States of matter 1 gases 1st- years class 11 notes fsc for kpk short question, long question, and excessive.

**Contents**show

## Short Questions Chemistry Class 11

**Q.2 i) Justify that 1 cm**^{3}** of H**_{2}** and 1 cm**^{3}** of CH**_{4}** at STP will have the same number of molecules although one molecule of CH**_{4}** is 8 times heavier than that of H**_{2}**.**

^{3}

_{2}

^{3}

_{4}

_{4}

_{2}

**Answer:**

One cm^{3} of H_{2} and one cm^{3} of CH_{4} have the same number of molecules irrespective of the fact that one molecule of CH_{4} is 8 times heavier than that of H_{2}. The reason is that the masses do not disturb the volume occupied because the molecules of the gases are widely separated from each other at STP. One molecule is approximately at a distance of 300 times its diameter from its immediate neighbour at STP.

**Q.2 ii) Do you think that some of the postulates of Kinetic Molecular Theory of gases are faulty? Write down these postulates.**

**Answer:**

Yes, some of the postulates of Kinetic Molecular Theory of gases are faulty and real gases deviate from ideal behaviour due to these faulty assumptions (postulates). These postulates are given below:

- The size of the gas molecules is very small as compared to the distance between them. Therefore, the actual volume of the gas molecules is negligible as compared to the total volume of the gas.
- The attractive forces among the gas molecules are negligible. Therefore, every gas molecule behaves independently.

**Q.2 iii) High pressure and low temperature make the gases non-ideal.**

**Answer:**

At high pressure, the molecules are very much close to each other and they occupy an appreciable part of the entire volume. This means that the volume of gas molecules does not remain negligible and they deviate from ideal behaviour. The low temperature deprives the molecules of kinetic energies and attractive forces start dominating among the gas molecules. This also makes the gases behave non-ideally. Hence, high pressure and low temperature make the gases non-ideal.

Read more: Stoichiometry Chemistry Class 11 Notes Chapter 1 for kpk

**Q.2 iv) Rapid expansion of gases causes cooling.**

**Answer:**

In a compressed gas, molecules are very close to each other and have attractive forces. When gas is expanded suddenly, molecules move away from each other. This process requires energy, which is obtained from the gas itself, hence it is cooled.

**Q.2 v) Lighter gases can diffuse more rapidly than heavier ones.**

**Answer:**

The rate of diffusion of gases depends on the molecular mass or density. Lower the density or mass, higher is the rate of diffusion and vice versa. It can also be explained by Graham’s law of diffusion which states that the rate of diffusion of the gases is inversely proportional to the square roots of their densities.

So, the lighter gases diffuse rapidly and dense gases diffuse slowly.

Read more: Physics Class 11 Notes Thermodynamics Chapter 10 for kpk

## Chemistry Notes for Class 11 Long Questions for kpk 2021

**Q.1 a) State Boyle’s law with its mathematical expression.**

**Answer:****Boyle’s law**

*“At constant pressure, volume of a fixed mass of a gas is inversely proportional to the pressure applied on it”.*

**Mathematical expression**Mathematically, Boyle’s law can be written as,

At constant temperature,

V ∝ 1 / P

PV = k

Where “k” is the proportionality constant.

Boyle’s law can also be expressed as ‘the product of pressure and volume for a fixed mass of gas remains constant at constant temperature’.

For initial state,

P

_{1}V

_{1}= K

For final state,

P

_{2}V

_{2}= K

Comparing both the equations,

P

_{1}V

_{1 }= P

_{2}V

_{2}

This equation is known as Boyle’s law equation.

**Q.1 b) A mass of gas is under a pressure of 760 torr and occupies volume of 525 cm**^{3}. If the pressure is doubled, what volume would the gas now occupy? Assume the temperature is constant.

^{3}. If the pressure is doubled, what volume would the gas now occupy? Assume the temperature is constant.

**Answer:**

P_{1} = 760 torr

V_{1} = 525 cm^{3}

P_{2} = 2P_{1} = 2(760) = 1520 torr

V_{2} = ?

According to the Boyle’s law equation,

P_{1}V_{1} = P_{2}V_{2}

Rearranging equation for V_{2},

V_{2} = P_{1}V_{1} / P_{2}

By putting the values,

V_{2} = (760 x 525) / 1520 = **262.5 cm**^{3}

**Q.2 a) State Charles’ law with its mathematical expression.**

**Answer:****Charles’ law**

*“The volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant pressure”.*

**Mathematical expression**

Mathematically, it can be written as,

At constant pressure,

V ∝ T

V = kT

V / T = k

Where “k” is the proportionality constant.

For initial state,

V_{1} / T_{1} = k

For final state,

V_{2} / T_{2} = k

Comparing both the equations,

V_{1} / T_{1} = V_{2} / T_{2}

This equation is known as Charles’s law equation.

**Q.2 b) Assume that one dm**^{3} of air near a thermonuclear explosion is heated from 0°C to 546000°C. To what volume does the air expand?

^{3}of air near a thermonuclear explosion is heated from 0°C to 546000°C. To what volume does the air expand?

**Answer:**

Initial volume of air = V_{1} = 1 dm^{3}

Initial temperature of air = T_{1} = 0°C = 0 + 273 = 273 K

Final temperature of air = T_{2} = 546000°C = 546000 + 273 = 546273 K

Final volume of air = V_{2} = ?

According to Charles’s law equation,

V_{1} / T_{1} = V_{2} / T_{2}

Rearranging for V_{2},

V_{2 }= V_{1} / T_{1 }x T_{2}

By putting the values,

V_{2 }= (1 / 273) x 546273

= **2001.0 dm ^{3}**

**Q.3 a) What is diffusion and effusion?**

**Answer:****Diffusion**

*“The spontaneous mixing of different non-reacting gases is called diffusion.”*

In this process, gas molecules spread by random motion and collisions to form a homogeneous mixture.*Example:*Spreading of the fragrance of rose is because of diffusion.

**Effusion**

*“The controlled mixing or passage of gas molecules through a small orifice (pinhole) is called effusion.”*

In this process, gas molecules spread one by one without collisions to form a homogeneous mixture.*Example:*

During puncturing of a vehicle tyre, gas molecules spread by escaping through a small hole. This phenomenon is effusion.

**Q.3 b) State Graham’s law of diffusion. Also prove its mathematical expression by an experiment.**

**Answer:**

Graham’s law of Diffusion

Thomas Graham proposed his law of diffusion of gases in 1829. It states that:

*“The rates of diffusion of two gases are inversely proportional to the square roots of their densities or molecular masses at the same temperature and pressure.”*

Mathematically, it is expressed as:

Where r_{1} and r_{2} are the rates of diffusion of two gases, d_{1} and d_{2} are the densities and M_{1} and M_{2} are the molecular masses of two gases.*Experimental Verification of Graham’s Law of Diffusion:*

Mathematical expression of Graham’s law of diffusion can be verified by a simple experimental setup. Take 100 cm long tube and plug a cotton swab soaked in HCl at one end and another soaked in NH_{3} at the other end of the tube simultaneously. The two gases are found to escape from their solutions and meet a distance of roughly 60 cm from NH_{3} plug and 40 cm from HCl plug where they react to form white smoke ring of NH_{4}Cl.

NH_{3} + HCl → NH_{4}Cl

It means that the rate of diffusion of NH_{3} is 1.5 times faster than the rate of diffusion of HCl.

Similarly,

Since,

Molecular mass of HCl = 36.5

Molecular mass of NH_{3} = 17

These results verify the mathematical expression of Graham’s law of diffusion.

**Q.3 c) Hydrogen gas diffuses through a porous plate at a rate of 500 cm per minute at 0°C. What is the rate of diffusion of oxygen gas through the same porous plate at 0°C?**

**Answer:**

Rate of diffusion of H_{2} = r_{H2} = 500 cm/min

Rate of diffusion of O_{2} = r_{O2} = ?

Molecular weight of H_{2} = 2.016 amu

Molecular weight of O_{2} = 32 amu

According to Graham’s law of diffusion,

**Q.3 d) **The rate of effusion of an unknown gas through a pinhole is found to be 0.279 times the rate of effusion of H_{2} through the same pinhole. Calculate the molecular mass of the unknown gas at STP.

**Answer:**

Let the unknown gas is X. As the rate of effusion of unknown gas is 0.279 times the effusion of H_{2}, we can write,

r_{x} / r_{H2} = 0.279

Molecular mass of H_{2} = MH_{2} = 2.016 amu

Molecular mass of X = M_{x} = ?

According to Graham’s law,

**Q.4 a) State Dalton’s law of partial pressure.**

**Answer:****Dalton’s Law of Partial Pressures**

*“The total pressure of a mixture of gases is the sum of partial pressures of all the gases present in it”.*

Mathematically, it is written as:

P_{t} = P_{1} + P_{2} + P_{3}

Where Pt is the total pressure of mixture of gases and P_{1}, P_{2} and P_{3} are the individual partial pressures of component gases.

**Q.5 a) State basic postulates of Kinetic Molecular Theory of gases.**

**Answer:****Postulates of Kinetic Molecular Theory**

Basic postulates of Kinetic Molecular Theory of gases are following;

- Gases consist of small particles called molecules.
- Gas molecules have greater kinetic energy than those of liquids and solids.
- Molecules are far away from one another and have large distance at ordinary conditions. That is why the volume of gas consists of mostly empty spaces.
- They move randomly. They collide with each other and with the walls of the container. Their collisions are perfectly elastic. The pressure of a gas is as a result of the collision of molecules with the walls of the container.
- The size of the gas molecule is very small as compared to the distance between them. Therefore, the actual volume of the gas molecules is negligible as compared to the total volume of the gas.
- The attractive forces among the gas molecules are negligible. Therefore, every gas molecule behaves independently.
- The average kinetic energy of the gas molecules is due to their motion. It varies directly with the absolute temperature.
- At the same temperature, molecules of all gases have the same average kinetic energy.
- The forces of gravity have almost no effect on gas molecules.

**Q.6 a) Derive an Ideal gas expression.**

**Answer:****Derivation of Ideal gas Expression**Ideal gas expression (equation) is derived by combining the Boyle’s law, Charles’ law and Avogadro’s law. According to Boyle’s law, volume of the fixed mass of a gas is inversely proportional to the applied pressure at constant temperature. That is,

V ∝ 1 / P (1)

According to Charles’ law, volume of the fixed mass of a gas is directly proportional to the absolute temperature at constant pressure. That is,

V ∝ T (2)

According to the Avogadro’s law, volume of a gas is directly proportional to its number of moles at constant temperature and pressure.

V ∝ n (3)

By combining the three equations (1), (2) and (3),

V ∝ 1 / P x T x n

V = R(T/P)n

By cross multiplication and re-arrangement, the equation becomes,

PV = nRT (4)

Equation (4) is called Ideal gas equation. Since standard moles for any gas is taken one i.e. n=1, then equation (4) becomes,

PV=RT

or,

PV / T = R (5)

For initial state, equation (5) can be written as,

P

_{1}V

_{1}/ T

_{1}= R (6)

For final state, equation (5) can be written as,

P

_{2}V

_{2}/ T

_{2}= R (7)

Comparing equation (6) and equation (7), we get:

P

_{1}V

_{1}/ T

_{1}= P

_{2}V

_{2}/ T

_{2}

Equation (8) is more generalized expression of Ideal as equation.

**Q.6 b) It was desired to obtain a volume of 1000 cm**^{3} of oxygen at 100º C and 640 mm of Hg. How many moles of oxygen would be required?

^{3}of oxygen at 100º C and 640 mm of Hg. How many moles of oxygen would be required?

**Answer:**

Volume of oxygen = V = 1000 cm^{3} = 1000/1000 dm^{3} = 1 dm^{3}

Temperature = T = 100^{o}C = 100 + 273 = 373 K

Pressure = P = 640 mmHg

R = 62.4 dm^{3} mmHg mol^{-1} K^{-1}

Moles of oxygen = n =?

According to general gas equation,

PV = nRT

Rearranging for number of moles (n),

n = PV / RT

By putting the values in above equation,

Hence, 0.0275 moles of oxygen gas are required to obtain the desired conditions.

**Q.6 c) A sample of Krypton with a volume of 6.25 dm**^{3} and a pressure of 765 torr and a temperature of 20°C is expanded to a volume of 9.55 dm^{3} and a pressure of 375 torr. What will be its final temperature?

^{3}and a pressure of 765 torr and a temperature of 20°C is expanded to a volume of 9.55 dm

^{3}and a pressure of 375 torr. What will be its final temperature?

**Answer:**

Initial volume of Krypton = V_{1} = 6.25 dm^{3}

Initial pressure of Krypton = P_{1} = 765 torr

Initial temperature of Krypton = 20^{o}C = 20 + 273 = 293 K

Final volume of Krypton = V_{2} = 9.55 dm^{3}

Final pressure of Krypton = P_{2} = 375 torr

Final temperature of Krypton = T_{2} =?

According to generalized form of Ideal gas equation,

P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2}

Rearranging for T_{2},

**Q.6 d) Calculate the density of CH**_{4} at 0°C and 1 atmosphere.

_{4}at 0°C and 1 atmosphere.

**Answer:**

Temperature of CH_{4} = T =0^{o}C = 0 + 273 = 273 K

Pressure of CH_{4} = P = 1 atm

Molar mass of CH_{4} = M = 16 g mol^{-1}

R = 0.0821 atm dm^{3} mol^{-1} K^{-1}

As we know that,

d = PM / RT

By putting the values,

**Q.7 a) What are Ideal and non-ideal gases? Why do real gases deviate from ideal behaviour? Explain with graph.**

**Answer:****Ideal and Non-Ideal Gases**

An ideal gas is defined as one which exactly follows the ideal gas equation (PV=nRT) at all conditions of temperature and pressure.

Non-ideal gas is one which does not follow ideal gas equation, it deviates from gas laws to an extent that depends upon the pressure, temperature and nature of the gas.**Deviation from the Ideal Behavior**

Under usual room conditions the volumes calculated from Boyle’s and Charles’ laws correspond fairly closely to those obtained experimentally. However, at high pressures or very low temperatures deviations become appreciable. The reasons of these deviations are explained as follows:**1.** Attractive forces which result from attraction between electrons of one atom and nuclei of adjacent atoms, exist between all molecules, but are effective only at very short range. They are called van der Waal’s forces. As the pressure on a gas is increased, molecules are forced closer together and van der Waal’s forces become stronger, drawing the molecules still closer and resulting in a greater volume decrease than Boyle’s law predicts.

This effect is even more pronounced at low temperature because the more slowly the molecules are moving the more effective the attractive force becomes.

This explanation suggests that the assumption of kinetic theory that the gas molecules do not have attractive forces between them is one reason of deviation of gases from ideal behavior. Molecules of real gases do exert force on each other, the condensation of every gas on cooling shows that attractive forces are always there among the molecules. These forces are not very important when the molecules are far apart (i.e. at low pressure) but they become noticeable at high pressures.**2. **Still another factor enters at high pressure to produce an error in the gas laws. Increase in pressure does not make the molecules any smaller but only brings them closer together. At very high pressure the molecules are so close together and there are so many in a given volume that that themselves occupy an appreciable part of the entire volume. Thus the space between them (the only part of the volume that can really decrease) is not as great as the total volume. The decrease in volume which results from pressure increase is therefore, less than it should be according to Boyle’s law.

This increases pressure volume product increasing PV/nRT ratio and a positive deviation from ideality is observed. By this phenomenon we observe another fault assumption in the kinetic theory that the volume occupied by the molecules is practically negligible.

From above discussion, it is concluded that real gases deviate from ideal behavior due to two faulty assumptions present in kinetic theory, these are following:

- No forces of attraction exist between molecules.
- Molecules have negligible or practically no volume.

*Explanation with graph*

For a fixed amount of an ideal gas at constant temperature and pressure the right side of ideal gas equation assumes a constant value. As PV = nRT i.e. product of pressure and volume is a constant quantity. For an ideal gas the value of factor PV/nRT is unity, called compressibility factor and is denoted by Z.

*Z = PV / nRT = 1*

If we plot PV/nRT against pressure for ideal gas, we get a straight line, parallel to x-axis for all values of pressure, as shown in the figure below.

Figure shows that PV for hydrogen increases continuously with increase of pressure. In case of nitrogen. methane and carbon dioxide however PV first decreases, passes through a minimum where it remains constant for a while and the rises.

**Q.7 b) Derive a non-ideal gas expression. (The van der waal’s equation).**

**Answer:****Derivation of Non-Ideal Gas Expression(Van der Waals’ equation)**

An ideal gas would obey the gas laws strictly but all real gases more or less disobey these laws. The nature and extent of deviation depends upon the conditions. For example, at high temperature and very low pressure, gases obey the gas laws almost perfectly whereas they do not do so at high pressure and low temperature.

In 1873, a Dutch scientist, J.D. Van der Waals put forward an equation for real gases (non-ideal), which is called van der Waals’ equation.

*1. Volume correction due to the finite size of the molecules*When a gas is compressed, the molecules become so close together that further increase in pressure will be opposed by the molecules themselves. This is only possible when the molecules of the gas have finite volume. The volume ‘V’ in the ideal gas equation, PV = RT, is the free volume V in which the molecules are effectively free to move about. But when the molecules do occupy an appreciable part of the total volume, then Vfree should be set equal to the difference between V and Vmolecules.

**V _{free} =V – V_{molecules}**

For an ideal gas, V_{molecules} = 0 and Vander Waals, therefore, proposed that V in the gas equation be changed to (V-b) where b is the effective volume of the gas molecules but has been theoretically shown to be roughly equal to 4 times their volume.

*2. Pressure correction due to the mutual attraction of the molecules*

The attractive forces between the molecules come in to play when molecules are brought closer together by compressing the gas. Consider a molecule A in the interior of the gas, which is completely surrounded by other gas molecules. The resultant attractive force experienced by the molecule A due to all the other molecules is zero.

However, as this molecule approaches the wall of the container it is subjected to an inward pull due to unbalanced molecular attraction. Thus when the molecule is about to strike the wall and contributes its share to the total pressure of the gas, the other molecules in the gas exert an attractive force tending to prevent it from doing so. The observed pressure P, consequently will be less than the ideal pressure Pi by P’.

P = Pi – P’

Thus the true kinetic pressure is

Pi = P + P’

Van der Waal’s argued that the part of the pressure used up against inter-molecular attraction should decrease as the volume increases. He suggested the following expression to account for the molecular attraction.

P’ = a / V^{2}

Where ‘a’ is co-efficient of attraction i.e. attraction per unit volume and is constant for a particular real gas. Thus the effective kinetic pressure is given by

Pi = P + a / V^{2}

Now making correction for both the pressure and volume; the ideal gas equation for one mole of the real gas becomes

(P + a / V^{2})(V – b) = RT

(P + a / V^{2}) = Kinetic pressure of a real gas

(V – b) = Free volume of a real gas

This equation was first put forward by van der Waal (1879) and is called van der Waal’s equation. It represents the behavior of real gases (non-ideal) over wide ranges of temperature and pressure more accurately than the ideal gas equation.

The van der Waals’ equation for n moles of a gas may be written as below:

**Q.8) **Write detailed notes on the following:

(a) Absolute zero on the basis of Charles’s law.

(b) Liquefaction of gases.

(c) Plasma, the fourth state of matter.

**Answer:****(a). Absolute zero on the basis of Charle’s law**

At absolute zero, the particles of the matter are at the lowest energy point. Charles derives a relation and proved that if fix volume of gas is heated at constant pressure, it results in the increase in the volume of gas. This increase in the volume of gas is directly proportional to the increase in temperature.

Charles law states that an increase in the volume of a given gas is directly proportional to the increase in temperature. A gradual increase in temperature elevates the energy status of atoms in the matter. At absolute zero particles have a minimum energy level. Increase in temperature increases the kinetic energy of atoms. At fixed pressure, a linear relationship between temperature and volume of a gas is observed.

(b) Liquefaction of gases

The conversion of a gas to its liquid state by the combined efforts of lowered temperature and increased pressure is called liquefaction of gas.

The increased pressure is obtained by a compressor. Thus, the gas molecules come close together. Since the expansion of gases causes cooling. Therefore the lowered temperature is obtained by releasing the compressed gas to expand.

It is impossible to liquify a gas by pressure alone if the required temperature is not obtained. Above this required temperature, attractive forces are not strong enough to condense a gas in to liquid.

“The temperature above which a gas cannot be liquefied by pressure alone is called Critical temperature”. It is written as Tc while the pressure at critical temperature is called critical pressure “Pc”.

The volume occupied by one mole of a gas at Tc and Pc is called Critical volume and written as Vc.

Example | Gas | Tc in ^{o}C | Pc in atm |

1 | O_{2} | -118.8 | 49.7 |

2 | N_{2} | -147.1 | 33.5 |

3 | H_{2} | -239.9 | 12.8 |

**Methods for Liquifaction of Gases:**

Various methods are used for the liquefaction of gases. These methods are generally based upon joul Thomson effect.** Joul Thomson Effect**When compressed gas is allowed to expand suddenly, it produces cooling. This is called Joul Thomson effect.

In compressed gas, molecules are very close to each other and have attractive forces. When gas is expanded suddenly molecules move away from each other. This process requires energy, which is obtained from the gas itself, hence it is cooled.

*Linde’s Method*It is based upon Joul Thomson effect. Lind liquefied air by this process. The compressed air at about 200 atm is passed through a water-cooled pipe where the heat of compression is removed.

This compressed air is then passed through a spiral tube having a jet at the end. When the gas comes out of jet into low-pressure area (1 atm) it is cooled down due to Joul Thomson effect. The cooled air moves up, cools the incoming gas of the jet, and then again enters into the compression pump, to be compressed again. By repeating compression and expansion process, again and again, air is liquefied. The apparatus is shown in the figure below:**(c) ****Plasma**

The fourth state of matter which is a mixture of neutral particles, positive ions and negative electrons is called Plasma.*Origin of Plasma State:*

This term was originally applied by Irving Langmuire in 1928 to an ionized gas. Since an ionized gas cannot exist at room temperature, that is why, it was observed for the first time in an electrical discharge at high temperature.*Formation of Plasma:*

On heating a solid, it is converted into liquid. On further heating, the liquid is converted into vapours. Thus the phase of matter changes from solid to liquid and then liquid to vapours. Now if vapours are further heated, some of them lose electrons and positive ions are formed. Hence, a mixture of neutral particles, positive ions and negative electrons, is formed. This is called Plasma. The ionization is produced by high temperature. It means this state of matter consists of an ionized substance but neutral in nature.

The free electric charges make the plasma electrically conductive so that it responds strongly to electromagnetic fields.*Occurrence*** of plasma:**

The plasma of matter is found in the region around sun and stars. Since sun and stars have far more matter, therefore, it is said that more matter (about 99%) is made up of plasma. The sun is a 1.5 million ball of plasma. It is heated by nuclear fission.

On earth plasma is limited to lightening bolts, flames, auroras and fluorescent lights, neon signs etc. When an electric current is passed through neon gas, it produces both plasma and light.

It has been found that plasma has a complex set of interactions. It is unique, fascinating and complex state of matter. Although it contains positive ions and negative electrons but their number is equal and hence plasma is macroscopically neutral.*Applications:*

Since plasma can respond to both electric and magnetic fields it can have many uses.

- A fluorescent bulb is different from regular bulbs. It consists of a long tube filled with gas say neon gas. When electricity is passed through the gas, it charges the gas. The charging and exciting of gas create glowing plasma inside the tube. The colour of the plasma depends upon the gas used.
- They are used for the plasma processing of semiconductors, sterilization of some medical products, lamps, lasers, diamond coated films and high power microwaves sources and pulsed power switches.
- It helps in working of computers and electronic equipment.