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This mock test of Test: Molecular Velocity for JEE helps you for every JEE entrance exam.
This contains 20 Multiple Choice Questions for JEE Test: Molecular Velocity (mcq) to study with solutions a complete question bank.
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QUESTION: 1

**Direction (Q. Nos. 1-15) This section contains 15 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Q. Pressure and temperature needed to confine 1.0 x 10 ^{25} gas molecules, each with a mass of 1.0x 10^{25} kg and root mean square velocity of 1.0x 10^{3}ms^{-1} in a 1.0 m^{3} container are **

Solution:

Applying the formula

PV = 1/3(mNV^{2}rms)

Where m = mass of each particle and N = no of particles

We get P = 3.33*10^{5} pascal or 3.3 bar

Applying the formula

V_{rms} = √(3RT/M)

We get T = 2405 K(approx)

So option a is correct.

QUESTION: 2

At 298 K, which of the following gases has the lowest average molecular speed?

Solution:

Under the same conditions the average kinetic energy of the gases should be equal.

speed of gas1/speed of gas 2 = square root of (molar mass of gas2/molar mass of gas 1)

So gases with small molar mass will move quicker

eg speed of N_{2}/speed of CO_{2} = sq. rt (44/28) = 1.25

speed of N_{2}/speed of F_{2} = sq rt. (38/28) = 1.16

So order is N_{2}, F_{2} and CO_{2}

then we concluded right ans. is (A)

QUESTION: 3

By what ratio will the average velocity of the molecules in a gas change when the temperature is raised from 50°C to 200°C?

Solution:

This is the correct answer. If we take v_{200}/v_{50}, then we will get option b. But that is not according to the question.

QUESTION: 4

The root mean square velocity of a monoatomic gas (molar mass = m g mol^{-1}) is u. Its kinetic energy per mole (E) is related to u by equation

Solution:

K.E = ½ × mass × V_{rms}^{2}

Dividing both sides by moles

K.E/Moles = ½ × mass × V_{rms}^{2} /Mole

K.E/mole = ½ V_{rms}2 × molar mass

Given, K.E/mole = E, molar mass = m and V_{rms} = u

E = ½ × m × u^{2}

u^{2} = 2E/m

u = √(2E/m)

QUESTION: 5

Root mean square velocity (u) is dependent on temperature. Value of is

Solution:

The correct answer is option B

Root mean square is v=under root 3RT/M

so we simply do differentiation with respect to temperature DU/DT=D /DT

So,

= 3R/2M

QUESTION: 6

The pressure needed to confine 1.00 mole of N_{2} molecules to 24.8 L flask is 1.0 bar. Thus, root mean square velocity of the gas assuming ideal behaviour is

Solution:

Vrms = √ 3RT/M = √ 3pV/M

= √ (3×105 × 24.8×10-3)/28

= 51.4 ms^{-1 }

So according to me, option A is correct

QUESTION: 7

The ratio between the root mean square speed of H_{2} gas at 50 K and that of O_{2} gas at 800 K is

**[IIT JEEE 1996]**

Solution:

QUESTION: 8

The rms velocity of hydrogen gas is √7 times of the rms velocity of nitrogen gas. If T is the temperature of the gas, then

**[IIT JEE 2000]**

Solution:

QUESTION: 9

For CO_{2}, given that average velocity at T, is equal to most probable velocity at T_{2}.

Thus, T_{1}/T_{2} is

Solution:

QUESTION: 10

For gaseous state, if most probable speed is denoted by C*, average speed by and mean square speed by C, then for a large number of molecules, the ratios of these speeds are

**[JEE Main 2013]**

Solution:

RMS velocity = √ (3RT/M) = 1.732k

Average velocity = √ (8RT/πM)

Most probable velocity = √ (2RT/M)

=√2:√8/π:√3=1:1.128:1.225

So, C∗:C:C=1:1.128:1.225

QUESTION: 11

Molar mass of a certain gas B is double that of A.

Also RMS velocity of gas A is 200 ms ^{-1} at a certain temperature. RMS velocity of gas B at the temperature half that of A is

Solution:

The correct answer is Option B.

Molar mass of B is double that of A.

2 MA = 8 MB

QUESTION: 12

A vessei of capacity 1 dm^{3} contains 1.0^{3} x 10^{23} H_{2} molecules exerting a pressure of 101.325 kPa. Calculate RMS speed and average speed.

Solution:

V = 1dm^{3} = 10^{-3}m^{3}

M.w = 2*10^{-3}g

P = 101.325kPa

= 1.03*10^{23} H_{2} mol

No of moles = number of molecules/NA

= 1.03/6.023

= 101.325 * 103 * 10-3 = 8.314 * 1.03/6.023 * T

T = 71.27K

u_{avg} = (8RT/πM)^{1/2}

= [(8*8.314*71.27)/(3.142*2*10-3)]^1/2

= 868.53 m/s

u_{rms} = (3RT/M)^{1/2}

[(3*8.314*71.27)/(2*10-3)]^{1/2}

= 942.5 m/s

QUESTION: 13

The molecular velocity of any gas is

**[AlEEE 2011]**

Solution:

The molecular velocity of any gas whether it is average velocity, root mean square velocity and most probable velocity. It must be proportional to the square to the square root of absolute temperature.

QUESTION: 14

Distribution of molecules with velocity is represented by the curve as shown Select the correct statement(s).

Solution:

*Multiple options can be correct

QUESTION: 15

Select the correct statement (s).

Solution:

*Answer can only contain numeric values

QUESTION: 16

**Direction (Q. Nos. 16 and 17) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)**.

**Q. Temperature of an ideal gas is changed from -173°C to 127°C. Thus average velocity changes by .......... times.**

Solution:

T_{1} = 273 - 173 = 100K

T_{2} = 273 + 127 = 400K

c_{1}/c_{2} = (T1/T2)^1/2

c_{1}/c_{2} = (100/400)^1/2

c_{1}/c_{2} = (1/4)^1/2

c_{1}/c_{2} = 1/2

c_{2} = 2c_{1}

*Answer can only contain numeric values

QUESTION: 17

At what temperature (in °C) root mean square velocity of O_{2} gas at 300 K is equal to most probable velocity of Ne(20 g mol^{-1})?

Solution:

V_{rms} = √(3RT/M)

V_{mp} 0= √(2RT/M)

Putting corresponding values,

3×300/32 = 2×T/20

On solving, we get T = 281.5K

T in °C = 281.5-273 = 8.1 °C = 8

QUESTION: 18

**Direction (Q. Nos. 18-20) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).**

**Distribution of molecular velocities was studied by Maxwell and is given by**

is also called probability P.

**Q.
Based on following curves, **a

Solution:

The correct answer is option (C)

The correct order of temperature is,

T_{1} < T_{2} < T_{3}

For a particular speed, the number of molecules having that speed is more at lower temperature than the number of molecules having that speed is more at higher temperature.

QUESTION: 19

**Distribution of molecular velocities was studied by Maxwell and is given by**

is also called probability P.

**Q. Following is the distribution of molecules at 300 K for gases X, Y and Z**

**Thus,**

Solution:

QUESTION: 20

**Distribution of molecular velocities was studied by Maxwell and is given by**

is also called probability P.

**Q.
As velocity increases, probability also increases. It is maximum at a particular velocity. This is called most probable velocity. At this point, variation of probability P with velocity u is **

**Thus, based on this, velocity is**

Solution:

The correct answer is Option B.

The speed of an individual gas particle is:

The most probable speed is the maximum value on the distribution plot.

This is established by finding the velocity when the derivative of Equation;

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