QBManager

Kinetic Theory of Gases

139298 The temperature of a gas is raised from $27^{\circ} C$ to $927^{\circ} C$ . The root mean square speed

1 is

\sqrt{(\frac{927}{27})}

times the earlier value

2 remains the same

3 gets halved

4 gets doubled

Explanation:

D Given that,
$T_{1} = 27^{\circ} C = 300 K$
$T_{2} = 927^{\circ} C = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 kT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{1200}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 {(v_{rms})}_{1}$
Hence, root mean sequare speed of gas gets doubled.

AIPMT-1994

Kinetic Theory of Gases

139299 The mean free path of molecules of a gas, (radius $r$ ) is inversely proportional to

1

r^{3}

2

r^{2}

3

r

4

\sqrt{r}

Explanation:

B We know that,
mean free path $(γ) = \frac{1}{\sqrt{2} π} d^{2} n$
$d =$ diameter of molecules
$d = 2 r$
$γ = \frac{1}{\sqrt{2} \cdot π (2 r)^{2} \cdot n}$
$γ = \frac{1}{π^{2}}$
Hence, $γ \propto \frac{1}{r^{2}}$
Mean Free path of molecules of a gas is inversely proportional to $r^{2}$ .

AIPMT-2014

Kinetic Theory of Gases

139300 The molecules of a given mass of a gas have r.m.s. velocity of $200 {ms}^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5}$ ${Nm}^{- 2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} C$ and $0.05 \times 10^{5} {Nm}^{- 2}$ the rms velocity of its molecules in ${ms}^{- 1}$ is

1

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

2

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

3

\frac{100}{3}

4

100 \sqrt{2}

Explanation:

A Given that, ${(v_{rms})}_{1} = 200 m / s$
$T_{1} = 27^{\circ} C$
$= 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C$
$= 127 + 273 = 400 K$
$P_{1} = 1.0 \times 10^{5} {Nm}^{- 2}$
$P_{2} = .05 \times 10^{5} {Nm}^{- 2}$
${(v_{rms})}_{2} = ?$
We know that,
$V_{rms} velocity of gas molecule is -$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Hence,
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}$
${(v_{rms})}_{2} = \frac{2}{\sqrt{3}} \times 200$
${(v_{rms})}_{2} = \frac{400}{\sqrt{3}} m / s .$

NEET-2016

Kinetic Theory of Gases

139301 The r.m.s speed of hydrogen gas molecules at a certain temperature is $200 {ms}^{- 1}$ . If the absolute temperature of the gas is doubled and the hydrogen gas dissociates into atomic hydrogen, then the r.m.s speed will become

1

200 {ms}^{- 1}

2

400 {ms}^{- 1}

3

600 {ms}^{- 1}

4

800 {ms}^{- 1}

Explanation:

B Given that,

${(v_{rmm})}_{1} = 200 {ms}^{- 1}$
$T_{1} = T$
$T_{2} = 2 T$
$M_{1} = 2 M$
$M_{2} = M$
$We know that,$
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$Hence,$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}} \times \frac{M_{2}}{M_{1}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{T}{2 T} \times \frac{M}{2 M}}$
${(v_{rms})}_{2} = 2 \times 200$
${(v_{rms})}_{2} = 400 m s^{- 1}$

AP EAMCET(Medical)-2016

Kinetic Theory of Gases

139302 When the temperature of an ideal gas is increased by $600 K$ , the velocity of sound in the gas becomes $\sqrt{3}$ times the initial velocity in it. The initial temperature of the gas is:

1

- 73^{\circ} C

2

27^{\circ} C

3

127^{\circ} C

4

327^{\circ} C

Explanation:

B Given that,

$v_{1} = v$
$v_{2} = \sqrt{3} \cdot v$
$T_{1} = T K$
$T_{2} = T + 600 K$
$We know that,$
$v \propto \sqrt{T}$
$Hence,$
$\frac{\sqrt{3} v}{v} = \sqrt{\frac{T + 600}{T}}$
$\sqrt{3} = \sqrt{\frac{T + 600}{T}}$
$3 = \frac{T + 600}{T}$
$3 T = T + 600$
$T = 300 K$
$T = 300 - 273$
$T = 27^{\circ} C$

AP EAMCET(Medical)-2000

Kinetic Theory of Gases

139298 The temperature of a gas is raised from $27^{\circ} C$ to $927^{\circ} C$ . The root mean square speed

1 is

\sqrt{(\frac{927}{27})}

times the earlier value

2 remains the same

3 gets halved

4 gets doubled

Explanation:

D Given that,
$T_{1} = 27^{\circ} C = 300 K$
$T_{2} = 927^{\circ} C = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 kT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{1200}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 {(v_{rms})}_{1}$
Hence, root mean sequare speed of gas gets doubled.

AIPMT-1994

Kinetic Theory of Gases

139299 The mean free path of molecules of a gas, (radius $r$ ) is inversely proportional to

1

r^{3}

2

r^{2}

3

r

4

\sqrt{r}

Explanation:

B We know that,
mean free path $(γ) = \frac{1}{\sqrt{2} π} d^{2} n$
$d =$ diameter of molecules
$d = 2 r$
$γ = \frac{1}{\sqrt{2} \cdot π (2 r)^{2} \cdot n}$
$γ = \frac{1}{π^{2}}$
Hence, $γ \propto \frac{1}{r^{2}}$
Mean Free path of molecules of a gas is inversely proportional to $r^{2}$ .

AIPMT-2014

Kinetic Theory of Gases

139300 The molecules of a given mass of a gas have r.m.s. velocity of $200 {ms}^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5}$ ${Nm}^{- 2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} C$ and $0.05 \times 10^{5} {Nm}^{- 2}$ the rms velocity of its molecules in ${ms}^{- 1}$ is

1

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

2

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

3

\frac{100}{3}

4

100 \sqrt{2}

Explanation:

A Given that, ${(v_{rms})}_{1} = 200 m / s$
$T_{1} = 27^{\circ} C$
$= 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C$
$= 127 + 273 = 400 K$
$P_{1} = 1.0 \times 10^{5} {Nm}^{- 2}$
$P_{2} = .05 \times 10^{5} {Nm}^{- 2}$
${(v_{rms})}_{2} = ?$
We know that,
$V_{rms} velocity of gas molecule is -$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Hence,
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}$
${(v_{rms})}_{2} = \frac{2}{\sqrt{3}} \times 200$
${(v_{rms})}_{2} = \frac{400}{\sqrt{3}} m / s .$

NEET-2016

Kinetic Theory of Gases

139301 The r.m.s speed of hydrogen gas molecules at a certain temperature is $200 {ms}^{- 1}$ . If the absolute temperature of the gas is doubled and the hydrogen gas dissociates into atomic hydrogen, then the r.m.s speed will become

1

200 {ms}^{- 1}

2

400 {ms}^{- 1}

3

600 {ms}^{- 1}

4

800 {ms}^{- 1}

Explanation:

B Given that,

${(v_{rmm})}_{1} = 200 {ms}^{- 1}$
$T_{1} = T$
$T_{2} = 2 T$
$M_{1} = 2 M$
$M_{2} = M$
$We know that,$
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$Hence,$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}} \times \frac{M_{2}}{M_{1}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{T}{2 T} \times \frac{M}{2 M}}$
${(v_{rms})}_{2} = 2 \times 200$
${(v_{rms})}_{2} = 400 m s^{- 1}$

AP EAMCET(Medical)-2016

Kinetic Theory of Gases

139302 When the temperature of an ideal gas is increased by $600 K$ , the velocity of sound in the gas becomes $\sqrt{3}$ times the initial velocity in it. The initial temperature of the gas is:

1

- 73^{\circ} C

2

27^{\circ} C

3

127^{\circ} C

4

327^{\circ} C

Explanation:

B Given that,

$v_{1} = v$
$v_{2} = \sqrt{3} \cdot v$
$T_{1} = T K$
$T_{2} = T + 600 K$
$We know that,$
$v \propto \sqrt{T}$
$Hence,$
$\frac{\sqrt{3} v}{v} = \sqrt{\frac{T + 600}{T}}$
$\sqrt{3} = \sqrt{\frac{T + 600}{T}}$
$3 = \frac{T + 600}{T}$
$3 T = T + 600$
$T = 300 K$
$T = 300 - 273$
$T = 27^{\circ} C$

AP EAMCET(Medical)-2000

Kinetic Theory of Gases

139298 The temperature of a gas is raised from $27^{\circ} C$ to $927^{\circ} C$ . The root mean square speed

1 is

\sqrt{(\frac{927}{27})}

times the earlier value

2 remains the same

3 gets halved

4 gets doubled

Explanation:

D Given that,
$T_{1} = 27^{\circ} C = 300 K$
$T_{2} = 927^{\circ} C = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 kT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{1200}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 {(v_{rms})}_{1}$
Hence, root mean sequare speed of gas gets doubled.

AIPMT-1994

Kinetic Theory of Gases

139299 The mean free path of molecules of a gas, (radius $r$ ) is inversely proportional to

1

r^{3}

2

r^{2}

3

r

4

\sqrt{r}

Explanation:

B We know that,
mean free path $(γ) = \frac{1}{\sqrt{2} π} d^{2} n$
$d =$ diameter of molecules
$d = 2 r$
$γ = \frac{1}{\sqrt{2} \cdot π (2 r)^{2} \cdot n}$
$γ = \frac{1}{π^{2}}$
Hence, $γ \propto \frac{1}{r^{2}}$
Mean Free path of molecules of a gas is inversely proportional to $r^{2}$ .

AIPMT-2014

Kinetic Theory of Gases

139300 The molecules of a given mass of a gas have r.m.s. velocity of $200 {ms}^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5}$ ${Nm}^{- 2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} C$ and $0.05 \times 10^{5} {Nm}^{- 2}$ the rms velocity of its molecules in ${ms}^{- 1}$ is

1

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

2

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

3

\frac{100}{3}

4

100 \sqrt{2}

Explanation:

A Given that, ${(v_{rms})}_{1} = 200 m / s$
$T_{1} = 27^{\circ} C$
$= 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C$
$= 127 + 273 = 400 K$
$P_{1} = 1.0 \times 10^{5} {Nm}^{- 2}$
$P_{2} = .05 \times 10^{5} {Nm}^{- 2}$
${(v_{rms})}_{2} = ?$
We know that,
$V_{rms} velocity of gas molecule is -$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Hence,
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}$
${(v_{rms})}_{2} = \frac{2}{\sqrt{3}} \times 200$
${(v_{rms})}_{2} = \frac{400}{\sqrt{3}} m / s .$

NEET-2016

Kinetic Theory of Gases

139301 The r.m.s speed of hydrogen gas molecules at a certain temperature is $200 {ms}^{- 1}$ . If the absolute temperature of the gas is doubled and the hydrogen gas dissociates into atomic hydrogen, then the r.m.s speed will become

1

200 {ms}^{- 1}

2

400 {ms}^{- 1}

3

600 {ms}^{- 1}

4

800 {ms}^{- 1}

Explanation:

B Given that,

${(v_{rmm})}_{1} = 200 {ms}^{- 1}$
$T_{1} = T$
$T_{2} = 2 T$
$M_{1} = 2 M$
$M_{2} = M$
$We know that,$
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$Hence,$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}} \times \frac{M_{2}}{M_{1}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{T}{2 T} \times \frac{M}{2 M}}$
${(v_{rms})}_{2} = 2 \times 200$
${(v_{rms})}_{2} = 400 m s^{- 1}$

AP EAMCET(Medical)-2016

Kinetic Theory of Gases

139302 When the temperature of an ideal gas is increased by $600 K$ , the velocity of sound in the gas becomes $\sqrt{3}$ times the initial velocity in it. The initial temperature of the gas is:

1

- 73^{\circ} C

2

27^{\circ} C

3

127^{\circ} C

4

327^{\circ} C

Explanation:

B Given that,

$v_{1} = v$
$v_{2} = \sqrt{3} \cdot v$
$T_{1} = T K$
$T_{2} = T + 600 K$
$We know that,$
$v \propto \sqrt{T}$
$Hence,$
$\frac{\sqrt{3} v}{v} = \sqrt{\frac{T + 600}{T}}$
$\sqrt{3} = \sqrt{\frac{T + 600}{T}}$
$3 = \frac{T + 600}{T}$
$3 T = T + 600$
$T = 300 K$
$T = 300 - 273$
$T = 27^{\circ} C$

AP EAMCET(Medical)-2000

Kinetic Theory of Gases

139298 The temperature of a gas is raised from $27^{\circ} C$ to $927^{\circ} C$ . The root mean square speed

1 is

\sqrt{(\frac{927}{27})}

times the earlier value

2 remains the same

3 gets halved

4 gets doubled

Explanation:

D Given that,
$T_{1} = 27^{\circ} C = 300 K$
$T_{2} = 927^{\circ} C = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 kT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{1200}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 {(v_{rms})}_{1}$
Hence, root mean sequare speed of gas gets doubled.

AIPMT-1994

Kinetic Theory of Gases

139299 The mean free path of molecules of a gas, (radius $r$ ) is inversely proportional to

1

r^{3}

2

r^{2}

3

r

4

\sqrt{r}

Explanation:

B We know that,
mean free path $(γ) = \frac{1}{\sqrt{2} π} d^{2} n$
$d =$ diameter of molecules
$d = 2 r$
$γ = \frac{1}{\sqrt{2} \cdot π (2 r)^{2} \cdot n}$
$γ = \frac{1}{π^{2}}$
Hence, $γ \propto \frac{1}{r^{2}}$
Mean Free path of molecules of a gas is inversely proportional to $r^{2}$ .

AIPMT-2014

Kinetic Theory of Gases

139300 The molecules of a given mass of a gas have r.m.s. velocity of $200 {ms}^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5}$ ${Nm}^{- 2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} C$ and $0.05 \times 10^{5} {Nm}^{- 2}$ the rms velocity of its molecules in ${ms}^{- 1}$ is

1

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

2

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

3

\frac{100}{3}

4

100 \sqrt{2}

Explanation:

A Given that, ${(v_{rms})}_{1} = 200 m / s$
$T_{1} = 27^{\circ} C$
$= 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C$
$= 127 + 273 = 400 K$
$P_{1} = 1.0 \times 10^{5} {Nm}^{- 2}$
$P_{2} = .05 \times 10^{5} {Nm}^{- 2}$
${(v_{rms})}_{2} = ?$
We know that,
$V_{rms} velocity of gas molecule is -$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Hence,
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}$
${(v_{rms})}_{2} = \frac{2}{\sqrt{3}} \times 200$
${(v_{rms})}_{2} = \frac{400}{\sqrt{3}} m / s .$

NEET-2016

Kinetic Theory of Gases

139301 The r.m.s speed of hydrogen gas molecules at a certain temperature is $200 {ms}^{- 1}$ . If the absolute temperature of the gas is doubled and the hydrogen gas dissociates into atomic hydrogen, then the r.m.s speed will become

1

200 {ms}^{- 1}

2

400 {ms}^{- 1}

3

600 {ms}^{- 1}

4

800 {ms}^{- 1}

Explanation:

B Given that,

${(v_{rmm})}_{1} = 200 {ms}^{- 1}$
$T_{1} = T$
$T_{2} = 2 T$
$M_{1} = 2 M$
$M_{2} = M$
$We know that,$
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$Hence,$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}} \times \frac{M_{2}}{M_{1}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{T}{2 T} \times \frac{M}{2 M}}$
${(v_{rms})}_{2} = 2 \times 200$
${(v_{rms})}_{2} = 400 m s^{- 1}$

AP EAMCET(Medical)-2016

Kinetic Theory of Gases

139302 When the temperature of an ideal gas is increased by $600 K$ , the velocity of sound in the gas becomes $\sqrt{3}$ times the initial velocity in it. The initial temperature of the gas is:

1

- 73^{\circ} C

2

27^{\circ} C

3

127^{\circ} C

4

327^{\circ} C

Explanation:

B Given that,

$v_{1} = v$
$v_{2} = \sqrt{3} \cdot v$
$T_{1} = T K$
$T_{2} = T + 600 K$
$We know that,$
$v \propto \sqrt{T}$
$Hence,$
$\frac{\sqrt{3} v}{v} = \sqrt{\frac{T + 600}{T}}$
$\sqrt{3} = \sqrt{\frac{T + 600}{T}}$
$3 = \frac{T + 600}{T}$
$3 T = T + 600$
$T = 300 K$
$T = 300 - 273$
$T = 27^{\circ} C$

AP EAMCET(Medical)-2000

Kinetic Theory of Gases

139298 The temperature of a gas is raised from $27^{\circ} C$ to $927^{\circ} C$ . The root mean square speed

1 is

\sqrt{(\frac{927}{27})}

times the earlier value

2 remains the same

3 gets halved

4 gets doubled

Explanation:

D Given that,
$T_{1} = 27^{\circ} C = 300 K$
$T_{2} = 927^{\circ} C = 1200 K$
We know that,
$v_{rms} = \sqrt{\frac{3 kT}{M}}$
$v_{rms} \propto \sqrt{T}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{1200}} = \frac{1}{2}$
${(v_{rms})}_{2} = 2 {(v_{rms})}_{1}$
Hence, root mean sequare speed of gas gets doubled.

AIPMT-1994

Kinetic Theory of Gases

139299 The mean free path of molecules of a gas, (radius $r$ ) is inversely proportional to

1

r^{3}

2

r^{2}

3

r

4

\sqrt{r}

Explanation:

B We know that,
mean free path $(γ) = \frac{1}{\sqrt{2} π} d^{2} n$
$d =$ diameter of molecules
$d = 2 r$
$γ = \frac{1}{\sqrt{2} \cdot π (2 r)^{2} \cdot n}$
$γ = \frac{1}{π^{2}}$
Hence, $γ \propto \frac{1}{r^{2}}$
Mean Free path of molecules of a gas is inversely proportional to $r^{2}$ .

AIPMT-2014

Kinetic Theory of Gases

139300 The molecules of a given mass of a gas have r.m.s. velocity of $200 {ms}^{- 1}$ at $27^{\circ} C$ and $1.0 \times 10^{5}$ ${Nm}^{- 2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} C$ and $0.05 \times 10^{5} {Nm}^{- 2}$ the rms velocity of its molecules in ${ms}^{- 1}$ is

1

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

2

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

3

\frac{100}{3}

4

100 \sqrt{2}

Explanation:

A Given that, ${(v_{rms})}_{1} = 200 m / s$
$T_{1} = 27^{\circ} C$
$= 27 + 273 = 300 K$
$T_{2} = 127^{\circ} C$
$= 127 + 273 = 400 K$
$P_{1} = 1.0 \times 10^{5} {Nm}^{- 2}$
$P_{2} = .05 \times 10^{5} {Nm}^{- 2}$
${(v_{rms})}_{2} = ?$
We know that,
$V_{rms} velocity of gas molecule is -$
$v_{rms} = \sqrt{\frac{3 R T}{M}}$
$v_{rms} \propto \sqrt{T}$
Hence,
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}$
${(v_{rms})}_{2} = \frac{2}{\sqrt{3}} \times 200$
${(v_{rms})}_{2} = \frac{400}{\sqrt{3}} m / s .$

NEET-2016

Kinetic Theory of Gases

139301 The r.m.s speed of hydrogen gas molecules at a certain temperature is $200 {ms}^{- 1}$ . If the absolute temperature of the gas is doubled and the hydrogen gas dissociates into atomic hydrogen, then the r.m.s speed will become

1

200 {ms}^{- 1}

2

400 {ms}^{- 1}

3

600 {ms}^{- 1}

4

800 {ms}^{- 1}

Explanation:

B Given that,

${(v_{rmm})}_{1} = 200 {ms}^{- 1}$
$T_{1} = T$
$T_{2} = 2 T$
$M_{1} = 2 M$
$M_{2} = M$
$We know that,$
$v_{rms} = \sqrt{\frac{3 RT}{M}}$
$Hence,$
$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}} \times \frac{M_{2}}{M_{1}}}$
$\frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{T}{2 T} \times \frac{M}{2 M}}$
${(v_{rms})}_{2} = 2 \times 200$
${(v_{rms})}_{2} = 400 m s^{- 1}$

AP EAMCET(Medical)-2016

Kinetic Theory of Gases

139302 When the temperature of an ideal gas is increased by $600 K$ , the velocity of sound in the gas becomes $\sqrt{3}$ times the initial velocity in it. The initial temperature of the gas is:

1

- 73^{\circ} C

2

27^{\circ} C

3

127^{\circ} C

4

327^{\circ} C

Explanation:

B Given that,

$v_{1} = v$
$v_{2} = \sqrt{3} \cdot v$
$T_{1} = T K$
$T_{2} = T + 600 K$
$We know that,$
$v \propto \sqrt{T}$
$Hence,$
$\frac{\sqrt{3} v}{v} = \sqrt{\frac{T + 600}{T}}$
$\sqrt{3} = \sqrt{\frac{T + 600}{T}}$
$3 = \frac{T + 600}{T}$
$3 T = T + 600$
$T = 300 K$
$T = 300 - 273$
$T = 27^{\circ} C$

AP EAMCET(Medical)-2000

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Question Bank Series

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