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Kinetic Theory of Gases

139226 Five gas molecules chosen at random are found to have speeds of $500, 600, 700, 800$ and 900 $m / s$ . Then which of the following statements is correct?

1 The root mean square speed and the average speed are the same.

2 The root mean square speed is

14 m / s

higher than the average speed.

3 The root mean square speed is

14 m / s

lower than the average speed.

4 The root mean square speed is

\sqrt{} 14 m / s

higher than the average speed.

Explanation:

B Given that, Speed of five gas molecule 500, 600, 700, 800, 900
So, average speed $= \frac{\sum v_{n}}{\sum n}$
$= \frac{500 + 600 + 700 + 800 + 900}{5}$
$= \frac{3500}{5}$
$V_{avg} = 700 m / \sec$ .
Rms velocity $= \sqrt{\frac{\sum (V)^{2}}{\sum n}}$
$= \sqrt{\frac{(500)^{2} + (600)^{2} + (700)^{2} + (800)^{2} + (900)^{2}}{5}}$
$= \sqrt{\frac{2550000}{5}}$
$= \sqrt{510000}$
$= 100 \sqrt{51}$
$= 100 \times 7.14$
$= 714 m / \sec$
Root means square velocity is greater than average velocity by $14 m / \sec$
$V_{rms} > V_{avg.}$
$714 > 700$

BITSAT-2017

Kinetic Theory of Gases

139228 If $m$ represents the mass of each molecule of a gas and $T$ , its absolute temperature, then the root mean square velocity of the gaseous molecule is proportional to

1

mT

(e)

{mT}^{- 1 / 2}

2

m^{1 / 2} T^{1 / 2}

3

m^{- 1 / 2} T

4

m^{- 1 / 2} T^{1 / 2}

Explanation:

D We know that, thermal equilibrium and kinetic interpretation of temperature-
$v_{rms} = \sqrt{\frac{3 RT}{m}}$
i.e., $v_{rms} \propto \sqrt{T}$
$v_{rms} \propto \frac{1}{\sqrt{m}}$
$v_{rms} = \sqrt{T} \times \frac{1}{\sqrt{m}} = T^{1 / 2} \frac{1}{m^{1 / 2}}$
$v_{rms} = m^{- 1 / 2} T^{1 / 2}$

Kerala CEE- 2014

Kinetic Theory of Gases

139232 During an experiment, an ideal gas is found to obey an additional law ${VP}^{2} =$ constant. The gas is initially at temperature $T$ and volume $V$ . What will be the temperature of the gas when it expand to a volume $2 V$ ?

1

\sqrt{3} T

2

T \sqrt{1 / 2}

3

T \sqrt{2}

4

T \sqrt{3}

Explanation:

C Given, $P^{2} V =$ constant
According to an ideal gas equation-
$PV = nRT$
$P = \frac{nRT}{V}$
Putting the value of $P$ in following equation,
${(\frac{nRT}{V})}^{2} \times V = constant$
$or, \frac{T^{2}}{V} = k$
$Since, \frac{T_{1}^{2}}{V_{1}} = \frac{T_{2}^{2}}{V_{2}}$
$T_{2}^{2} = \frac{V_{2} T_{1}^{2}}{V_{1}}$
$T_{2} = \sqrt{\frac{T_{1}^{2} V_{2}}{V_{1}}} (∵ T_{1} = T & V_{2} = 2 V_{1})$
$T_{2} = T \sqrt{2}$

UPSEE - 2013

Kinetic Theory of Gases

139233 The molecules of a given mass of gas have a root mean square velocity of $200 m s^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5} N m^{- 2}$ pressure. When the temperature is $127^{\circ} C$ and the pressure is $0.5 \times$ $10^{5} {Nm}^{- 2}$ , the root mean square velocity in ${ms}^{-}$ 1 , is

1

\frac{400}{\sqrt{3}}

2

100 \sqrt{2}

3

\frac{100 \sqrt{2}}{3}

4

\frac{100}{3}

Explanation:

A Given that,
$v_{rms} = 200 m / \sec .$
Temperature $(T_{1}) = 27^{\circ} C + 273 = 300 K$
Pressure $(P_{1}) = 1.0 \times 10^{5} N / m^{2}$
$T_{2} = 127^{\circ} C = 273 + 127 = 400 K$
Pressure $(P_{2}) = 0.5 \times 10^{5} N / m^{2}$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Then, $\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{v_{2}} = \sqrt{\frac{300}{400}}$
$\frac{200}{v_{2}} = \sqrt{\frac{3}{4}}$
$\frac{200}{v_{2}} = \frac{\sqrt{3}}{2}$
$v_{2} = \frac{400}{\sqrt{3}}$

JCECE-2010

Kinetic Theory of Gases

139226 Five gas molecules chosen at random are found to have speeds of $500, 600, 700, 800$ and 900 $m / s$ . Then which of the following statements is correct?

1 The root mean square speed and the average speed are the same.

2 The root mean square speed is

14 m / s

higher than the average speed.

3 The root mean square speed is

14 m / s

lower than the average speed.

4 The root mean square speed is

\sqrt{} 14 m / s

higher than the average speed.

Explanation:

B Given that, Speed of five gas molecule 500, 600, 700, 800, 900
So, average speed $= \frac{\sum v_{n}}{\sum n}$
$= \frac{500 + 600 + 700 + 800 + 900}{5}$
$= \frac{3500}{5}$
$V_{avg} = 700 m / \sec$ .
Rms velocity $= \sqrt{\frac{\sum (V)^{2}}{\sum n}}$
$= \sqrt{\frac{(500)^{2} + (600)^{2} + (700)^{2} + (800)^{2} + (900)^{2}}{5}}$
$= \sqrt{\frac{2550000}{5}}$
$= \sqrt{510000}$
$= 100 \sqrt{51}$
$= 100 \times 7.14$
$= 714 m / \sec$
Root means square velocity is greater than average velocity by $14 m / \sec$
$V_{rms} > V_{avg.}$
$714 > 700$

BITSAT-2017

Kinetic Theory of Gases

139228 If $m$ represents the mass of each molecule of a gas and $T$ , its absolute temperature, then the root mean square velocity of the gaseous molecule is proportional to

1

mT

(e)

{mT}^{- 1 / 2}

2

m^{1 / 2} T^{1 / 2}

3

m^{- 1 / 2} T

4

m^{- 1 / 2} T^{1 / 2}

Explanation:

D We know that, thermal equilibrium and kinetic interpretation of temperature-
$v_{rms} = \sqrt{\frac{3 RT}{m}}$
i.e., $v_{rms} \propto \sqrt{T}$
$v_{rms} \propto \frac{1}{\sqrt{m}}$
$v_{rms} = \sqrt{T} \times \frac{1}{\sqrt{m}} = T^{1 / 2} \frac{1}{m^{1 / 2}}$
$v_{rms} = m^{- 1 / 2} T^{1 / 2}$

Kerala CEE- 2014

Kinetic Theory of Gases

139232 During an experiment, an ideal gas is found to obey an additional law ${VP}^{2} =$ constant. The gas is initially at temperature $T$ and volume $V$ . What will be the temperature of the gas when it expand to a volume $2 V$ ?

1

\sqrt{3} T

2

T \sqrt{1 / 2}

3

T \sqrt{2}

4

T \sqrt{3}

Explanation:

C Given, $P^{2} V =$ constant
According to an ideal gas equation-
$PV = nRT$
$P = \frac{nRT}{V}$
Putting the value of $P$ in following equation,
${(\frac{nRT}{V})}^{2} \times V = constant$
$or, \frac{T^{2}}{V} = k$
$Since, \frac{T_{1}^{2}}{V_{1}} = \frac{T_{2}^{2}}{V_{2}}$
$T_{2}^{2} = \frac{V_{2} T_{1}^{2}}{V_{1}}$
$T_{2} = \sqrt{\frac{T_{1}^{2} V_{2}}{V_{1}}} (∵ T_{1} = T & V_{2} = 2 V_{1})$
$T_{2} = T \sqrt{2}$

UPSEE - 2013

Kinetic Theory of Gases

139233 The molecules of a given mass of gas have a root mean square velocity of $200 m s^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5} N m^{- 2}$ pressure. When the temperature is $127^{\circ} C$ and the pressure is $0.5 \times$ $10^{5} {Nm}^{- 2}$ , the root mean square velocity in ${ms}^{-}$ 1 , is

1

\frac{400}{\sqrt{3}}

2

100 \sqrt{2}

3

\frac{100 \sqrt{2}}{3}

4

\frac{100}{3}

Explanation:

A Given that,
$v_{rms} = 200 m / \sec .$
Temperature $(T_{1}) = 27^{\circ} C + 273 = 300 K$
Pressure $(P_{1}) = 1.0 \times 10^{5} N / m^{2}$
$T_{2} = 127^{\circ} C = 273 + 127 = 400 K$
Pressure $(P_{2}) = 0.5 \times 10^{5} N / m^{2}$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Then, $\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{v_{2}} = \sqrt{\frac{300}{400}}$
$\frac{200}{v_{2}} = \sqrt{\frac{3}{4}}$
$\frac{200}{v_{2}} = \frac{\sqrt{3}}{2}$
$v_{2} = \frac{400}{\sqrt{3}}$

JCECE-2010

Kinetic Theory of Gases

139226 Five gas molecules chosen at random are found to have speeds of $500, 600, 700, 800$ and 900 $m / s$ . Then which of the following statements is correct?

1 The root mean square speed and the average speed are the same.

2 The root mean square speed is

14 m / s

higher than the average speed.

3 The root mean square speed is

14 m / s

lower than the average speed.

4 The root mean square speed is

\sqrt{} 14 m / s

higher than the average speed.

Explanation:

B Given that, Speed of five gas molecule 500, 600, 700, 800, 900
So, average speed $= \frac{\sum v_{n}}{\sum n}$
$= \frac{500 + 600 + 700 + 800 + 900}{5}$
$= \frac{3500}{5}$
$V_{avg} = 700 m / \sec$ .
Rms velocity $= \sqrt{\frac{\sum (V)^{2}}{\sum n}}$
$= \sqrt{\frac{(500)^{2} + (600)^{2} + (700)^{2} + (800)^{2} + (900)^{2}}{5}}$
$= \sqrt{\frac{2550000}{5}}$
$= \sqrt{510000}$
$= 100 \sqrt{51}$
$= 100 \times 7.14$
$= 714 m / \sec$
Root means square velocity is greater than average velocity by $14 m / \sec$
$V_{rms} > V_{avg.}$
$714 > 700$

BITSAT-2017

Kinetic Theory of Gases

139228 If $m$ represents the mass of each molecule of a gas and $T$ , its absolute temperature, then the root mean square velocity of the gaseous molecule is proportional to

1

mT

(e)

{mT}^{- 1 / 2}

2

m^{1 / 2} T^{1 / 2}

3

m^{- 1 / 2} T

4

m^{- 1 / 2} T^{1 / 2}

Explanation:

D We know that, thermal equilibrium and kinetic interpretation of temperature-
$v_{rms} = \sqrt{\frac{3 RT}{m}}$
i.e., $v_{rms} \propto \sqrt{T}$
$v_{rms} \propto \frac{1}{\sqrt{m}}$
$v_{rms} = \sqrt{T} \times \frac{1}{\sqrt{m}} = T^{1 / 2} \frac{1}{m^{1 / 2}}$
$v_{rms} = m^{- 1 / 2} T^{1 / 2}$

Kerala CEE- 2014

Kinetic Theory of Gases

139232 During an experiment, an ideal gas is found to obey an additional law ${VP}^{2} =$ constant. The gas is initially at temperature $T$ and volume $V$ . What will be the temperature of the gas when it expand to a volume $2 V$ ?

1

\sqrt{3} T

2

T \sqrt{1 / 2}

3

T \sqrt{2}

4

T \sqrt{3}

Explanation:

C Given, $P^{2} V =$ constant
According to an ideal gas equation-
$PV = nRT$
$P = \frac{nRT}{V}$
Putting the value of $P$ in following equation,
${(\frac{nRT}{V})}^{2} \times V = constant$
$or, \frac{T^{2}}{V} = k$
$Since, \frac{T_{1}^{2}}{V_{1}} = \frac{T_{2}^{2}}{V_{2}}$
$T_{2}^{2} = \frac{V_{2} T_{1}^{2}}{V_{1}}$
$T_{2} = \sqrt{\frac{T_{1}^{2} V_{2}}{V_{1}}} (∵ T_{1} = T & V_{2} = 2 V_{1})$
$T_{2} = T \sqrt{2}$

UPSEE - 2013

Kinetic Theory of Gases

139233 The molecules of a given mass of gas have a root mean square velocity of $200 m s^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5} N m^{- 2}$ pressure. When the temperature is $127^{\circ} C$ and the pressure is $0.5 \times$ $10^{5} {Nm}^{- 2}$ , the root mean square velocity in ${ms}^{-}$ 1 , is

1

\frac{400}{\sqrt{3}}

2

100 \sqrt{2}

3

\frac{100 \sqrt{2}}{3}

4

\frac{100}{3}

Explanation:

A Given that,
$v_{rms} = 200 m / \sec .$
Temperature $(T_{1}) = 27^{\circ} C + 273 = 300 K$
Pressure $(P_{1}) = 1.0 \times 10^{5} N / m^{2}$
$T_{2} = 127^{\circ} C = 273 + 127 = 400 K$
Pressure $(P_{2}) = 0.5 \times 10^{5} N / m^{2}$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Then, $\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{v_{2}} = \sqrt{\frac{300}{400}}$
$\frac{200}{v_{2}} = \sqrt{\frac{3}{4}}$
$\frac{200}{v_{2}} = \frac{\sqrt{3}}{2}$
$v_{2} = \frac{400}{\sqrt{3}}$

JCECE-2010

Kinetic Theory of Gases

139226 Five gas molecules chosen at random are found to have speeds of $500, 600, 700, 800$ and 900 $m / s$ . Then which of the following statements is correct?

1 The root mean square speed and the average speed are the same.

2 The root mean square speed is

14 m / s

higher than the average speed.

3 The root mean square speed is

14 m / s

lower than the average speed.

4 The root mean square speed is

\sqrt{} 14 m / s

higher than the average speed.

Explanation:

B Given that, Speed of five gas molecule 500, 600, 700, 800, 900
So, average speed $= \frac{\sum v_{n}}{\sum n}$
$= \frac{500 + 600 + 700 + 800 + 900}{5}$
$= \frac{3500}{5}$
$V_{avg} = 700 m / \sec$ .
Rms velocity $= \sqrt{\frac{\sum (V)^{2}}{\sum n}}$
$= \sqrt{\frac{(500)^{2} + (600)^{2} + (700)^{2} + (800)^{2} + (900)^{2}}{5}}$
$= \sqrt{\frac{2550000}{5}}$
$= \sqrt{510000}$
$= 100 \sqrt{51}$
$= 100 \times 7.14$
$= 714 m / \sec$
Root means square velocity is greater than average velocity by $14 m / \sec$
$V_{rms} > V_{avg.}$
$714 > 700$

BITSAT-2017

Kinetic Theory of Gases

139228 If $m$ represents the mass of each molecule of a gas and $T$ , its absolute temperature, then the root mean square velocity of the gaseous molecule is proportional to

1

mT

(e)

{mT}^{- 1 / 2}

2

m^{1 / 2} T^{1 / 2}

3

m^{- 1 / 2} T

4

m^{- 1 / 2} T^{1 / 2}

Explanation:

D We know that, thermal equilibrium and kinetic interpretation of temperature-
$v_{rms} = \sqrt{\frac{3 RT}{m}}$
i.e., $v_{rms} \propto \sqrt{T}$
$v_{rms} \propto \frac{1}{\sqrt{m}}$
$v_{rms} = \sqrt{T} \times \frac{1}{\sqrt{m}} = T^{1 / 2} \frac{1}{m^{1 / 2}}$
$v_{rms} = m^{- 1 / 2} T^{1 / 2}$

Kerala CEE- 2014

Kinetic Theory of Gases

139232 During an experiment, an ideal gas is found to obey an additional law ${VP}^{2} =$ constant. The gas is initially at temperature $T$ and volume $V$ . What will be the temperature of the gas when it expand to a volume $2 V$ ?

1

\sqrt{3} T

2

T \sqrt{1 / 2}

3

T \sqrt{2}

4

T \sqrt{3}

Explanation:

C Given, $P^{2} V =$ constant
According to an ideal gas equation-
$PV = nRT$
$P = \frac{nRT}{V}$
Putting the value of $P$ in following equation,
${(\frac{nRT}{V})}^{2} \times V = constant$
$or, \frac{T^{2}}{V} = k$
$Since, \frac{T_{1}^{2}}{V_{1}} = \frac{T_{2}^{2}}{V_{2}}$
$T_{2}^{2} = \frac{V_{2} T_{1}^{2}}{V_{1}}$
$T_{2} = \sqrt{\frac{T_{1}^{2} V_{2}}{V_{1}}} (∵ T_{1} = T & V_{2} = 2 V_{1})$
$T_{2} = T \sqrt{2}$

UPSEE - 2013

Kinetic Theory of Gases

139233 The molecules of a given mass of gas have a root mean square velocity of $200 m s^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5} N m^{- 2}$ pressure. When the temperature is $127^{\circ} C$ and the pressure is $0.5 \times$ $10^{5} {Nm}^{- 2}$ , the root mean square velocity in ${ms}^{-}$ 1 , is

1

\frac{400}{\sqrt{3}}

2

100 \sqrt{2}

3

\frac{100 \sqrt{2}}{3}

4

\frac{100}{3}

Explanation:

A Given that,
$v_{rms} = 200 m / \sec .$
Temperature $(T_{1}) = 27^{\circ} C + 273 = 300 K$
Pressure $(P_{1}) = 1.0 \times 10^{5} N / m^{2}$
$T_{2} = 127^{\circ} C = 273 + 127 = 400 K$
Pressure $(P_{2}) = 0.5 \times 10^{5} N / m^{2}$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Then, $\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{v_{2}} = \sqrt{\frac{300}{400}}$
$\frac{200}{v_{2}} = \sqrt{\frac{3}{4}}$
$\frac{200}{v_{2}} = \frac{\sqrt{3}}{2}$
$v_{2} = \frac{400}{\sqrt{3}}$

JCECE-2010