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PHXI13:KINETIC THEORY

360234 The temperature of a gas is $- 50^{\circ} C$ . To what temeprature the gas should be heated so that the rms speed is increased by 3 times?

1

3295^{\circ} C

2

3097 K

3

223 K

4

669^{\circ} C

Explanation:

$v_{r m s} \propto \sqrt{T}$
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
Let initial speed be $v$
As speed is increased by 3 times so final speed become $4 v$ .
$\frac{v}{4 v} = \sqrt{\frac{273 - 50}{T}}$
$\Rightarrow \sqrt{\frac{223}{T}} = \frac{1}{4}$
$\Rightarrow T = 16 \times 223$
$= 3568 K$
$\Rightarrow t = (3568 - 273)^{\circ} C = 3295^{\circ} C$ .
Correct option is (1).

NEET - 2023

PHXI13:KINETIC THEORY

360235 In the isothermal expansion of $10 g$ of gas from volume $V$ to $2 V$ the work done by the gas is $575 J$ . What is the root mean square speed of the molecules of the gas at that temperature?

1

398 m / s

2

520 m / s

3

499 m / s

4

532 m / s

Explanation:

In isothermal process temperature remains constant.
$v_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P V}{M n}} = \sqrt{\frac{3 P V}{m}} (1)$
Isothermal work done is
$W = n R T \ln \frac{V_{2}}{V_{1}} = P V \ln 2$
Given that $W = 575 J = P V \ln 2$
$\Rightarrow P V = \frac{575}{\ln (2)} (2)$
From eq's (1) & (2) we get
$v_{r m s} = \sqrt{\frac{3 \times 575}{m (\ln 2)}} = \sqrt{\frac{3 \times 575}{10^{- 2} \times 0.693}} = 499 m / s$

JEE - 2013

PHXI13:KINETIC THEORY

360236 Given molecular weight of hydrogen molecule is $M = 2.016 \times 10^{- 3} k g / m o l$ . Calculate the root mean square speed of hydrogen molecules $(H_{2})$ at $373.15 K (100^{\circ} C)$ ,

1

2.15 k m / s

2

3.25 k m / s

3

4.22 k m / s

4

1.25 k m / s

Explanation:

$V_{r . m . s .} = \sqrt{\frac{3 R T}{M}}$
$C = \sqrt{\frac{3 \times 8.314 \times 373.15}{2.016 \times 10^{- 3}}} = \sqrt{4.617 \times 10^{6}}$
$= 2.148 k m / s$

PHXI13:KINETIC THEORY

360237 A gas molecule of mass $M$ at the surface of the Earth has kinetic energy equivalent to $0^{\circ} C$ . If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. ( $k_{B}$ is Boltzmann constant).

1 0

2

\frac{273 k_{B}}{2 M g}

3

\frac{546 k_{B}}{3 M g}

4

\frac{819 k_{B}}{2 M g}

Explanation:

Kinetic energy of each molecule,
$K . E = \frac{3}{2} K_{B} T$
Given that $T = 273 K$
Height attained by the gas molecule, $h = ?$
$\begin{aligned} K \cdot E = \frac{3}{2} K_{B} (273) = \frac{819 K_{B}}{2} \\ K \cdot E = P \cdot E \\ \Rightarrow \frac{819 K_{B}}{2} = M g h \Rightarrow h = \frac{819 K_{B}}{2 M g} \end{aligned}$

JEE - 2014

PHXI13:KINETIC THEORY

360234 The temperature of a gas is $- 50^{\circ} C$ . To what temeprature the gas should be heated so that the rms speed is increased by 3 times?

1

3295^{\circ} C

2

3097 K

3

223 K

4

669^{\circ} C

Explanation:

$v_{r m s} \propto \sqrt{T}$
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
Let initial speed be $v$
As speed is increased by 3 times so final speed become $4 v$ .
$\frac{v}{4 v} = \sqrt{\frac{273 - 50}{T}}$
$\Rightarrow \sqrt{\frac{223}{T}} = \frac{1}{4}$
$\Rightarrow T = 16 \times 223$
$= 3568 K$
$\Rightarrow t = (3568 - 273)^{\circ} C = 3295^{\circ} C$ .
Correct option is (1).

NEET - 2023

PHXI13:KINETIC THEORY

360235 In the isothermal expansion of $10 g$ of gas from volume $V$ to $2 V$ the work done by the gas is $575 J$ . What is the root mean square speed of the molecules of the gas at that temperature?

1

398 m / s

2

520 m / s

3

499 m / s

4

532 m / s

Explanation:

In isothermal process temperature remains constant.
$v_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P V}{M n}} = \sqrt{\frac{3 P V}{m}} (1)$
Isothermal work done is
$W = n R T \ln \frac{V_{2}}{V_{1}} = P V \ln 2$
Given that $W = 575 J = P V \ln 2$
$\Rightarrow P V = \frac{575}{\ln (2)} (2)$
From eq's (1) & (2) we get
$v_{r m s} = \sqrt{\frac{3 \times 575}{m (\ln 2)}} = \sqrt{\frac{3 \times 575}{10^{- 2} \times 0.693}} = 499 m / s$

JEE - 2013

PHXI13:KINETIC THEORY

360236 Given molecular weight of hydrogen molecule is $M = 2.016 \times 10^{- 3} k g / m o l$ . Calculate the root mean square speed of hydrogen molecules $(H_{2})$ at $373.15 K (100^{\circ} C)$ ,

1

2.15 k m / s

2

3.25 k m / s

3

4.22 k m / s

4

1.25 k m / s

Explanation:

$V_{r . m . s .} = \sqrt{\frac{3 R T}{M}}$
$C = \sqrt{\frac{3 \times 8.314 \times 373.15}{2.016 \times 10^{- 3}}} = \sqrt{4.617 \times 10^{6}}$
$= 2.148 k m / s$

PHXI13:KINETIC THEORY

360237 A gas molecule of mass $M$ at the surface of the Earth has kinetic energy equivalent to $0^{\circ} C$ . If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. ( $k_{B}$ is Boltzmann constant).

1 0

2

\frac{273 k_{B}}{2 M g}

3

\frac{546 k_{B}}{3 M g}

4

\frac{819 k_{B}}{2 M g}

Explanation:

Kinetic energy of each molecule,
$K . E = \frac{3}{2} K_{B} T$
Given that $T = 273 K$
Height attained by the gas molecule, $h = ?$
$\begin{aligned} K \cdot E = \frac{3}{2} K_{B} (273) = \frac{819 K_{B}}{2} \\ K \cdot E = P \cdot E \\ \Rightarrow \frac{819 K_{B}}{2} = M g h \Rightarrow h = \frac{819 K_{B}}{2 M g} \end{aligned}$

JEE - 2014

PHXI13:KINETIC THEORY

360234 The temperature of a gas is $- 50^{\circ} C$ . To what temeprature the gas should be heated so that the rms speed is increased by 3 times?

1

3295^{\circ} C

2

3097 K

3

223 K

4

669^{\circ} C

Explanation:

$v_{r m s} \propto \sqrt{T}$
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
Let initial speed be $v$
As speed is increased by 3 times so final speed become $4 v$ .
$\frac{v}{4 v} = \sqrt{\frac{273 - 50}{T}}$
$\Rightarrow \sqrt{\frac{223}{T}} = \frac{1}{4}$
$\Rightarrow T = 16 \times 223$
$= 3568 K$
$\Rightarrow t = (3568 - 273)^{\circ} C = 3295^{\circ} C$ .
Correct option is (1).

NEET - 2023

PHXI13:KINETIC THEORY

360235 In the isothermal expansion of $10 g$ of gas from volume $V$ to $2 V$ the work done by the gas is $575 J$ . What is the root mean square speed of the molecules of the gas at that temperature?

1

398 m / s

2

520 m / s

3

499 m / s

4

532 m / s

Explanation:

In isothermal process temperature remains constant.
$v_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P V}{M n}} = \sqrt{\frac{3 P V}{m}} (1)$
Isothermal work done is
$W = n R T \ln \frac{V_{2}}{V_{1}} = P V \ln 2$
Given that $W = 575 J = P V \ln 2$
$\Rightarrow P V = \frac{575}{\ln (2)} (2)$
From eq's (1) & (2) we get
$v_{r m s} = \sqrt{\frac{3 \times 575}{m (\ln 2)}} = \sqrt{\frac{3 \times 575}{10^{- 2} \times 0.693}} = 499 m / s$

JEE - 2013

PHXI13:KINETIC THEORY

360236 Given molecular weight of hydrogen molecule is $M = 2.016 \times 10^{- 3} k g / m o l$ . Calculate the root mean square speed of hydrogen molecules $(H_{2})$ at $373.15 K (100^{\circ} C)$ ,

1

2.15 k m / s

2

3.25 k m / s

3

4.22 k m / s

4

1.25 k m / s

Explanation:

$V_{r . m . s .} = \sqrt{\frac{3 R T}{M}}$
$C = \sqrt{\frac{3 \times 8.314 \times 373.15}{2.016 \times 10^{- 3}}} = \sqrt{4.617 \times 10^{6}}$
$= 2.148 k m / s$

PHXI13:KINETIC THEORY

360237 A gas molecule of mass $M$ at the surface of the Earth has kinetic energy equivalent to $0^{\circ} C$ . If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. ( $k_{B}$ is Boltzmann constant).

1 0

2

\frac{273 k_{B}}{2 M g}

3

\frac{546 k_{B}}{3 M g}

4

\frac{819 k_{B}}{2 M g}

Explanation:

Kinetic energy of each molecule,
$K . E = \frac{3}{2} K_{B} T$
Given that $T = 273 K$
Height attained by the gas molecule, $h = ?$
$\begin{aligned} K \cdot E = \frac{3}{2} K_{B} (273) = \frac{819 K_{B}}{2} \\ K \cdot E = P \cdot E \\ \Rightarrow \frac{819 K_{B}}{2} = M g h \Rightarrow h = \frac{819 K_{B}}{2 M g} \end{aligned}$

JEE - 2014

PHXI13:KINETIC THEORY

360234 The temperature of a gas is $- 50^{\circ} C$ . To what temeprature the gas should be heated so that the rms speed is increased by 3 times?

1

3295^{\circ} C

2

3097 K

3

223 K

4

669^{\circ} C

Explanation:

$v_{r m s} \propto \sqrt{T}$
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
Let initial speed be $v$
As speed is increased by 3 times so final speed become $4 v$ .
$\frac{v}{4 v} = \sqrt{\frac{273 - 50}{T}}$
$\Rightarrow \sqrt{\frac{223}{T}} = \frac{1}{4}$
$\Rightarrow T = 16 \times 223$
$= 3568 K$
$\Rightarrow t = (3568 - 273)^{\circ} C = 3295^{\circ} C$ .
Correct option is (1).

NEET - 2023

PHXI13:KINETIC THEORY

360235 In the isothermal expansion of $10 g$ of gas from volume $V$ to $2 V$ the work done by the gas is $575 J$ . What is the root mean square speed of the molecules of the gas at that temperature?

1

398 m / s

2

520 m / s

3

499 m / s

4

532 m / s

Explanation:

In isothermal process temperature remains constant.
$v_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P V}{M n}} = \sqrt{\frac{3 P V}{m}} (1)$
Isothermal work done is
$W = n R T \ln \frac{V_{2}}{V_{1}} = P V \ln 2$
Given that $W = 575 J = P V \ln 2$
$\Rightarrow P V = \frac{575}{\ln (2)} (2)$
From eq's (1) & (2) we get
$v_{r m s} = \sqrt{\frac{3 \times 575}{m (\ln 2)}} = \sqrt{\frac{3 \times 575}{10^{- 2} \times 0.693}} = 499 m / s$

JEE - 2013

PHXI13:KINETIC THEORY

360236 Given molecular weight of hydrogen molecule is $M = 2.016 \times 10^{- 3} k g / m o l$ . Calculate the root mean square speed of hydrogen molecules $(H_{2})$ at $373.15 K (100^{\circ} C)$ ,

1

2.15 k m / s

2

3.25 k m / s

3

4.22 k m / s

4

1.25 k m / s

Explanation:

$V_{r . m . s .} = \sqrt{\frac{3 R T}{M}}$
$C = \sqrt{\frac{3 \times 8.314 \times 373.15}{2.016 \times 10^{- 3}}} = \sqrt{4.617 \times 10^{6}}$
$= 2.148 k m / s$

PHXI13:KINETIC THEORY

360237 A gas molecule of mass $M$ at the surface of the Earth has kinetic energy equivalent to $0^{\circ} C$ . If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. ( $k_{B}$ is Boltzmann constant).

1 0

2

\frac{273 k_{B}}{2 M g}

3

\frac{546 k_{B}}{3 M g}

4

\frac{819 k_{B}}{2 M g}

Explanation:

Kinetic energy of each molecule,
$K . E = \frac{3}{2} K_{B} T$
Given that $T = 273 K$
Height attained by the gas molecule, $h = ?$
$\begin{aligned} K \cdot E = \frac{3}{2} K_{B} (273) = \frac{819 K_{B}}{2} \\ K \cdot E = P \cdot E \\ \Rightarrow \frac{819 K_{B}}{2} = M g h \Rightarrow h = \frac{819 K_{B}}{2 M g} \end{aligned}$

JEE - 2014

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