volume of a fixed amount of gas from 87.5 to 118 cm3 while holding the pressure constant. What would be the final temperature if the initial temperature is 23 °C.
Answer:
Given that
V1 = 87.5 cm3 T1 = 23oC
= 23 + 273 = 296 K
V2 = 118 cm3 T2 =?
Formula applied
Putting values
= 399 – 273 = 126oC
Q.6) A sample of gas is cooled at constant pressure from 30 °C to 10 °C. Comment:
a) Will the volume of the gas decrease to one third of its original volume?
b) If not, then by what ratio will the volume decrease?
Answer:
Given that
Suppose V1 = V T1 = 30oC
= 30 + 273 = 303 K
V2 =? T2 = 10oC
= 10 + 273 = 283 K
Formula applied
Putting values
a)
Volume of the gas does not decrease to one third of its original volume.
b)
V1 : V2
1 : 0.93
The volume will decrease by ratio 1 : 0.93
Q.7) A balloon that contains 1.6 dm3 of air at standard temperature (0°C) and (1atm) pressure is taken under water to a depth at which its pressure increases to 3.0 atm. Suppose that temperature remains unchanged, what would be the new volume of the balloon. Does it contract or expand?
Answer:
Given that
P1 = 1 atm V1 = 1.6 dm3
P2 = 3.0 atm V2 =?
Formula applied
P1V1 = P2V2
Putting values
1 x 1.6 = 3.0 x V2
The balloon contracts to almost 1/3 of the initial volume.
Q.8) A sample of neon gas occupies a volume of 75.0 cm3 at very low pressure of 0.4 atm. Assuming temperature remains constant, what would be the volume at 1.0 atm pressure?
Answer:
Given that
P1 = 0.4 atm V1 = 75.0 cm3
P2 = 1 atm V2 =?
Formula applied
P1V1 = P2V2
Putting values
Q.9) A gas occupies a volume of 35.0 dm3 at 17 °C. If the gas temperature rises to 34°C at constant pressure, would you expect the volume to double? If not, calculate the new volume.
Answer:
Given that
V1 = 35 dm3 T1 = 17oC
= 17 + 273 = 290 K
V2 =? T2 = 34oC
= 34 + 273 = 307 K
Formula applied
Putting values
Volume will be doubled if absolute temperature is doubled from 290 K to 580 K. Increase in temperature from 290 K to merely 307 K cannot be expected to double the volume.
Q.10) The largest moon of Saturn, is Titan. It has an atmospheric pressure of 1.6 x l05 Pa. What is the atmospheric pressure in atm? Is it higher than earth’s atmospheric pressure?
Answer:
We know that
101325 Pa = 1 atm
= 1.58 atm
Yes, Titan has higher atmospheric pressure than Earth’s.
