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This mock test of Kinetic Theory Of Gases MCQ Level – 2 (part - 1) for IIT JAM helps you for every IIT JAM entrance exam.
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QUESTION: 1

The temperature at which the average speed of H_{2} equals that of O_{2} at 320K.

Select one:

Solution:

Therefore

which gives, T = 20 K

The correct answer is: 20 K

QUESTION: 2

The critical constant for water are 647 K, 22.09 MPa and 0.0566 dm^{3} mol^{-1}, the value of a and b are respectively.

Select one:

Solution:

We have

T_{c} = 647 K

p_{c} = 22.09 MPa = 22.09 × 10^{3} kPa

V_{c} = 0.0566 dm^{-3} mol^{-1}

Thus,

= 0.0189 dm^{3} mol^{-1}

= 212.3 kPa dm^{6} mol^{-2}

The correct answer is: = 212.3 kPa dm^{6} mol^{-2}, 0.0189 dm^{3} mol^{-1}

QUESTION: 3

Calculate molecular diameter d of helium from its van der Waals constant b = 24 cm^{3} mol^{-1}

Select one:

Solution:

Since, b = 4 × volume occupied by the molecule in 1 mole of a gas

or

therefore

= 1.335 × 10^{-8} cm = 133.5 pm

d = 2r = 267 pm

The correct answer is: 267 pm

QUESTION: 4

The kinetic energy of 0.5 mol of an ideal gas at 273 K

Select one:

Solution:

Total kinetic energy;

K.E. = (0.5 mol)

= 1702 J

The correct answer is: 1702 J

QUESTION: 5

The mean free path of the molecule of a certain gas at 300K is 2.6×10^{-5}m. The collision diameter of the molecule is 0.26 nm then the number of molecule per unit volume of the gas.

Select one:

Solution:

Here

T = 300K

Since,

= 1.281 × 10^{23} m^{-3}

The correct answer is: 1.281 × 10^{23} m^{-3}

QUESTION: 6

Using vander Waals equation, the pressure exerted by 22g of carbon dioxide in 0.5 dm^{3} at 298.15 K is (Given : a = 363.76 kPa dm^{6} mol^{-2} and b = 42.67 cm^{3} mol^{-1} )

Select one:

Solution:

Vander Waals equation

= nRT

and

So,

= 2589.31 kPa – 363.76 kPa

= 2225.55 kPa.

The correct answer is: 2222.55 kPa

QUESTION: 7

Calculate the fraction of N_{2} molecule at 101.325 kPa and 300 K whose speed are in the range of

Select one:

Solution:

Most probable speed,

Since, 0.005 u_{mp} = (0.005) (422.09 ms^{-1})

= 2.11 ms^{-1}

Therefore

4.22 ms^{-1}

According to Maxwell distribution law,

= exp (–1.000) = 0.3679

Hence,

The correct answer is: 8.303 × 10^{-3}

QUESTION: 8

The average speed of H_{2} molecules, if the density of the gas at 101.325 kPa is 0.09gdm^{-3}

Select one:

Solution:

Density ρ = 009 gdm^{-3} = 0.09 kgm^{-3}

Average speed,

Also,

= 1694 ms^{-1}

The correct answer is: 1694 ms^{-1}

QUESTION: 9

What is the ratio of the number of molecule having speed in the range of 2u_{mp} and 2u_{mp} + du to the number of molecule having speeds in the range of u_{mp} + du.

Select one:

Solution:

If dN_{1} is the number of molecule in the speed range u_{mp} to u_{mp} + du and dN_{2} is the corresponding number in the speed range 2u_{mp} to 2u_{mp} + du, then according to the Maxwell distribution, we have,

and

therefore

Now, Since,

therefore

The correct answer is: 0.199

QUESTION: 10

A bulb of capacity 1 dm^{3} contain 1.0^{3} × 10^{23} gaseous hydrogen molecules and the pressure exerted by these molecules is 101.325 kPa. Calculate the average square molecular speed.

Select one:

Solution:

We have

V = 1dm^{3}

N = 1.3 × 10^{23}

p = 101.325 kPa

We known, number of moles

n = N/N_{A}

n =

T = pV/nR

=

= 71.27 K

= 8.888 × 10 ^{5}Jkg^{-1}

= 8.888 × 10^{5} (ms^{-1} )^{2}

The correct answer is: 8.888 × 10^{5} (ms^{-1} )^{2}

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