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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - JEE MCQ


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8 Questions MCQ Test - Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced

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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 1

When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules : (1984 - 1 Mark)

Detailed Solution for Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 1

No work is required to tear apart the molecules due to the absence of attractive forces in an ideal gas.

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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 2

If a gas is expanded at constant temperature :

Detailed Solution for Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 2

At constant temp., when gas expands the K.E. of the molecules remains the same, but the pressure
decreases. 

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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 3

Equal weights of ethane and hydrogen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by hydrogen is (1993 - 1 Mark)

Detailed Solution for Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 3

Pressure exerted by H2 is proportional to its mole fraction.
Mole fraction of H2 =

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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 4

According to Graham’s law, at a given temperature the ratio of the rates of diffusion rA/rB of gases A and B is given by (1998 - 2 Marks)

(Where P and M are pressures and molecular weights of gases A and B respectively.)

Detailed Solution for Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 4

According to Graham’s law of diffusion for two gases undergoing diffusion at different pressures through same hole

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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 5

5. Refer to the figure given : (2006 - 5M; –1) Which of the following statements is wrong?

Detailed Solution for Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 5

For gas A, a = 0, implies Z varies linearly
with pressure.

For gas B, b = 0, . Hence, Z does not vary
linearly with pressure.
Given the intersection data for gas C, it is possible to find the values of ‘a’ and ‘b’. All vander Waal gases, like gas C, give positive slope at high pressures.

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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 6

A gas described by van der Waals equation –      (2008- 1 Mark)

Detailed Solution for Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 6

Vander Waals equation is

 = nRT  [For n moles of a gas)

a, b are vander Waals constants The ideal gas equation is PV = nRT [For n moles of a gas] where P is pressure excerted by ideal gas and V is volume occupied by ideal gas.
In vander Waals equation the term   represents the pressure exerted by the gas and (V– nb) the volume occupied by the gas. At low pressure, when the gas occoupies large volume the intermolecular distance between gaseous moleculas is quite large and in such  case there is no significant role played by intermolecular forces and thus the gas behaves like an ideal gas thus (a) is correct NOTE : Under high pressure the intermolecular distance decreases and the intermolecular forces play a significant role and the gas shows a devation from ideal behaviour.
Thus (b) is not correct. a, b i.e. the vander Waals coefficients defined on the nature of gas and are independent of temperature so (c) is correct.
The pressure  is not lower than P so (d) is
not correct.
Hence the correct anser is (a, c).

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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 7

According to kinetic theory of gases (2011)

Detailed Solution for Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 7

As per the postulates of the kinetic theory of gases, the molecules of gas collide with each other and with the walls of the container. These collisions are perfectly elastic in nature.
The gas molecule with mass m and speed u collides with the walls of the container to transfer the momentum Δp=−2mu. Thus heavier molecules transfer more momentum to the walls of the container. But this is not the postulate of the kinetic theory of gases.
As per Maxwell -Boltzmann distribution of molecular speed, very few molecules move with very high or very low speed. Most of the molecules move with average speed. Hence, only a small number of molecules have very high velocity.
According to the kinetic theory of gases, a gas molecule moves in straight line with constant velocity. When it collides with another molecule, it changes direction. Between collisions, the molecules move in straight lines with constant velocities.

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Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 8

One mole of a monoatomic real gas satisfies the equation p(V – b) = RT where b is a constant. The relationship of interatomic potential V(r) and interatomic distance r for the gas is given by (JEE Adv. 2015)

Detailed Solution for Test: MCQs (One or More Correct Option): States of Matter | JEE Advanced - Question 8

P(V–b) = RT ⇒ PV – Pb = RT

Hence Z > 1 at all pressures.
This means , repulsive tendencies will be dominant when interatomic distance are small.
This means, interatomic potential is never negative but becomes positive at small interatomic distances.

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