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Kinetic Theory of Gases

139213 Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules :

1 decreases

2 increases

3 remains same

4 increase of decreases depending on the nature of gas

Explanation:

C We know that, K.E $= \frac{3}{2} KT$
And temperature $T =$ Constant
But K.E only depends on temperature
So, Kinetic energy remains same.

Karnataka CET-2018

Kinetic Theory of Gases

139229 Mean free path of a gas molecule is

1 inversely proportional to number of molecules per unit volume

2 inversely proportional to diameter of the molecule

3 directly proportional to the square root of the absolute temperature

4 directly proportional to the molecular mass

5 independent of temperature

Explanation:

A We know that, mean free path $(λ) = \frac{kT}{\sqrt{2} π {nd}^{2}}$
Where, $n =$ number of molecules per unit volume
$d =$ diameter of the molecule.
Mean free path of a gas molecule is inversely proportional to number of molecules per unit volume.

Kerala CEE - 2009

Kinetic Theory of Gases

139230 The temperature at which oxygen molecules have the same root mean square speed as that of hydrogen molecules at $300 K$ is

1

600 K

2

2400 K

3

1200 K

4

300 K

5

4800 K

Explanation:

E Given that, ${(v_{rms})}_{O_{2}} = {(v_{rms})}_{H_{2}}$
We know that, $v_{rms} = \sqrt{\frac{3 RT}{M}}$
Then, $\sqrt{\frac{3 RT}{M_{O_{2}}}} = \sqrt{\frac{3 R (300)}{M_{H_{2}}}}$
Or, $T = 300 \times \frac{M_{O_{2}}}{M_{H_{2}}} = 300 \times \frac{32}{2} = 300 \times 16 = 4800 K$

Kerala CEE - 2009

Kinetic Theory of Gases

139231 The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{- 14} J$ is (Boltzmann constant $= 1.38 \times 10^{- 23} {JK}^{- 1}$ )

1

2 \times 10^{9} K

2

10^{9} K

3

6 \times 10^{9} K

4

3 \times 10^{9} K

5

4.5 \times 10^{9} K

Explanation:

A Given that, Boltzmann constant $(k) = 1.38 \times$ $10^{- 23} {JK}^{- 1}$ , Energy of proton gas $= 4.14 \times 10^{- 14} J$ We know,
Energy of proton gas $(E) = \frac{3}{2} k \cdot T$
Putting these value, we get-
$4.14 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times 10^{- 23} \times T$
Or $T = \frac{2 \times 4.14 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}} = \frac{8.28 \times 10^{- 14}}{4.14 \times 10^{- 23}} = 2 \times 10^{9} K$

Kerala CEE - 2009

Kinetic Theory of Gases

139213 Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules :

1 decreases

2 increases

3 remains same

4 increase of decreases depending on the nature of gas

Explanation:

C We know that, K.E $= \frac{3}{2} KT$
And temperature $T =$ Constant
But K.E only depends on temperature
So, Kinetic energy remains same.

Karnataka CET-2018

Kinetic Theory of Gases

139229 Mean free path of a gas molecule is

1 inversely proportional to number of molecules per unit volume

2 inversely proportional to diameter of the molecule

3 directly proportional to the square root of the absolute temperature

4 directly proportional to the molecular mass

5 independent of temperature

Explanation:

A We know that, mean free path $(λ) = \frac{kT}{\sqrt{2} π {nd}^{2}}$
Where, $n =$ number of molecules per unit volume
$d =$ diameter of the molecule.
Mean free path of a gas molecule is inversely proportional to number of molecules per unit volume.

Kerala CEE - 2009

Kinetic Theory of Gases

139230 The temperature at which oxygen molecules have the same root mean square speed as that of hydrogen molecules at $300 K$ is

1

600 K

2

2400 K

3

1200 K

4

300 K

5

4800 K

Explanation:

E Given that, ${(v_{rms})}_{O_{2}} = {(v_{rms})}_{H_{2}}$
We know that, $v_{rms} = \sqrt{\frac{3 RT}{M}}$
Then, $\sqrt{\frac{3 RT}{M_{O_{2}}}} = \sqrt{\frac{3 R (300)}{M_{H_{2}}}}$
Or, $T = 300 \times \frac{M_{O_{2}}}{M_{H_{2}}} = 300 \times \frac{32}{2} = 300 \times 16 = 4800 K$

Kerala CEE - 2009

Kinetic Theory of Gases

139231 The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{- 14} J$ is (Boltzmann constant $= 1.38 \times 10^{- 23} {JK}^{- 1}$ )

1

2 \times 10^{9} K

2

10^{9} K

3

6 \times 10^{9} K

4

3 \times 10^{9} K

5

4.5 \times 10^{9} K

Explanation:

A Given that, Boltzmann constant $(k) = 1.38 \times$ $10^{- 23} {JK}^{- 1}$ , Energy of proton gas $= 4.14 \times 10^{- 14} J$ We know,
Energy of proton gas $(E) = \frac{3}{2} k \cdot T$
Putting these value, we get-
$4.14 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times 10^{- 23} \times T$
Or $T = \frac{2 \times 4.14 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}} = \frac{8.28 \times 10^{- 14}}{4.14 \times 10^{- 23}} = 2 \times 10^{9} K$

Kerala CEE - 2009

Kinetic Theory of Gases

139213 Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules :

1 decreases

2 increases

3 remains same

4 increase of decreases depending on the nature of gas

Explanation:

C We know that, K.E $= \frac{3}{2} KT$
And temperature $T =$ Constant
But K.E only depends on temperature
So, Kinetic energy remains same.

Karnataka CET-2018

Kinetic Theory of Gases

139229 Mean free path of a gas molecule is

1 inversely proportional to number of molecules per unit volume

2 inversely proportional to diameter of the molecule

3 directly proportional to the square root of the absolute temperature

4 directly proportional to the molecular mass

5 independent of temperature

Explanation:

A We know that, mean free path $(λ) = \frac{kT}{\sqrt{2} π {nd}^{2}}$
Where, $n =$ number of molecules per unit volume
$d =$ diameter of the molecule.
Mean free path of a gas molecule is inversely proportional to number of molecules per unit volume.

Kerala CEE - 2009

Kinetic Theory of Gases

139230 The temperature at which oxygen molecules have the same root mean square speed as that of hydrogen molecules at $300 K$ is

1

600 K

2

2400 K

3

1200 K

4

300 K

5

4800 K

Explanation:

E Given that, ${(v_{rms})}_{O_{2}} = {(v_{rms})}_{H_{2}}$
We know that, $v_{rms} = \sqrt{\frac{3 RT}{M}}$
Then, $\sqrt{\frac{3 RT}{M_{O_{2}}}} = \sqrt{\frac{3 R (300)}{M_{H_{2}}}}$
Or, $T = 300 \times \frac{M_{O_{2}}}{M_{H_{2}}} = 300 \times \frac{32}{2} = 300 \times 16 = 4800 K$

Kerala CEE - 2009

Kinetic Theory of Gases

139231 The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{- 14} J$ is (Boltzmann constant $= 1.38 \times 10^{- 23} {JK}^{- 1}$ )

1

2 \times 10^{9} K

2

10^{9} K

3

6 \times 10^{9} K

4

3 \times 10^{9} K

5

4.5 \times 10^{9} K

Explanation:

A Given that, Boltzmann constant $(k) = 1.38 \times$ $10^{- 23} {JK}^{- 1}$ , Energy of proton gas $= 4.14 \times 10^{- 14} J$ We know,
Energy of proton gas $(E) = \frac{3}{2} k \cdot T$
Putting these value, we get-
$4.14 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times 10^{- 23} \times T$
Or $T = \frac{2 \times 4.14 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}} = \frac{8.28 \times 10^{- 14}}{4.14 \times 10^{- 23}} = 2 \times 10^{9} K$

Kerala CEE - 2009

Kinetic Theory of Gases

139213 Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules :

1 decreases

2 increases

3 remains same

4 increase of decreases depending on the nature of gas

Explanation:

C We know that, K.E $= \frac{3}{2} KT$
And temperature $T =$ Constant
But K.E only depends on temperature
So, Kinetic energy remains same.

Karnataka CET-2018

Kinetic Theory of Gases

139229 Mean free path of a gas molecule is

1 inversely proportional to number of molecules per unit volume

2 inversely proportional to diameter of the molecule

3 directly proportional to the square root of the absolute temperature

4 directly proportional to the molecular mass

5 independent of temperature

Explanation:

A We know that, mean free path $(λ) = \frac{kT}{\sqrt{2} π {nd}^{2}}$
Where, $n =$ number of molecules per unit volume
$d =$ diameter of the molecule.
Mean free path of a gas molecule is inversely proportional to number of molecules per unit volume.

Kerala CEE - 2009

Kinetic Theory of Gases

139230 The temperature at which oxygen molecules have the same root mean square speed as that of hydrogen molecules at $300 K$ is

1

600 K

2

2400 K

3

1200 K

4

300 K

5

4800 K

Explanation:

E Given that, ${(v_{rms})}_{O_{2}} = {(v_{rms})}_{H_{2}}$
We know that, $v_{rms} = \sqrt{\frac{3 RT}{M}}$
Then, $\sqrt{\frac{3 RT}{M_{O_{2}}}} = \sqrt{\frac{3 R (300)}{M_{H_{2}}}}$
Or, $T = 300 \times \frac{M_{O_{2}}}{M_{H_{2}}} = 300 \times \frac{32}{2} = 300 \times 16 = 4800 K$

Kerala CEE - 2009

Kinetic Theory of Gases

139231 The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{- 14} J$ is (Boltzmann constant $= 1.38 \times 10^{- 23} {JK}^{- 1}$ )

1

2 \times 10^{9} K

2

10^{9} K

3

6 \times 10^{9} K

4

3 \times 10^{9} K

5

4.5 \times 10^{9} K

Explanation:

A Given that, Boltzmann constant $(k) = 1.38 \times$ $10^{- 23} {JK}^{- 1}$ , Energy of proton gas $= 4.14 \times 10^{- 14} J$ We know,
Energy of proton gas $(E) = \frac{3}{2} k \cdot T$
Putting these value, we get-
$4.14 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times 10^{- 23} \times T$
Or $T = \frac{2 \times 4.14 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}} = \frac{8.28 \times 10^{- 14}}{4.14 \times 10^{- 23}} = 2 \times 10^{9} K$

Kerala CEE - 2009

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