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Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - NEET MCQ


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10 Questions MCQ Test Daily Test for NEET Preparation - Test: Molecular Nature of Matter & Behaviour of Gases (September 20)

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Molecular Nature of Matter & Behaviour of Gases (September 20) questions and answers have been prepared according to the NEET exam syllabus.The Test: Molecular Nature of Matter & Behaviour of Gases (September 20) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) below.
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Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 1
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In the case of real gases, the equation of state, PV = RT (where P, V and T are respectively the pressure, volume and absolute temperature), is strictly satisfied only if corrections are applied to the measured pressure P and the measured volume V. The corrections for P and V arise respectively due to

Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 1

According to kinetic theory of gases there is no intermolecular force of attraction between the molecule of gases but its not true as gases liquifise, at high pressure molecular interaction starts due to which pressure on the walls decreases as molecule dragged back so pressure exerted by real gas is less than ideal gas, due to repulsive forces molecule behave as small spheres and hence volume decreases

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 2
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Which one of the following quantities can be zero on an average for the molecules of an ideal gas in equilibrium?

Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 2

In case of ideal gases the average velocity is always zero. Hence the average momentum is zero. 
Whereas average speed is non- zero so the kinetic energy is also non-zero,  as these two are scalar quantities.

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Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 3
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Two specific heats of a perfect gas are related by:

Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 3

When cp & cv are given calorie & R in Joule then Cp – Cv = (R/J)

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 4
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One mole of any substance at any temperature, pressure or volume always contains ________ molecules.
Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 4

1 mole = 6.02 x 1023

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 5
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Oxygen and nitrogen in two enclosures have the same mass, volume and pressure. The ratio of the temperature of oxygen to that of nitrogen is:
Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 5

For same mass the ratio of moles of oxygen to that of nitrogen is 14:16 = 7:8
And we know that PV = nRT
Hence as V and P are also same, ratio of temperature of oxygen to that of nitrogen is inverse of the ratio of moles that is 8:7

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 6
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When the temperature goes up, the pressure inside a rigid container will _____.
Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 6

The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. This can be easily understood by visualising the particles of gas in the container moving with a greater energy when the temperature is increased.

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 7
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Four moles of an ideal diatomic gas is heated at constant volume from 20° C to 30° C. The molar specific heat of the gas at constant pressure (Cp) is 30.3 Jmol-1K-1 and the universal gas constant (R) is 8.3 Jmol-1K-1. The increase in internal energy of the gas is
Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 7

The value of Cp is 30.3
and as Cp-Cv = R(8.3)
hence Cv = 30.3-8.3
Cv is 22
change in internal energy = no of moles × Cv × change in temperature
hence
change in internal energy = 22 × 4 × 10
= 880j
Hence Option D is correct.
 

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 8
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A region of the earth’s atmosphere contains n molecules (treated as ideal gas molecules) per unit volume. The temperature of air in the region is T. If k represents Boltzmann’s constant and R represents universal gas constant, the pressure of air in the region is
Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 8

PV = nRT
Where n = number of moles = m/NA
So, P = (m/V)(R/NA)T
Also, we know that R/NA = k
So, P = nkT

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 9
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The average kinetic energy of translation of a molecule of an ideal gas at temperature T is:
Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 9

Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 10
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Three moles of an ideal monoatomic gas is initially in the state A shown in the adjoining pressure-temperature graph. It is taken to state B without changing its pressure. If R is the universal gas constant, the work done by the gas in this process is 
Detailed Solution for Test: Molecular Nature of Matter & Behaviour of Gases (September 20) - Question 10

The work done by the gas in taking it from state A to state B = PΔV where ΔW is the increase in volume at constant pressure P. 

We have PV = μRT where p is the number of moles in the sample of the gas and R is the universal gas constant. 

Therefore we have PΔV = μR ΔT = 3 xR(450 - 250) = 600R 

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