Kinetic Theory of Gases Obj-1 HC Verma Solutions Vol-2 Chapter-24

Kinetic Theory of Gases Obj-1 HC Verma Solutions Vol-2 Chapter-24 Concept of Physics for Class-12. Step by Step Solutions of Objective -1 (MCQ-1) Questions of Chapter-24 Kinetic Theory of Gases (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Kinetic Theory of Gases Obj-1 HC Verma Solutions Vol-2 Chapter-24 Concept of Physics

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Kinetic Theory of Gases Objective-1 (MCQ-1) Questions

HC Verma Solutions of Ch-24  Vol-2 Concept of Physics for Class-12

(Page-33)

Question-1 :-

Answer-1 :-

Temperature is defined as the average kinetic energy of the molecules in a gas sample. Average is same for all the molecules of the sample. So, kinetic energy is the same for all.
Hence, correct answer is d.

Question-2 :-  (Kinetic Theory of Gases Obj-1 HC Verma Solutions)

(a) low pressure and low temperature

(b) low pressure and high temperature

(c) high pressure and low temperature

(d) high pressure and high temperature.

Answer-2 :-

At low pressure, the concentration of gas molecules is very low. Hence, the kinetic assumption that the size of the molecules can be neglected compared to the volume of the container applies.
At high temperature, molecules move very fast. So, they tend to collide elastically and ​forces of interaction between the molecules minimise. This is the required idea condition.

Question-3 :-

The pressure of an ideal gas is written as P=2E/3V . Here E refers to

(a) translational kinetic energy

(b) rotational kinetic energy

(c) vibrational kinetic energy

(d) total kinetic energy.

Answer-3 :-

According to the kinetic theory, molecules show straight line in motion (translational).  So, the kinetic energy is essentially transitional.

Question-4 :-

The energy of a given sample of an ideal gas depends only on its

(a) volume

(b) pressure

(c) density

(d) temperature

Answer-4 :-

The option (d) temperature is correct

Explanation:

Temperature of a gas is directly proportional to its kinetic energy. Thus, energy of an ideal gas depends only on its temperature.

Question-5 :-

Which of the following gases has maximum hms speed at a given temperature?

(a) hydrogen

(b) nitrogen

(c) oxygen

(d) carbon dioxide.

Answer-5 :-

The option (a) hydrogen is correct

Explanation:

The has speed of a gas is given by √​3RT/Mo . Since hydrogen has the lowest M_o compared to other molecules, it will have the highest rms speed.

Question-6 :-

Figure 24-Q1 shows graphs of pressure vs density for an ideal gas at two temperatures T₁ and T₂.

Figure 24-Q1

(a) T₁ > T₂

(b) T₁ = T₂

(c) T₁ < T₂

(d) Any of the three is possible.

Answer-6 :-

The option (a) T₁ > T₂ is correct

Explanation:

_{The straight line T1 has greater slope than T​2. This means P/ρ ratio is greater for T1 than T2. Now, rms velocity of a gas is given by √3P/ρ . This means rms velocity of gas with T1 molecules is greater than T2 molecules. Again, gas with higher temperature has higher rms} velocity.
_{So, T1 > T2.}

Question-7 :-

The mean square speed of the molecules of a gas at absolute temperature T is proportional to

(a) 1/T

(b) √T

(c) T

(d) T²

Answer-7 :-

The option (c) T is correct

Explanation:

Root mean squared velocity is given by

Question-8 :-

Suppose a container is evacuated to leave just one molecule of a gas in it. Let v_a and v_rms represent the average speed and the rms speed of the gas.

(a) v_a > v_rms

(b) v_a < v_rms

(c) v_a = v_rms

(d) v_rms is undefined.

Answer-8 :-

The option (c) v_a = v_rms is correct

Explanation:

Speed is constant and same for a single molecule. Thus, rms speed will be equal to its average speed.

Question-9 :-

The rms speed of oxygen at room temperature is about 500 m/s. The rms speed of hydrogen at the same temperature is about

(a) 125 m s⁻¹

(b) 2000 m s⁻¹

(c) 8000 m s⁻¹

(d) 31 m s⁻¹.

Answer-9 :-

The option (b) 2000 m s⁻¹ is correct

Explanation:

Given,
Molecular mass of hydrogen, M_H = 2
Molecular mass of oxygen, M_o = 32
RMS speed is given by,

Question-10 :-  (Kinetic Theory of Gases Obj-1 HC Verma Solutions)

The pressure of a gas kept in an isothermal container is 200 kPa. If half the gas is removed from it, the pressure will be

(a) 100 kPa

(b) 200 kPa

(c) 400 kPa

(d) 800 kPa

Answer-10 :-

The option (a) 100 kPa is correct

Explanation:

Let the number of moles in the gas be n.
Applying equation of state, we get

When  half  of  the  gas  is  removed,   number  of  moles  left  behind =n/2

Let  the  pressure  be  P’ .

Question-11 :-

The rms speed of oxygen molecules in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become

(a) v

(b) v√2

(c) 2v

(d) 4v

Answer-11 :-

The option (c) 2v is correct

Explanation:

Question-12 :-

The quantity pV/kT represents

(a) mass of the gas

(b) kinetic energy of the gas

(c) number of moles of the gas

(d) number of molecules in the gas.

Answer-12 :-

The option (d) number of molecules in the gas. is correct

Explanation:

Here ,

PV=nRT…(1)

Also ,

Question-13 :-

The process on an ideal gas, shown in figure, is

(a) isothermal

(b) isobaric

(c) isochoric

(d) none of these

Answer-13 :-

The option (c) isochoric is correct

Explanation:

According to the graph, P is directly proportional to T.
Applying the equation of state, we get
PV = nRT

So,   V  is  also  a  constant .

Constant V implies the process is isochoric.

Question-14 :-

There is some liquid in a closed bottle. The amount of liquid is continuously decreasing. The vapour in the remaining part

(a) must be saturated

(b) must be unsaturated

(c) may be saturated

(d) there will be no vapour.

Answer-14 :-

The option (b) must be unsaturated is correct

Explanation:

As the liquid is decreasing, the liquid is vaporised. We know that vaporisation cannot occur in saturated air and there cannot be any liquid with no vapour at all. So, the vapour in the remaining part is unsaturated.

Question-15 :-

There is some liquid in a closed bottle. The amount of liquid remains constant as time passes. The vapour in the remaining part

(a) must be saturated

(b) must be unsaturated

(c) may be unsaturated

(d) there will be no vapour.

Answer-15 :-

The option (a) must be saturated is correct

Explanation:

Since the amount of liquid is constant, there is no vaporisation of the liquid inside the bottle. Also, since there cannot be a liquid with no vapours at all and vaporisation cannot take place in the remaining saturated part, the remaining part must be saturated with the vapours of the liquid.

Question-16 :-

Vapour is injected at a uniform rate in a closed vessel which was initially evacuated. The pressure in the vessel

(a) increases continuously

(b) decreases continuously

(c) first increases and then decreases

(d) first increases and then becomes constant.

Answer-16 :-

The option (d) first increases and then becomes constant. is correct

Explanation:

As the vapour is injected, the pressure of the chamber increases. But when the pressure becomes equal to the saturated vapour pressure, it condenses. So, if more vapour is injected beyond the saturated vapour pressure, the vapour will condense and thus the vapour pressure will be constant.

Question-17 :-

A vessel A has volume V and a vessel B has volume 2V. Both contain some water which has a constant volume. The pressure in the space above water is p_a for vessel A and p_bfor vessel B.

(a) p_a = p_b

(b) p_a = 2p_b

(c) p_b = 2p_a

(d) p_b = 4p_a

Answer-17 :-

The option (a) p_a = p_b is correct

Explanation:

The maximum pressure attainable above the water will be saturated vapour pressure at that temperature. Since saturated vapour pressure does not depend upon volume,  both the vessels will have same pressure.

—: End of Kinetic Theory of Gases Obj-1 (mcq-1) HC Verma Solutions Vol-2 Chapter-24 :–

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