Gases deviate from ideal behavior at high pressure. Which of the following statement(s) is correct for non-ideality? (a) at high pressure the collision between the gas molecule become enormous (b) at high pressure the gas molecules move only in one direction (c) at high pressure, the volume of gas become insignificant (d) at high pressure the intermolecular interactions become significant

(d) at high pressure the intermolecular interactions become significant

Rate of diffusion of a gas is

(d) inversely proportional to the square root of its molecular weight

Which of the following is the correct expression for the equation of state of van der Waals gas? (a) [P + \frac{a}{n^{2} V^{2}}](V – nb) = nRT (b) [P + \frac{n a}{n^{2} V^{2}}](V – nb) = nRT (c) [P + \frac{a n^{2}}{V^{2}}](V – nb) = nRT (d) [P + \frac{n^{2} a^{2}}{V^{2}}(V – nb) = nRT]

(c) [P + \frac{a n^{2}}{V^{2}}](V – nb) = nRT

When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules

(b) exert no attractive forces on each other

A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be

(b) Near the hydrogen chloride bottle

The value of the universal gas constant depends upon

(d) units of pressure and volume

Samacheer Class 11 Chemistry Gaseous State Book Back Answers & Solutions

Q.26State Boyle’s law.v

Solution

Boyle’s law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
V α \(\frac {1}{P}\);
where T and n are fixed or PV = Constant = k

Answer:

Boyle’s law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
V α \(\frac {1}{P}\);
where T and n are fixed or PV = Constant = k

Q.27A balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law.v

Solution

Yes, a balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law. The volume of balloon decreases when the temperature reduced from room temperature to low temperature. When cooled, the kinetic energy of the gas molecules decreases, so that the volume of the balloon also decreases.

Answer:

Yes, a balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law. The volume of balloon decreases when the temperature reduced from room temperature to low temperature. When cooled, the kinetic energy of the gas molecules decreases, so that the volume of the balloon also decreases.

Q.28Name two items that can serve as a model for ‘Gay Lusaac’ law and explain.v

Solution

Firing a bullet:
When gunpowder burns, it creates; a significant amount of superheated gas. The high pressure of the hot gas behind the bullet forces it out, of the barrel of the gun.
Heating food in an oven:
When you keep food in an oven for heating, the air inside the oven is heated, thus pressurized.

Answer:

Firing a bullet:
When gunpowder burns, it creates; a significant amount of superheated gas. The high pressure of the hot gas behind the bullet forces it out, of the barrel of the gun.
Heating food in an oven:
When you keep food in an oven for heating, the air inside the oven is heated, thus pressurized.

Q.29Give the mathematical expression that relates gas volume and moles. Describe in words what j the mathematical expression means.v

Solution

* The mathematical relationship betwêen the volume of a gas and the number of moles is V α n
* \(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant
Where V 1 and n 1 are the volume and number of moles of a gas and V 2 and n 2 are the values of volume and number of moles of same gas at a different set of conditions.
* If the volume of the gas increase then the number of moles of the gas also increases.
* At a certain temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas.

Answer:

* The mathematical relationship betwêen the volume of a gas and the number of moles is V α n
* \(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant
Where V 1 and n 1 are the volume and number of moles of a gas and V 2 and n 2 are the values of volume and number of moles of same gas at a different set of conditions.
* If the volume of the gas increase then the number of moles of the gas also increases.
* At a certain temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas.

Q.30What are ideal gases? In what way real gases differ from ideal gases?v

Solution

The kinetic theory of gases which is the basis for the gas equation (PV = nRT), assumes that the individual gas molecules occupy negligible volume when compared to the total volume of the gas and there is no attractive force between the gas molecules. Gases whose behaviour is consistent with these assumptions under all conditions are called ideal gases.
But in practice both these assumptions are not valid under all conditions. For example, the fact that gases can be liquefied shows that the attractive force exists among molecules. Hence, there is no gas which behaves ideally under all conditions. The non-ideal gases are called real gases. The real gases f tend to approach the ideal behaviour under certain conditions.

Answer:

The kinetic theory of gases which is the basis for the gas equation (PV = nRT), assumes that the individual gas molecules occupy negligible volume when compared to the total volume of the gas and there is no attractive force between the gas molecules. Gases whose behaviour is consistent with these assumptions under all conditions are called ideal gases.
But in practice both these assumptions are not valid under all conditions. For example, the fact that gases can be liquefied shows that the attractive force exists among molecules. Hence, there is no gas which behaves ideally under all conditions. The non-ideal gases are called real gases. The real gases f tend to approach the ideal behaviour under certain conditions.

Q.31Can a Vander Waals gas with a = 0 be liquefied? Explain.v

Solution

a = 0 for a Van der Waals gas i.e. for a real gas. Van der Waals constant a = 0. It cannot be liquefied.
If a = 0, there is a very less interaction between the molecules of gas.
‘a’ is the measure of strength of Van der Waals force of attraction between the molecules of the gas.
If a is equal to zero, the Van der Waals force of attraction is very less and the gas cannot be liquefied.

Answer:

a = 0 for a Van der Waals gas i.e. for a real gas. Van der Waals constant a = 0. It cannot be liquefied.
If a = 0, there is a very less interaction between the molecules of gas.
‘a’ is the measure of strength of Van der Waals force of attraction between the molecules of the gas.
If a is equal to zero, the Van der Waals force of attraction is very less and the gas cannot be liquefied.

Q.32Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?v

Solution

Gaseous pressure is developed by the continuous bombardment of the molecules of the gas among themselves and also with the walls of the container. When the molecules hit the sticky area of the container, the number of molecules decreases and hence, the pressure decreases. Therefore, pressure is less than the ordinary area of walls.

Answer:

Gaseous pressure is developed by the continuous bombardment of the molecules of the gas among themselves and also with the walls of the container. When the molecules hit the sticky area of the container, the number of molecules decreases and hence, the pressure decreases. Therefore, pressure is less than the ordinary area of walls.

Q.33Explain the following observations (a) Aerated water bottles are kept underwater during summer (b) Liquid ammonia bottle is cooled before opening the seal (c) The tyre of an automobile is inflated to slightly lesser pressure in summer than in winter (d) The size of a weather balloon becomes larger and larger as it ascends up to a larger altitude.v

Solution

(a) In aerated water bottles, CO 2 gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more gas will be present above the liquid surface in the glass bottle.
In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept underwater. As a result, the temperature is likely to decrease and the solubility of CO 2 is likely to increase in aqueous solution resulting in decreased pressure.
(b) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes an accident.
However, if the bottle is cooled under tap water for some time, there will be a decrease in the volume of a gas to a large extent. if the seal is opened now, the gas will come out of the bottle at a slower rate, reduces the chances of an accident.
(c) The pressure of air is directly proportional to the temperature. Since the temperature is higher in summer than in higher, the pressure of the air in the tube of the lyre is likely to be quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer, Therefore, it is advisable to inflate the types to lesser pressure in summer than in winter.
(d) The volume of the gas is inversely proportional to the pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

Answer:

(a) In aerated water bottles, CO 2 gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more gas will be present above the liquid surface in the glass bottle.
In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept underwater. As a result, the temperature is likely to decrease and the solubility of CO 2 is likely to increase in aqueous solution resulting in decreased pressure.
(b) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes an accident.
However, if the bottle is cooled under tap water for some time, there will be a decrease in the volume of a gas to a large extent. if the seal is opened now, the gas will come out of the bottle at a slower rate, reduces the chances of an accident.
(c) The pressure of air is directly proportional to the temperature. Since the temperature is higher in summer than in higher, the pressure of the air in the tube of the lyre is likely to be quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer, Therefore, it is advisable to inflate the types to lesser pressure in summer than in winter.
(d) The volume of the gas is inversely proportional to the pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

Q.34Give a suitable explanation for the following facts about gases. (a) Gases don’t settle at the bottom of a container.v

Solution

According to kinetic theory, gas molecules are moving continuously at random. Hence, they do not settle at the bottom of a container.
(b) Gases diffuse through all the space available to them.
Gases have a tendency to occupy all the available space. The gas molecules migrate from a region of higher concentration to a region of lower concentration. Hence, gases diffuse through all the space available to them.

Answer:

According to kinetic theory, gas molecules are moving continuously at random. Hence, they do not settle at the bottom of a container.
(b) Gases diffuse through all the space available to them.
Gases have a tendency to occupy all the available space. The gas molecules migrate from a region of higher concentration to a region of lower concentration. Hence, gases diffuse through all the space available to them.

Q.35Suggest why there is no hydrogen in our atmosphere. Why does the moon have no atmosphere?v

Solution

1. Hydrogen is the lightest element thus when produced in a free state, it rises above all the other gases to the top of the atmosphere, where it is open to cosmic storms and solar flares. There is literally leaks from the atmosphere to the empty space. Hydrogen easily gains the velocity required to escape Earth’s magnetic field. Hydrogen is very reactive in nature. So it would have reacted with O 2, in its way to produce H 2 O. So majority portion of H 2 reacts and a very less amount of it present in the upper level of the atmosphere and gains velocity to escape the atmosphere.
2. Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on the surface of the moon is small. Therefore, the value of escape velocity on the surface of the moon is very small. The molecule of the atmospheric gases on the surface of the moon has thermal velocities greater than the escape velocity. That’s why all the molecules of gases have escaped and there is no atmosphere on the moon. The moon has insufficient gravity to retain an atmosphere. So we conclude that the moon has no atmosphere.

Answer:

1. Hydrogen is the lightest element thus when produced in a free state, it rises above all the other gases to the top of the atmosphere, where it is open to cosmic storms and solar flares. There is literally leaks from the atmosphere to the empty space. Hydrogen easily gains the velocity required to escape Earth’s magnetic field. Hydrogen is very reactive in nature. So it would have reacted with O 2, in its way to produce H 2 O. So majority portion of H 2 reacts and a very less amount of it present in the upper level of the atmosphere and gains velocity to escape the atmosphere.
2. Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on the surface of the moon is small. Therefore, the value of escape velocity on the surface of the moon is very small. The molecule of the atmospheric gases on the surface of the moon has thermal velocities greater than the escape velocity. That’s why all the molecules of gases have escaped and there is no atmosphere on the moon. The moon has insufficient gravity to retain an atmosphere. So we conclude that the moon has no atmosphere.

Q.36Explain whether a gas approaches ideal behavior or deviates from ideal behaviour if (a) it is compressed to a smaller volume at f constant pressure.v

Solution

When the gas is compressed to a smaller volume, the compressibility factor (Z) decreases. Hence, the gas deviates from ideal behavior
(b) the temperature is raised while keeping the volume constant.
When the temperature is increased, the compressibility factor approaches unity. Hence, the gas behaves ideally.
(c) More gas is introduced into the same volume and at the same temperature.
When more gas is introduced into a container of the same volume and at the same temperature, the compressibility factor tend to unity. Hence, the gas behaves ideally.

Answer:

When the gas is compressed to a smaller volume, the compressibility factor (Z) decreases. Hence, the gas deviates from ideal behavior
(b) the temperature is raised while keeping the volume constant.
When the temperature is increased, the compressibility factor approaches unity. Hence, the gas behaves ideally.
(c) More gas is introduced into the same volume and at the same temperature.
When more gas is introduced into a container of the same volume and at the same temperature, the compressibility factor tend to unity. Hence, the gas behaves ideally.

Q.37Which of the following gases would you expect to deviate from ideal behavior under conditions of low-temperature F 2, Cl 2, or Br 2 ? Explain.v

Solution

1. Bromine deviates (Br 2 ) from the ideal gas maximum than Cl 2 and F 2. Because Br 2 has the biggest size (atomic weight 79.9) provides maximum attraction between bromine molecules which is directly proportional to the size of the molecule and the boiling point of the liquid made from those molecules.
2. Br 2 deviates from ideal behaviour because it has the largest atomic radii compared to Cl 2 and F 2. So it contains more electrons than the other two, and the Vander Waals forces are stronger in Br 2 than in Cl 2 and F 2. So Br 2 deviates from ideal behaviour.

Answer:

1. Bromine deviates (Br 2 ) from the ideal gas maximum than Cl 2 and F 2. Because Br 2 has the biggest size (atomic weight 79.9) provides maximum attraction between bromine molecules which is directly proportional to the size of the molecule and the boiling point of the liquid made from those molecules.
2. Br 2 deviates from ideal behaviour because it has the largest atomic radii compared to Cl 2 and F 2. So it contains more electrons than the other two, and the Vander Waals forces are stronger in Br 2 than in Cl 2 and F 2. So Br 2 deviates from ideal behaviour.

Q.38Distinguish between diffusion and effusion.v

Solution

Diffusion
Effusion
1. The property of gas which involves the movement of the gas molecules through another gas is called diffusion.
It is the property in which a gas escapes from a container through a very small hole.
2. It is the ability of gases to mix with each other
It is the ability of gas to travel through a small hole.
3. The rate of diffusion of a gas depends on how fast the gas molecules are moving
The rate that this happens depends on how many gas molecules “collide” with the pore.
4. e.g., Smell of perfume diffuses into the air
e.g., Air escaping slowly through the pinhole in a tire.

Answer:

Diffusion
Effusion
1. The property of gas which involves the movement of the gas molecules through another gas is called diffusion.
It is the property in which a gas escapes from a container through a very small hole.
2. It is the ability of gases to mix with each other
It is the ability of gas to travel through a small hole.
3. The rate of diffusion of a gas depends on how fast the gas molecules are moving
The rate that this happens depends on how many gas molecules “collide” with the pore.
4. e.g., Smell of perfume diffuses into the air
e.g., Air escaping slowly through the pinhole in a tire.

Q.39Aerosol cans carry a clear warning of heating of the can. Why?v

Solution

If aerosol cans are heated, then they will produce more vapour inside the can, which will make the pressure rise very quickly. The rise in temperature can double the pressure inside. Even though the cans are tested, they will burst if the pressure goes up too far. A bursting can could be dangerous.

Answer:

If aerosol cans are heated, then they will produce more vapour inside the can, which will make the pressure rise very quickly. The rise in temperature can double the pressure inside. Even though the cans are tested, they will burst if the pressure goes up too far. A bursting can could be dangerous.

Q.40When the driver of an automobile applies the brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward the back of the car. Upon forward acceleration, the passengers are pushed toward the front of the car. Why?v

Solution

1. When the driver of an automobile applies the brake, the passengers are pushed toward the front of the car due to the inertia of the body, but a helium balloon pushed toward the back of the car. A helium balloon responds to the air around it. Helium molecules are lighter than the air of our atmosphere, and so they move toward the back by gravity as a result of the accelerating frame.
2. Upon forwarding acceleration, the passenger’s arc pushed toward the front of the car, because the body in motion tends to stay in motion until acted upon by an outside force. A helium balloon is going to move opposite to this pseudo gravitational force.

Answer:

1. When the driver of an automobile applies the brake, the passengers are pushed toward the front of the car due to the inertia of the body, but a helium balloon pushed toward the back of the car. A helium balloon responds to the air around it. Helium molecules are lighter than the air of our atmosphere, and so they move toward the back by gravity as a result of the accelerating frame.
2. Upon forwarding acceleration, the passenger’s arc pushed toward the front of the car, because the body in motion tends to stay in motion until acted upon by an outside force. A helium balloon is going to move opposite to this pseudo gravitational force.

Q.41Would it be easier to drink water with a straw on the top of Mount Everest?v

Solution

It would be harder on the top of the mountain; because the external pressure pushing on the liquid to force it up the straw is less.

Answer:

It would be harder on the top of the mountain; because the external pressure pushing on the liquid to force it up the straw is less.

Q.42Write the Van der Waals equation for a real gas. Explain the correction term for pressure and Volume.v

Solution

Vander Waals equation for a real gas is given by
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Where n is the number of moles of gas and V is the volume of the container
⇒ P ∝ \(\frac{n^{2}}{V^{2}}\)
⇒ P = \(\frac{a n^{2}}{V^{2}}\)
Where a is proportionality constant and depends on the nature of gas
Therefore, P = P + \(\frac{a n^{2}}{V^{2}}\)
Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\) π(2r) 3 = 8V m
where V m is a volume of a single molecule.
Excluded volume for single molecule = \(\frac{8 V_{m}}{2}\) = 4V m
Excluded volume for n molecule
= n(4V m ) = nb
Where b is van der Waals constant which is equal to 4V m
⇒ V’ = nb
V ideal = V – nb
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas.

Answer:

Vander Waals equation for a real gas is given by
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
Pressure Correction:
The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.
Where n is the number of moles of gas and V is the volume of the container
⇒ P ∝ \(\frac{n^{2}}{V^{2}}\)
⇒ P = \(\frac{a n^{2}}{V^{2}}\)
Where a is proportionality constant and depends on the nature of gas
Therefore, P = P + \(\frac{a n^{2}}{V^{2}}\)
Volume Correction:
As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor V to this effect. Let us calculate the correction term by considering gas molecules as spheres.
V = excluded volume
Excluded volume for two molecules = \(\frac{4}{3}\) π(2r) 3 = 8V m
where V m is a volume of a single molecule.
Excluded volume for single molecule = \(\frac{8 V_{m}}{2}\) = 4V m
Excluded volume for n molecule
= n(4V m ) = nb
Where b is van der Waals constant which is equal to 4V m
⇒ V’ = nb
V ideal = V – nb
Replacing the corrected pressure and volume in the ideal gas equation PV = nRT we get the van der Waals equation of state for real gases as below,
(P + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT
The constants a and b are van der Waals constants and their values vary with the nature of the gas.

Q.43Derive the values of van der Waals equation constants in terms of critical constants.v

Solution

The Van der walls equation for n moles is
(p + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT ……… (1)
For 1 mole
(P + \(\frac{n^{2}}{V^{2}}\)) (V – b) = RT ……………(2)
From the equation we can derive the values of critical constants P c, V c, and T c in terms of a and b, the van der Waals constants, On expanding the above equation,
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0 ………….(3)
Multiply equation (3) by \(\frac{V^{2}}{P}\)
\(\frac{V^{2}}{P}\) (PV +\(\frac{a}{V}\) – pb – \(\frac{a b}{V^{2}}\) – RT) = 0
V 3 + \(\frac{a}{P}\)V – bV 2 – \(\frac{a b}{P}\) – \(\frac{R T V^{2}}{P}\) = 0 ………..(4)
When the above equation is rearranged in powers of V
V 3 – [\(\frac{R T}{P}\) + b]V 2 + \(\frac{a}{P}\)V – \(\frac{a b}{P}\) = 0 ……………..(5)
The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V c. The pressure and temperature becomes P c and T c respectively
V = V c
V – V c = 0
(V – V c ) 3 = 0
V 3 – 3V c V 2 + 3V c 2 V – V c 3 = 0 …………..(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V 2, V and constant terms in (5) and (6).
-3V c V 2 = –[\(\frac{R T_{c}}{P_{c}}\) + b] V 2
3V c = \(\frac{R T_{c}}{P_{c}}\) + b ………………(7)
3V c 2 = \(\frac{a}{P_{c}}\) ……………(8)
V c 3 = \(\frac{ab}{P_{c}}\) …………….(9)
The critical constants can be calculated using the values of van der waals constant of a gas and vice versa.
a = 3V c 2 P c and
b = \(\frac{V_{c}}{3}\)

Answer:

The Van der walls equation for n moles is
(p + \(\frac{a n^{2}}{V^{2}}\))(V – nb) = nRT ……… (1)
For 1 mole
(P + \(\frac{n^{2}}{V^{2}}\)) (V – b) = RT ……………(2)
From the equation we can derive the values of critical constants P c, V c, and T c in terms of a and b, the van der Waals constants, On expanding the above equation,
PV + \(\frac{a}{V}\) – Pb – \(\frac{a b}{V^{2}}\) – RT = 0 ………….(3)
Multiply equation (3) by \(\frac{V^{2}}{P}\)
\(\frac{V^{2}}{P}\) (PV +\(\frac{a}{V}\) – pb – \(\frac{a b}{V^{2}}\) – RT) = 0
V 3 + \(\frac{a}{P}\)V – bV 2 – \(\frac{a b}{P}\) – \(\frac{R T V^{2}}{P}\) = 0 ………..(4)
When the above equation is rearranged in powers of V
V 3 – [\(\frac{R T}{P}\) + b]V 2 + \(\frac{a}{P}\)V – \(\frac{a b}{P}\) = 0 ……………..(5)
The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V c. The pressure and temperature becomes P c and T c respectively
V = V c
V – V c = 0
(V – V c ) 3 = 0
V 3 – 3V c V 2 + 3V c 2 V – V c 3 = 0 …………..(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V 2, V and constant terms in (5) and (6).
-3V c V 2 = –[\(\frac{R T_{c}}{P_{c}}\) + b] V 2
3V c = \(\frac{R T_{c}}{P_{c}}\) + b ………………(7)
3V c 2 = \(\frac{a}{P_{c}}\) ……………(8)
V c 3 = \(\frac{ab}{P_{c}}\) …………….(9)
The critical constants can be calculated using the values of van der waals constant of a gas and vice versa.
a = 3V c 2 P c and
b = \(\frac{V_{c}}{3}\)

Q.44Why do astronauts have to wear protective suits when they are on the surface of the moon?v

Solution

In space, there is no pressure, if we do wear a pressurized suit, our body will die. In space, we have to wear a pressurized suit, otherwise, our body will continue to push out and blow up like a balloon. It would look cool, but we will be dead. So the astronauts in space must wear a pressurized suit (protective suits).

Answer:

In space, there is no pressure, if we do wear a pressurized suit, our body will die. In space, we have to wear a pressurized suit, otherwise, our body will continue to push out and blow up like a balloon. It would look cool, but we will be dead. So the astronauts in space must wear a pressurized suit (protective suits).

Q.45When ammonia combines with HCl, NH 4 Cl is formed as white dense fumes. Why do more fumes appear near HCl?v

Solution

HCl and NH 4 Cl molecules diffuse through the air towards each other. When they meet, they reactto
form a white powder called ammonium chloride, NH 4 Cl.
HCl (g) + NH 3(g) ⇌ NHC1
Hydrogen chloride + ammonia ⇌ ammonium chloride.
The ring of white powder is closer to the HCl than
the NH 3. This is because the NH 3 molecules are lighter (smaller) and have diffùsed more quickly through the air in the tube.

Answer:

HCl and NH 4 Cl molecules diffuse through the air towards each other. When they meet, they reactto
form a white powder called ammonium chloride, NH 4 Cl.
HCl (g) + NH 3(g) ⇌ NHC1
Hydrogen chloride + ammonia ⇌ ammonium chloride.
The ring of white powder is closer to the HCl than
the NH 3. This is because the NH 3 molecules are lighter (smaller) and have diffùsed more quickly through the air in the tube.

Q.46A sample of gas at 15°C at 1 atm. has a volume of 2.58 dm 3. When the temperature is raised to 38°C at 1 atm does the volume of the gas Increase? If so, calculate the final volume.v

Solution

T 1 = 15°C + 273;
T 2 = 38 + 273
T 1 = 228;
T 2 = 311K
V 1 = 2.58dm 3;
V 2 = ?
(P = 1 atom constant)
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V 2 = \(\left[\frac{V_{1}}{T_{1}}\right]\) × T 2
= \(\frac{2.58 d m^{3}}{288 K}\) × 311K
V 2 = 2.78 dm 3 i.e, volume increased from 2.58 dm 3 to 2.78 dm 3

Answer:

T 1 = 15°C + 273;
T 2 = 38 + 273
T 1 = 228;
T 2 = 311K
V 1 = 2.58dm 3;
V 2 = ?
(P = 1 atom constant)
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V 2 = \(\left[\frac{V_{1}}{T_{1}}\right]\) × T 2
= \(\frac{2.58 d m^{3}}{288 K}\) × 311K
V 2 = 2.78 dm 3 i.e, volume increased from 2.58 dm 3 to 2.78 dm 3

Q.47Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen In a vessel of volume of 37.6 dm 3 at 298K, and sample B is in a vessel of volume 16.5 dm 3 at 298K. Calculate the number of moles in sample B.v

Solution

n A = 1.5mol; n B = ?
V A = 37.6 dm 3; V B = 16.5 dm 3
(T = 298K constant)
\(\frac{V_{A}}{n_{A}}=\frac{V_{B}}{n_{B}}\)
n A = \(\left[\frac{n_{A}}{n_{B}}\right] V_{B}\)

Answer:

n A = 1.5mol; n B = ?
V A = 37.6 dm 3; V B = 16.5 dm 3
(T = 298K constant)
\(\frac{V_{A}}{n_{A}}=\frac{V_{B}}{n_{B}}\)
n A = \(\left[\frac{n_{A}}{n_{B}}\right] V_{B}\)

Q.48Sulphur hexafluoride Is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas In a steel vessel of volume 5.43 dm3 at 69.5°C, assumIng Ideal gas behaviour.v

Solution

n = 1.82 mole
V = 5.43 dm 3
T = 69.5 + 273 = 342.5
P =?
PV = nRT
P = nRT/V
P =
P = 94.25 atm

Answer:

n = 1.82 mole
V = 5.43 dm 3
T = 69.5 + 273 = 342.5
P =?
PV = nRT
P = nRT/V
P =
P = 94.25 atm

Q.49Argon is an Inert gas used In light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C Is heated to 85°C at constant volume. Calculate its final pressure in atm.v

Solution

P 1 = 1.2 atm
T 1 = 180°C + 273 = 291 K
T 2 = 850°C + 273 = 358 K
P 2 =?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
P 2 = \(\left[\frac{P_{1}}{T_{1}}\right] \times T_{2}\)
P 2 = \(\frac{1.2 a t m}{291 K}\) × 358 K
P 2 = 1.48 atm

Answer:

P 1 = 1.2 atm
T 1 = 180°C + 273 = 291 K
T 2 = 850°C + 273 = 358 K
P 2 =?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
P 2 = \(\left[\frac{P_{1}}{T_{1}}\right] \times T_{2}\)
P 2 = \(\frac{1.2 a t m}{291 K}\) × 358 K
P 2 = 1.48 atm

Q.50A small bubble rises from the bottom of a lake where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure Is I arm. Calculate the final volume in (mL) of the bubble, If its initial volume 1.5 mL.v

Solution

T 1 = 6°C + 273 = 279 K
P 1 = 4 atm; V 2 = 1.5 ml
T 2 = 25°C + 273 = 298 K
P 2 = 1 atm; V 2 =?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
= \(\frac{4 a t m \times 1.5 m l \times 298 K}{279 K \times 1 a t m}\) = 6.41 mol

Answer:

T 1 = 6°C + 273 = 279 K
P 1 = 4 atm; V 2 = 1.5 ml
T 2 = 25°C + 273 = 298 K
P 2 = 1 atm; V 2 =?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
= \(\frac{4 a t m \times 1.5 m l \times 298 K}{279 K \times 1 a t m}\) = 6.41 mol

Q.51Hydrochloric acid Is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 × 10 -3 dm 3 of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?v

Solution

V = 154.4 × 10 -3 dm 3
P = 742 mm of Hg
T = 298K;
m =?
m = \(\frac{P V}{R T}\)
= \(\frac{742 m m H g \times 154.4 \times 10^{-3} L}{62 m m H g L K^{-1} m o l^{-1} \times 298 K}\)
n = \(\frac{\text { Mass }}{\text { Molar Mass }}\)
Mass = n × Molar Mass
= 0.0006 × 2.016
= 0.0121 g = 12.1 mg

Answer:

V = 154.4 × 10 -3 dm 3
P = 742 mm of Hg
T = 298K;
m =?
m = \(\frac{P V}{R T}\)
= \(\frac{742 m m H g \times 154.4 \times 10^{-3} L}{62 m m H g L K^{-1} m o l^{-1} \times 298 K}\)
n = \(\frac{\text { Mass }}{\text { Molar Mass }}\)
Mass = n × Molar Mass
= 0.0006 × 2.016
= 0.0121 g = 12.1 mg

Q.52It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What Is the molar mass of the unknown gas?v

Solution

A gas’s partial pressure formula is the gas pressure exerted if that gas were alone.

Answer:

A gas’s partial pressure formula is the gas pressure exerted if that gas were alone.

Q.54Combustible gas is stored in a metal tank at a pressure of 2.98 atm at 25°C. The tank can withstand a maximum pressure of 12 atm after which It will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal 1100 K).v

Solution

Pressure of the gas in the tank at its melting point
T = 298 K;
P 1 = 2.98 atom;
T 2 = 1100 K;
P 2 =?
\(\frac{P_{1} P_{2}}{T_{1} T_{2}}\) =????
⇒ P 2 = \(\frac{P}{T_{1}}\) × T 2
= \(\frac{2.98 \text { atm }}{298 K}\) × 1100 K = 11 atm
At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

Answer:

Pressure of the gas in the tank at its melting point
T = 298 K;
P 1 = 2.98 atom;
T 2 = 1100 K;
P 2 =?
\(\frac{P_{1} P_{2}}{T_{1} T_{2}}\) =????
⇒ P 2 = \(\frac{P}{T_{1}}\) × T 2
= \(\frac{2.98 \text { atm }}{298 K}\) × 1100 K = 11 atm
At 1100 K the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.