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PHXI13:KINETIC THEORY

360310 The mean free path of collision of gas molecules varies with its diameter $(d)$ of the molecules as

1

d^{- 1}

2

d^{- 2}

3

d^{- 3}

4

d^{- 4}

Explanation:

Mean free path of gas molecules,
$λ = \frac{λ_{1} + λ_{2} + \dots + λ_{n}}{n}$
$λ = \frac{1}{\sqrt{2} π n d^{2}}$
$\Rightarrow λ \propto d^{- 2}$

PHXI13:KINETIC THEORY

360311 Assertion :
Mean free path of a gas molecule varies inversely as density of the gas.
Reason :
Mean free path varies inversely as temperature of the gas.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The mean free path of a gas molecule is the average distance between two successive collisions. It is represented by $λ$ .
$λ = \frac{k_{B} T}{\sqrt{2} π σ^{2} ρ}$
Here, $σ =$ diameter of molecule $k_{B} =$ Boltzmann's constant
Hence, mean free path varies inversely as density of the gas. The mean free path varies directly as the temperature and inversely as the pressure of the gas. So option (3) is correct.

PHXI13:KINETIC THEORY

360312 The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as:

1

\frac{1}{\sqrt{2} n π d^{2}}

2

\frac{1}{\sqrt{2} n^{2} π d^{2}}

3

\frac{1}{\sqrt{2} n^{2} π^{2} d^{2}}

4

\frac{1}{\sqrt{2} n π d}

Explanation:

Mean free path for a gas sample
$λ_{m} = \frac{1}{\sqrt{2} π d^{2} n}$
where $d$ is diameter of a gas molecule and $n$ is molecular density

NEET - 2020

PHXI13:KINETIC THEORY

360313 Modern vaccum pumps can evacuate a vessel down to a pressure of $4.0 \times 10^{- 15} a t m$ . At room temperature (300 K). Taking $R = 8.0 J K^{- 1} m o l e^{- 1}$ , $1 a t m = 10^{5} P a$ and $N_{A v o g a d r o} = 6 \times 10^{23} m o l e^{- 1}$ , the mean distance between molecules of gas in an evacuated vessel will be of the order of:

1

0.2 μ m

2

0.2 m m

3

0.2 c m

4

0.2 n m

Explanation:

From the given information we calculate the average volume occupied by each molecule (not the volume of molecule). The average volume occupied or share of each molecule is equal to the ratio of total volume divides by total number of molecules. This volume can be approximated as a cube of side length $(λ)$ . The cube root of the average volume per molecule is equal to $(λ)$ . Here the value $(λ)$ is also called as mean free path and it is equal to average distance between two molecules of a gas as shown in the figure.
supporting img

Ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T$
Given that
$R = 8 \frac{J}{K . m o l e}, T = 300 K$
$N_{A} = 6 \times 10^{23} m o l e^{- 1}$
$P = 4.0 \times 10^{- 15} \times 10^{5} N / m^{2}$
$\frac{V}{N} = \frac{R T}{P N_{A}} = \frac{8 \times 300}{4.0 \times 10^{- 15} \times 10^{5} \times 6 \times 10^{23}}$
$λ^{3} = \frac{V}{N} = 10^{- 11} m^{3}$
$λ = 2.15 \times 10^{- 4} = 0.215 m m$

JEE - 2014

PHXI13:KINETIC THEORY

360310 The mean free path of collision of gas molecules varies with its diameter $(d)$ of the molecules as

1

d^{- 1}

2

d^{- 2}

3

d^{- 3}

4

d^{- 4}

Explanation:

Mean free path of gas molecules,
$λ = \frac{λ_{1} + λ_{2} + \dots + λ_{n}}{n}$
$λ = \frac{1}{\sqrt{2} π n d^{2}}$
$\Rightarrow λ \propto d^{- 2}$

PHXI13:KINETIC THEORY

360311 Assertion :
Mean free path of a gas molecule varies inversely as density of the gas.
Reason :
Mean free path varies inversely as temperature of the gas.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The mean free path of a gas molecule is the average distance between two successive collisions. It is represented by $λ$ .
$λ = \frac{k_{B} T}{\sqrt{2} π σ^{2} ρ}$
Here, $σ =$ diameter of molecule $k_{B} =$ Boltzmann's constant
Hence, mean free path varies inversely as density of the gas. The mean free path varies directly as the temperature and inversely as the pressure of the gas. So option (3) is correct.

PHXI13:KINETIC THEORY

360312 The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as:

1

\frac{1}{\sqrt{2} n π d^{2}}

2

\frac{1}{\sqrt{2} n^{2} π d^{2}}

3

\frac{1}{\sqrt{2} n^{2} π^{2} d^{2}}

4

\frac{1}{\sqrt{2} n π d}

Explanation:

Mean free path for a gas sample
$λ_{m} = \frac{1}{\sqrt{2} π d^{2} n}$
where $d$ is diameter of a gas molecule and $n$ is molecular density

NEET - 2020

PHXI13:KINETIC THEORY

360313 Modern vaccum pumps can evacuate a vessel down to a pressure of $4.0 \times 10^{- 15} a t m$ . At room temperature (300 K). Taking $R = 8.0 J K^{- 1} m o l e^{- 1}$ , $1 a t m = 10^{5} P a$ and $N_{A v o g a d r o} = 6 \times 10^{23} m o l e^{- 1}$ , the mean distance between molecules of gas in an evacuated vessel will be of the order of:

1

0.2 μ m

2

0.2 m m

3

0.2 c m

4

0.2 n m

Explanation:

From the given information we calculate the average volume occupied by each molecule (not the volume of molecule). The average volume occupied or share of each molecule is equal to the ratio of total volume divides by total number of molecules. This volume can be approximated as a cube of side length $(λ)$ . The cube root of the average volume per molecule is equal to $(λ)$ . Here the value $(λ)$ is also called as mean free path and it is equal to average distance between two molecules of a gas as shown in the figure.
supporting img

Ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T$
Given that
$R = 8 \frac{J}{K . m o l e}, T = 300 K$
$N_{A} = 6 \times 10^{23} m o l e^{- 1}$
$P = 4.0 \times 10^{- 15} \times 10^{5} N / m^{2}$
$\frac{V}{N} = \frac{R T}{P N_{A}} = \frac{8 \times 300}{4.0 \times 10^{- 15} \times 10^{5} \times 6 \times 10^{23}}$
$λ^{3} = \frac{V}{N} = 10^{- 11} m^{3}$
$λ = 2.15 \times 10^{- 4} = 0.215 m m$

JEE - 2014

PHXI13:KINETIC THEORY

360310 The mean free path of collision of gas molecules varies with its diameter $(d)$ of the molecules as

1

d^{- 1}

2

d^{- 2}

3

d^{- 3}

4

d^{- 4}

Explanation:

Mean free path of gas molecules,
$λ = \frac{λ_{1} + λ_{2} + \dots + λ_{n}}{n}$
$λ = \frac{1}{\sqrt{2} π n d^{2}}$
$\Rightarrow λ \propto d^{- 2}$

PHXI13:KINETIC THEORY

360311 Assertion :
Mean free path of a gas molecule varies inversely as density of the gas.
Reason :
Mean free path varies inversely as temperature of the gas.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The mean free path of a gas molecule is the average distance between two successive collisions. It is represented by $λ$ .
$λ = \frac{k_{B} T}{\sqrt{2} π σ^{2} ρ}$
Here, $σ =$ diameter of molecule $k_{B} =$ Boltzmann's constant
Hence, mean free path varies inversely as density of the gas. The mean free path varies directly as the temperature and inversely as the pressure of the gas. So option (3) is correct.

PHXI13:KINETIC THEORY

360312 The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as:

1

\frac{1}{\sqrt{2} n π d^{2}}

2

\frac{1}{\sqrt{2} n^{2} π d^{2}}

3

\frac{1}{\sqrt{2} n^{2} π^{2} d^{2}}

4

\frac{1}{\sqrt{2} n π d}

Explanation:

Mean free path for a gas sample
$λ_{m} = \frac{1}{\sqrt{2} π d^{2} n}$
where $d$ is diameter of a gas molecule and $n$ is molecular density

NEET - 2020

PHXI13:KINETIC THEORY

360313 Modern vaccum pumps can evacuate a vessel down to a pressure of $4.0 \times 10^{- 15} a t m$ . At room temperature (300 K). Taking $R = 8.0 J K^{- 1} m o l e^{- 1}$ , $1 a t m = 10^{5} P a$ and $N_{A v o g a d r o} = 6 \times 10^{23} m o l e^{- 1}$ , the mean distance between molecules of gas in an evacuated vessel will be of the order of:

1

0.2 μ m

2

0.2 m m

3

0.2 c m

4

0.2 n m

Explanation:

From the given information we calculate the average volume occupied by each molecule (not the volume of molecule). The average volume occupied or share of each molecule is equal to the ratio of total volume divides by total number of molecules. This volume can be approximated as a cube of side length $(λ)$ . The cube root of the average volume per molecule is equal to $(λ)$ . Here the value $(λ)$ is also called as mean free path and it is equal to average distance between two molecules of a gas as shown in the figure.
supporting img

Ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T$
Given that
$R = 8 \frac{J}{K . m o l e}, T = 300 K$
$N_{A} = 6 \times 10^{23} m o l e^{- 1}$
$P = 4.0 \times 10^{- 15} \times 10^{5} N / m^{2}$
$\frac{V}{N} = \frac{R T}{P N_{A}} = \frac{8 \times 300}{4.0 \times 10^{- 15} \times 10^{5} \times 6 \times 10^{23}}$
$λ^{3} = \frac{V}{N} = 10^{- 11} m^{3}$
$λ = 2.15 \times 10^{- 4} = 0.215 m m$

JEE - 2014

PHXI13:KINETIC THEORY

360310 The mean free path of collision of gas molecules varies with its diameter $(d)$ of the molecules as

1

d^{- 1}

2

d^{- 2}

3

d^{- 3}

4

d^{- 4}

Explanation:

Mean free path of gas molecules,
$λ = \frac{λ_{1} + λ_{2} + \dots + λ_{n}}{n}$
$λ = \frac{1}{\sqrt{2} π n d^{2}}$
$\Rightarrow λ \propto d^{- 2}$

PHXI13:KINETIC THEORY

360311 Assertion :
Mean free path of a gas molecule varies inversely as density of the gas.
Reason :
Mean free path varies inversely as temperature of the gas.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The mean free path of a gas molecule is the average distance between two successive collisions. It is represented by $λ$ .
$λ = \frac{k_{B} T}{\sqrt{2} π σ^{2} ρ}$
Here, $σ =$ diameter of molecule $k_{B} =$ Boltzmann's constant
Hence, mean free path varies inversely as density of the gas. The mean free path varies directly as the temperature and inversely as the pressure of the gas. So option (3) is correct.

PHXI13:KINETIC THEORY

360312 The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as:

1

\frac{1}{\sqrt{2} n π d^{2}}

2

\frac{1}{\sqrt{2} n^{2} π d^{2}}

3

\frac{1}{\sqrt{2} n^{2} π^{2} d^{2}}

4

\frac{1}{\sqrt{2} n π d}

Explanation:

Mean free path for a gas sample
$λ_{m} = \frac{1}{\sqrt{2} π d^{2} n}$
where $d$ is diameter of a gas molecule and $n$ is molecular density

NEET - 2020

PHXI13:KINETIC THEORY

360313 Modern vaccum pumps can evacuate a vessel down to a pressure of $4.0 \times 10^{- 15} a t m$ . At room temperature (300 K). Taking $R = 8.0 J K^{- 1} m o l e^{- 1}$ , $1 a t m = 10^{5} P a$ and $N_{A v o g a d r o} = 6 \times 10^{23} m o l e^{- 1}$ , the mean distance between molecules of gas in an evacuated vessel will be of the order of:

1

0.2 μ m

2

0.2 m m

3

0.2 c m

4

0.2 n m

Explanation:

From the given information we calculate the average volume occupied by each molecule (not the volume of molecule). The average volume occupied or share of each molecule is equal to the ratio of total volume divides by total number of molecules. This volume can be approximated as a cube of side length $(λ)$ . The cube root of the average volume per molecule is equal to $(λ)$ . Here the value $(λ)$ is also called as mean free path and it is equal to average distance between two molecules of a gas as shown in the figure.
supporting img

Ideal gas equation is
$P V = n R T = \frac{N}{N_{A}} R T$
Given that
$R = 8 \frac{J}{K . m o l e}, T = 300 K$
$N_{A} = 6 \times 10^{23} m o l e^{- 1}$
$P = 4.0 \times 10^{- 15} \times 10^{5} N / m^{2}$
$\frac{V}{N} = \frac{R T}{P N_{A}} = \frac{8 \times 300}{4.0 \times 10^{- 15} \times 10^{5} \times 6 \times 10^{23}}$
$λ^{3} = \frac{V}{N} = 10^{- 11} m^{3}$
$λ = 2.15 \times 10^{- 4} = 0.215 m m$

JEE - 2014