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Q1. The degree of freedom for tri atomic gas is:

• 3
• 4
• 5
• 6

Solution

The degrees of freedom of the system is given by:- f = 3 N - KWhere, f = degrees of freedomN = Number of Particles in the system.K = Independent relation among the particles.For a tri atomic gas; N = 3 and K = 3;f = 3 X 3 - 3 = 6

Q2. The K.E of one mole of an ideal gas is E=5/2 RT. Then C_ρ will be

• 0.4 R
• 3.5 R
• 0.3 R
• 2.5 R

Solution

E= C_ν 1.T= 5/2 RT Therefore C_ν = 5/2 RT C_ρ = C_ν +R= 5/2R + R= 7/2 R= 3.5R

Q3. When the temperature goes up, the pressure inside a rigid container will _____.

• go down
• cause particles to cool
• go up
• remain unchanged

Solution

When the temperature goes up, the pressure inside a rigid container will go up.

Q4. Which one of the following quantities can be zero on an average for the molecules of an ideal gas in equilibrium?

• Kinetic energy
• Density
• Momentum
• Speed

Solution

Momentum can be zero on an average for the molecules of an ideal gas in equilibrium, since the molecules travel in random directions.

Q5. The degree of freedom for diatomic gas is:

• 3
• 4
• 5
• 6

Solution

The degrees of freedom of the system is given by:- f = 3 N - KWhere, f = degrees of freedomN = Number of Particles in the system.K = Independent relation among the particles.For a diatomic gas ; N = 2 and K = 1;f = 3 X 2 - 1 = 5

Q6. The kinetic energy of one mole of an ideal gas is E=3/2 RT. Then C_ρ will be

• 1.5 R
• 0.5R
• 2.2 R
• 2.5R

Solution

 

Q7. Value of gas constant, R for one mole of a gas is independent of the

• Mass of gas
• Atomicity of the gas
• Distance between two molecules of gas at 273 K
• Volume of gas

Solution

Value of gas constant, R for one mole of a gas is independent of the 'Atomicity of the gas'.

Q8. The average kinetic energy of a molecule in an ideal gas is

• proportional to the absolute temperature of the gas
• proportional to the pressure
• proportional to the volume
• depends on the nature of the ideal gas.

Solution

the average kinetic energy,

Q9. Four moles of O₂ are mixed with two moles of Helium. What will be the ratio of specific heats of constant pressure and constant volume for the mixture?

• 6.5
• 4
• 1.5
• 1.79

Solution

For a monoatomic gas like helium γ_he= 5/3   For a diatomic gas like oxygen γ_o= 7/5 For γ of the mixture, one has to take weighted average of the two gases  

Q10. 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

• intermolecular attraction and the size of molecules
• size of molecules and expansion of the container
• expansion of the container and intermolecular attraction
• kinetic energy of molecules and collision of molecules

Solution

In kinetic theory of gases it is assumed that there is no force between molecules But there is actually intermolecular attraction which reduces the pressure. So the correction for P arises due to intermolecular attraction.The entire volume V of the container is not available for the molecules since the molecules have a finite size. The assumption (in kinetic theory) that the molecules are point masses without appreciable volume is incorrect. So the correction for V arises due to the size of molecules.

Q11. 1calorie = ?

• 41.48 Joules
• 4.148 Joules
• 414.8 Joules
• 4148 Joules

Solution

1 calorie = 4.148 Joules  

Q12. 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

• nT/k
• nkT
• RT/n
• nRkT

Solution

Q13. 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:

• 7/8
• 8/7
• 1
• 49/64

Solution

Q14. Pressure exerted by an ideal gas molecule is given by the expression

• P=MC²/V
• P=MC²/3V
• P=M/VC
• P=M²/CV²

Solution

Pressure exerted by an ideal gas molecule is given as P = MC²/3V where, M is the mass of the gas, C is the r.m.s velocity of the gas molecules, and V is the volume of the gas

Q15. What is the law of equipartition of energy?

• The total energy is unequally distributed in all possible energy modes, with each mode having an average energy equal to 2kBT. This is known as the law of equipartition of energy.
• The total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to kBT. This is known as the law of equipartition of energy
• The total energy is unequally distributed in all possible energy modes, with each mode having an average energy equal to ½ kBT. This is known as the law of equipartition of energy.
• The total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to ½ kBT. This is known as the law of equipartition of energy.

Solution

The total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to ½ kBT. This is known as the law of equipartition of energy.

Q16. The perfect gas equation is PV = nRT where n is the

• number of moles
• number of molecules
• number of atoms
• mass of molecules

Solution

n refers to number of moles.

Q17. If the molecule has 'f' degrees of freedom, the average kinetic energy per molecule is

• (f/2)kT
• 2fkT
• (3f/2)kt
• f²kt

Solution

The average kinetic energy per molecule is (f/2)kT.

Q18. An ideal gas at 27 ⁰C is heated at constant pressure so as to double its volume. The temperature of the gas will be

• 300 ⁰C
• 54 ⁰C
• 600 ⁰C
• 327 ⁰C

Solution

At constant pressure, V₁/V₂ = T₁/T₂ where T is in kelvin. When V is doubled, T also doubles.

Q19. If a gas has n degree of freedom, ratio of principal specific heats of the gas is

• 1+ 2/n
• 2n
• 1-n/2
• -2n

Solution

The ratio of specific heats of gas is related to degree of freedom n as

Q20. On which of the following factors, does the average kinetic energy of gas molecules depend?

• Pressure
• Absolute temperature
• Velocity of gas molecules
• Volume of the gas

Solution

Average kinetic energy depends only upon the absolute temperature and is directly proportional to it.

Q21. The degree of freedom for monoatomic gas is:

• 2
• 3
• 5
• 6

Solution

The degrees of freedom of the system is given by:- f = 3 N - KWhere, f = degrees of freedomN = Number of Particles in the system.K = Independent relation among the particles.For a monatomic gas; N = 1 and K = 0;f = 3 X 1 - 0 = 3

Q22. The average kinetic energy of translation of a molecule of an ideal gas at temperature T is:

• (1/2) kT
• (3/2) kT
• (5/7) kT
• (7/2) kT

Solution

The average kinetic energy of translation of a molecule of an ideal gas at temperature T is (3/2) kT

Q23. The absolute zero may be defined as that temperature at which

• Temperature is 273 K
• Root mean square velocity of the gas molecule reduces to zero
• Mass of molecules of gas is zero
• Volume is maximum

Solution

The absolute zero may be defined as that temperature at which the root mean square velocity of the gas molecule reduces to zero.

Q24. According to kinetic theory of gases, 0 kelvin is the temperature at which for an ideal gas

• pressure is not zero
• volume is not zero
• internal energy is zero
• nearly all molecular motion becomes very rapid

Solution

According to the kinetic theory of gases, the internal energy of an ideal gas is zero at 0 kelvin.

Q25. Charles's Law is about the relationship between _______ in flexible containers.

• pressure and volume
• pressure, volume and temperature
• temperature and volume
• temperature and pressure

Solution

Charles's Law is about the relationship between temperature and volume in flexible containers

Q26. For a diatomic gas, the number of translational degrees of freedom is

• 3
• 4
• 5
• 6

Solution

3, this is because the diatomic molecule can move along say, the x , y and z axes.

Q27. A rigid diatomic molecule has _______ degrees of freedom

• 3
• 5
• 7
• 9

Solution

It has 3 translational and 2 rotational degrees of freedom.

Q28. Four molecules of a gas have speeds 2,3,4,6 km/s respectively. Calculate their root mean square speed

• 2.01 km/s
• 3.3 km/s
• 1.9 km/s
• 2.5 km/s

Solution

Q29. How many degrees of freedom do non linear triatomic gas molecules has?

• Two
• Three
• Six
• five

Solution

In non-linear tri-atomic molecule, there are three atoms present at the three vertices of a triangle. There are three fixed distances amongst the three atoms. Here, A = 3 and R = 3 Therefore, N = 3 × 3 - 3 =6 Hence, a non linear triatomic gas molecule has six degrees of freedom.

Q30. Four moles of an ideal diatomic gas is heated at constant volume from 20 ^oC to 30 ^oC. The molar specific heat of the gas at constant pressure (C_p) 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:

• 80.3 J
• 303 J
• 332 J
• 880 J