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Kinetic Theory of Gases

139210 The temperature of an ideal gas is increased from $100 K$ to $400 K$ . If the rms speed of the gas molecule is $v$ at $100 K$ , then at $400 K$ it becomes

1

2 v

2

4 v

3

0.5 v

4

0.25 v

5

v

Explanation:

A $T_{1} = 100 K, T_{2} = 400 K$
${(v_{1})}_{rms} = v$
$∴ v_{rms} = \sqrt{\frac{3 K_{b} T}{m}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{1})}_{rms}}{{(v_{2})}_{rms}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{v}{{(v_{2})}_{rms}} = \sqrt{\frac{100}{400}}$
${(v_{2})}_{rms} = 2 v$

Kerala CEE -2018

Kinetic Theory of Gases

139211 For a molecule of an ideal gas, the number density is $2 \sqrt{2} \times 10^{8} {cm}^{- 3}$ and the mean free path is $\frac{10^{- 2}}{π} cm$ . The diameter of the gas molecule is

1

5 \times 10^{- 4} cm

}$

2

0.5 \times 10^{- 4} cm

3 $2.5 \times 10^{-4} \mathrm{~cm

4

4 \times 10^{- 4} cm

Explanation:

A Number density $(n) = 2 \sqrt{2} \times 10^{8} {cm}^{- 3}$
Mean free path $(λ) = \frac{10^{- 2}}{π} cm$
$λ = \frac{1}{\sqrt{2} π {nd}^{2}}$
$d^{2} = \frac{1}{\sqrt{2} π n λ}$
$d^{2} = \frac{π}{\sqrt{2} \times π \times 2 \sqrt{2} \times 10^{8} \times 10^{- 2}}$
$d^{2} = \frac{1}{4 \times 10^{6}}$
$d = \frac{1}{2 \times 10^{3}} = 0.5 \times 10^{- 3} cm$
$d = 5 \times 10^{- 4} cm$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139214 The root mean square (rms) velocity of an ideal gas at temperature $T$ is $v$ . If the temperature is increased to $4 T$ , the $rms$ velocity of the gas is

1

\sqrt{3} v

2

\sqrt{2} v

3

2 v

4

3 v

Explanation:

C Given that,
${(v_{rms})}_{1} = v$
$T_{1} = T$
$T_{2} = 4 T$
We know that,
$(v_{rms}) = \sqrt{\frac{3 RT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \frac{\sqrt{T}}{\sqrt{4 T}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 . {(v_{rms})}_{1}$
$v_{2} = 2 v$

TS- EAMCET-05.05.2018

Kinetic Theory of Gases

139215 The mass of $H_{2}$ molecule is $3.32 \times 10^{- 24} g$ . If $10^{23}$ hydrogen molecules per second strike $2 {cm}^{2}$ of wall at an angle of $45^{\circ}$ with the normal, while moving with a speed of $10^{5} cm / s$ , the pressure exterted on the wall is nearly.

1

1350 N / m^{2}

2

2350 N / m^{2}

3

3320 N / m^{2}

4

1660 N / m^{2}

Explanation:

B Given that,
Mass of hydrogen $= 3.32 \times 10^{- 24} gm$
Angle with normal plane $= 45^{\circ}$
$= 3.32 \times 10^{- 27} kg$
Speed $= 10^{5} cm / s = 10^{3} m / s$
Area $= 2 \times 10^{- 4} m^{2}$

Change in momentum along y-axis
$Δ p_{y} = mn (u \cos 45^{\circ} - (- u \cos 45^{\circ}))$
$Δ p_{y} = mn (u \cos 45^{\circ} + u \cos 45^{\circ})$
$Δ p_{y} = 2 mn (u \cos 45^{\circ})$
Force exerted by molecules on the surface of wall-
$F =$ Rate of change of momentum $= \frac{Δ p}{Δ t}$
$F = \frac{2 mnu \cos 45^{\circ}}{1}$
$F = 2 \times 3.32 \times 10^{- 27} \times 10^{23} \times 10^{3} \times \frac{1}{\sqrt{2}}$
$F = 0.47 N$
Pressure $= \frac{F}{A} = \frac{0.47}{2 \times 10^{- 4}} N / m^{2}$
Pressure $= 2350 N / m^{2}$

BITSAT-2018

Kinetic Theory of Gases

139210 The temperature of an ideal gas is increased from $100 K$ to $400 K$ . If the rms speed of the gas molecule is $v$ at $100 K$ , then at $400 K$ it becomes

1

2 v

2

4 v

3

0.5 v

4

0.25 v

5

v

Explanation:

A $T_{1} = 100 K, T_{2} = 400 K$
${(v_{1})}_{rms} = v$
$∴ v_{rms} = \sqrt{\frac{3 K_{b} T}{m}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{1})}_{rms}}{{(v_{2})}_{rms}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{v}{{(v_{2})}_{rms}} = \sqrt{\frac{100}{400}}$
${(v_{2})}_{rms} = 2 v$

Kerala CEE -2018

Kinetic Theory of Gases

139211 For a molecule of an ideal gas, the number density is $2 \sqrt{2} \times 10^{8} {cm}^{- 3}$ and the mean free path is $\frac{10^{- 2}}{π} cm$ . The diameter of the gas molecule is

1

5 \times 10^{- 4} cm

}$

2

0.5 \times 10^{- 4} cm

3 $2.5 \times 10^{-4} \mathrm{~cm

4

4 \times 10^{- 4} cm

Explanation:

A Number density $(n) = 2 \sqrt{2} \times 10^{8} {cm}^{- 3}$
Mean free path $(λ) = \frac{10^{- 2}}{π} cm$
$λ = \frac{1}{\sqrt{2} π {nd}^{2}}$
$d^{2} = \frac{1}{\sqrt{2} π n λ}$
$d^{2} = \frac{π}{\sqrt{2} \times π \times 2 \sqrt{2} \times 10^{8} \times 10^{- 2}}$
$d^{2} = \frac{1}{4 \times 10^{6}}$
$d = \frac{1}{2 \times 10^{3}} = 0.5 \times 10^{- 3} cm$
$d = 5 \times 10^{- 4} cm$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139214 The root mean square (rms) velocity of an ideal gas at temperature $T$ is $v$ . If the temperature is increased to $4 T$ , the $rms$ velocity of the gas is

1

\sqrt{3} v

2

\sqrt{2} v

3

2 v

4

3 v

Explanation:

C Given that,
${(v_{rms})}_{1} = v$
$T_{1} = T$
$T_{2} = 4 T$
We know that,
$(v_{rms}) = \sqrt{\frac{3 RT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \frac{\sqrt{T}}{\sqrt{4 T}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 . {(v_{rms})}_{1}$
$v_{2} = 2 v$

TS- EAMCET-05.05.2018

Kinetic Theory of Gases

139215 The mass of $H_{2}$ molecule is $3.32 \times 10^{- 24} g$ . If $10^{23}$ hydrogen molecules per second strike $2 {cm}^{2}$ of wall at an angle of $45^{\circ}$ with the normal, while moving with a speed of $10^{5} cm / s$ , the pressure exterted on the wall is nearly.

1

1350 N / m^{2}

2

2350 N / m^{2}

3

3320 N / m^{2}

4

1660 N / m^{2}

Explanation:

B Given that,
Mass of hydrogen $= 3.32 \times 10^{- 24} gm$
Angle with normal plane $= 45^{\circ}$
$= 3.32 \times 10^{- 27} kg$
Speed $= 10^{5} cm / s = 10^{3} m / s$
Area $= 2 \times 10^{- 4} m^{2}$

Change in momentum along y-axis
$Δ p_{y} = mn (u \cos 45^{\circ} - (- u \cos 45^{\circ}))$
$Δ p_{y} = mn (u \cos 45^{\circ} + u \cos 45^{\circ})$
$Δ p_{y} = 2 mn (u \cos 45^{\circ})$
Force exerted by molecules on the surface of wall-
$F =$ Rate of change of momentum $= \frac{Δ p}{Δ t}$
$F = \frac{2 mnu \cos 45^{\circ}}{1}$
$F = 2 \times 3.32 \times 10^{- 27} \times 10^{23} \times 10^{3} \times \frac{1}{\sqrt{2}}$
$F = 0.47 N$
Pressure $= \frac{F}{A} = \frac{0.47}{2 \times 10^{- 4}} N / m^{2}$
Pressure $= 2350 N / m^{2}$

BITSAT-2018

Kinetic Theory of Gases

139210 The temperature of an ideal gas is increased from $100 K$ to $400 K$ . If the rms speed of the gas molecule is $v$ at $100 K$ , then at $400 K$ it becomes

1

2 v

2

4 v

3

0.5 v

4

0.25 v

5

v

Explanation:

A $T_{1} = 100 K, T_{2} = 400 K$
${(v_{1})}_{rms} = v$
$∴ v_{rms} = \sqrt{\frac{3 K_{b} T}{m}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{1})}_{rms}}{{(v_{2})}_{rms}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{v}{{(v_{2})}_{rms}} = \sqrt{\frac{100}{400}}$
${(v_{2})}_{rms} = 2 v$

Kerala CEE -2018

Kinetic Theory of Gases

139211 For a molecule of an ideal gas, the number density is $2 \sqrt{2} \times 10^{8} {cm}^{- 3}$ and the mean free path is $\frac{10^{- 2}}{π} cm$ . The diameter of the gas molecule is

1

5 \times 10^{- 4} cm

}$

2

0.5 \times 10^{- 4} cm

3 $2.5 \times 10^{-4} \mathrm{~cm

4

4 \times 10^{- 4} cm

Explanation:

A Number density $(n) = 2 \sqrt{2} \times 10^{8} {cm}^{- 3}$
Mean free path $(λ) = \frac{10^{- 2}}{π} cm$
$λ = \frac{1}{\sqrt{2} π {nd}^{2}}$
$d^{2} = \frac{1}{\sqrt{2} π n λ}$
$d^{2} = \frac{π}{\sqrt{2} \times π \times 2 \sqrt{2} \times 10^{8} \times 10^{- 2}}$
$d^{2} = \frac{1}{4 \times 10^{6}}$
$d = \frac{1}{2 \times 10^{3}} = 0.5 \times 10^{- 3} cm$
$d = 5 \times 10^{- 4} cm$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139214 The root mean square (rms) velocity of an ideal gas at temperature $T$ is $v$ . If the temperature is increased to $4 T$ , the $rms$ velocity of the gas is

1

\sqrt{3} v

2

\sqrt{2} v

3

2 v

4

3 v

Explanation:

C Given that,
${(v_{rms})}_{1} = v$
$T_{1} = T$
$T_{2} = 4 T$
We know that,
$(v_{rms}) = \sqrt{\frac{3 RT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \frac{\sqrt{T}}{\sqrt{4 T}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 . {(v_{rms})}_{1}$
$v_{2} = 2 v$

TS- EAMCET-05.05.2018

Kinetic Theory of Gases

139215 The mass of $H_{2}$ molecule is $3.32 \times 10^{- 24} g$ . If $10^{23}$ hydrogen molecules per second strike $2 {cm}^{2}$ of wall at an angle of $45^{\circ}$ with the normal, while moving with a speed of $10^{5} cm / s$ , the pressure exterted on the wall is nearly.

1

1350 N / m^{2}

2

2350 N / m^{2}

3

3320 N / m^{2}

4

1660 N / m^{2}

Explanation:

B Given that,
Mass of hydrogen $= 3.32 \times 10^{- 24} gm$
Angle with normal plane $= 45^{\circ}$
$= 3.32 \times 10^{- 27} kg$
Speed $= 10^{5} cm / s = 10^{3} m / s$
Area $= 2 \times 10^{- 4} m^{2}$

Change in momentum along y-axis
$Δ p_{y} = mn (u \cos 45^{\circ} - (- u \cos 45^{\circ}))$
$Δ p_{y} = mn (u \cos 45^{\circ} + u \cos 45^{\circ})$
$Δ p_{y} = 2 mn (u \cos 45^{\circ})$
Force exerted by molecules on the surface of wall-
$F =$ Rate of change of momentum $= \frac{Δ p}{Δ t}$
$F = \frac{2 mnu \cos 45^{\circ}}{1}$
$F = 2 \times 3.32 \times 10^{- 27} \times 10^{23} \times 10^{3} \times \frac{1}{\sqrt{2}}$
$F = 0.47 N$
Pressure $= \frac{F}{A} = \frac{0.47}{2 \times 10^{- 4}} N / m^{2}$
Pressure $= 2350 N / m^{2}$

BITSAT-2018

Kinetic Theory of Gases

139210 The temperature of an ideal gas is increased from $100 K$ to $400 K$ . If the rms speed of the gas molecule is $v$ at $100 K$ , then at $400 K$ it becomes

1

2 v

2

4 v

3

0.5 v

4

0.25 v

5

v

Explanation:

A $T_{1} = 100 K, T_{2} = 400 K$
${(v_{1})}_{rms} = v$
$∴ v_{rms} = \sqrt{\frac{3 K_{b} T}{m}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{1})}_{rms}}{{(v_{2})}_{rms}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{v}{{(v_{2})}_{rms}} = \sqrt{\frac{100}{400}}$
${(v_{2})}_{rms} = 2 v$

Kerala CEE -2018

Kinetic Theory of Gases

139211 For a molecule of an ideal gas, the number density is $2 \sqrt{2} \times 10^{8} {cm}^{- 3}$ and the mean free path is $\frac{10^{- 2}}{π} cm$ . The diameter of the gas molecule is

1

5 \times 10^{- 4} cm

}$

2

0.5 \times 10^{- 4} cm

3 $2.5 \times 10^{-4} \mathrm{~cm

4

4 \times 10^{- 4} cm

Explanation:

A Number density $(n) = 2 \sqrt{2} \times 10^{8} {cm}^{- 3}$
Mean free path $(λ) = \frac{10^{- 2}}{π} cm$
$λ = \frac{1}{\sqrt{2} π {nd}^{2}}$
$d^{2} = \frac{1}{\sqrt{2} π n λ}$
$d^{2} = \frac{π}{\sqrt{2} \times π \times 2 \sqrt{2} \times 10^{8} \times 10^{- 2}}$
$d^{2} = \frac{1}{4 \times 10^{6}}$
$d = \frac{1}{2 \times 10^{3}} = 0.5 \times 10^{- 3} cm$
$d = 5 \times 10^{- 4} cm$

AP EAMCET (22.04.2018) Shift-1

Kinetic Theory of Gases

139214 The root mean square (rms) velocity of an ideal gas at temperature $T$ is $v$ . If the temperature is increased to $4 T$ , the $rms$ velocity of the gas is

1

\sqrt{3} v

2

\sqrt{2} v

3

2 v

4

3 v

Explanation:

C Given that,
${(v_{rms})}_{1} = v$
$T_{1} = T$
$T_{2} = 4 T$
We know that,
$(v_{rms}) = \sqrt{\frac{3 RT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \frac{\sqrt{T}}{\sqrt{4 T}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 . {(v_{rms})}_{1}$
$v_{2} = 2 v$

TS- EAMCET-05.05.2018

Kinetic Theory of Gases

139215 The mass of $H_{2}$ molecule is $3.32 \times 10^{- 24} g$ . If $10^{23}$ hydrogen molecules per second strike $2 {cm}^{2}$ of wall at an angle of $45^{\circ}$ with the normal, while moving with a speed of $10^{5} cm / s$ , the pressure exterted on the wall is nearly.

1

1350 N / m^{2}

2

2350 N / m^{2}

3

3320 N / m^{2}

4

1660 N / m^{2}

Explanation:

B Given that,
Mass of hydrogen $= 3.32 \times 10^{- 24} gm$
Angle with normal plane $= 45^{\circ}$
$= 3.32 \times 10^{- 27} kg$
Speed $= 10^{5} cm / s = 10^{3} m / s$
Area $= 2 \times 10^{- 4} m^{2}$

Change in momentum along y-axis
$Δ p_{y} = mn (u \cos 45^{\circ} - (- u \cos 45^{\circ}))$
$Δ p_{y} = mn (u \cos 45^{\circ} + u \cos 45^{\circ})$
$Δ p_{y} = 2 mn (u \cos 45^{\circ})$
Force exerted by molecules on the surface of wall-
$F =$ Rate of change of momentum $= \frac{Δ p}{Δ t}$
$F = \frac{2 mnu \cos 45^{\circ}}{1}$
$F = 2 \times 3.32 \times 10^{- 27} \times 10^{23} \times 10^{3} \times \frac{1}{\sqrt{2}}$
$F = 0.47 N$
Pressure $= \frac{F}{A} = \frac{0.47}{2 \times 10^{- 4}} N / m^{2}$
Pressure $= 2350 N / m^{2}$

BITSAT-2018

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