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CHXI06:STATES OF MATTER

314308 For two gases, $A$ and $B$ with molecular masses $M_{A}$ , and $M_{B}$ , it is observed that at a certain temperature $T$ the mean velocity of $A$ is equal to the root mean square velocity of $B$ . Thus, the mean velocity of $A$ can be made equal to the mean velocity of $B$ , if

1 A is at temperature T, and B at

T_{2}, T_{2} > T

2 A is lowered to a temperature

T_{2} < T

while

B

is at

T

3 Both A and B are raised to a higher temperature.

4 Both A and B are placed at lower temperature

Explanation:

$U_{A V} = \sqrt{\frac{8 R T}{π M}}; U_{r m s} = \sqrt{\frac{3 R T}{M}}$
Average velocity of "A" at temperature $T, U_{A V} (A) = \sqrt{\frac{8 R T}{π M_{A}}}$
RMS velocity of "B" at temperature
$T, U_{r m s} (B) = \sqrt{\frac{3 R T}{M_{B}}}$
$∴ \sqrt{\frac{8 R T}{π M_{A}}} = \sqrt{\frac{3 R T}{M_{B}}}$
on solving, $M_{A} = \frac{56}{66} M_{B}$
Let the temperature of $B$ be maintained at $T$ and that of $A$ be changed to $T_{2}$
Now, rms of both A and B should be equal
$\sqrt{\frac{3 R T_{2}}{M_{A}}} = \sqrt{\frac{3 R T}{M_{B}}}$
$\frac{T_{2}}{M_{A}} = \frac{T}{M_{B}}$
$M_{A} = \frac{56}{66} M_{B}$
$\frac{66 T_{2}}{56 M_{B}} = \frac{T}{M_{B}}$
$66 T_{2} = 56 T$
$T_{2} = \frac{56 T}{66}; T_{2} < T$

CHXI06:STATES OF MATTER

314309 Which of the following statement is incorrect?

1 RMS velocity depends on the molecular weight of the gas.

2 R.M.S. velocity depends on temperature of the gas.

3 R.M.S. velocity depends on the number of molecules present in unit volume.

4 R.M.S velocity is greater than average velocity and most probable velocity.

CHXI06:STATES OF MATTER

314310 At what temperature will the average speed of ${CH}_{4}$ molecules have the same value as $O_{2}$ gas at $300 K$ ?

1

1200 K

2

150 K

3

600 K

4

300 K

Explanation:

Given ,
Average speed of ${CH}_{4}$ molecules = Average speed of $O_{2}$ molecules
${(\sqrt{\frac{8 R T}{π M}})}_{C H_{4}} = {(\sqrt{\frac{8 R T}{π M}})}_{O_{2}}$
${(\frac{T}{M})}_{C H_{4}} = {(\frac{T}{M})}_{O_{2}}$
${(\frac{T}{16})}_{C H_{4}} = {(\frac{300}{32})}_{O_{2}}$
$T_{C H_{4}} = 150 K$

CHXI06:STATES OF MATTER

314311 At what temperature will average speed of the molecule of the second member of the series $C_{n} H_{2 n}$ be the same of ${Cl}_{2}$ at $627^{\circ} C$ ?

1

259.4 K

2

400 K

3

532.4 K

4 None of these

Explanation:

Second member of
$C_{n} H_{2 n}$ series $= C_{3} H_{6} = 42$
${(U_{A V})}_{C_{3} H_{6}} = {(U_{A V})}_{C_{2}}$
$\sqrt{\frac{8 R T_{1}}{π M_{1}}} = \sqrt{\frac{8 R T_{2}}{π M_{2}}}$
$\frac{900}{71} = \frac{T_{2}}{42}$
$T_{2} = 532.4 K$

CHXI06:STATES OF MATTER

314308 For two gases, $A$ and $B$ with molecular masses $M_{A}$ , and $M_{B}$ , it is observed that at a certain temperature $T$ the mean velocity of $A$ is equal to the root mean square velocity of $B$ . Thus, the mean velocity of $A$ can be made equal to the mean velocity of $B$ , if

1 A is at temperature T, and B at

T_{2}, T_{2} > T

2 A is lowered to a temperature

T_{2} < T

while

B

is at

T

3 Both A and B are raised to a higher temperature.

4 Both A and B are placed at lower temperature

Explanation:

$U_{A V} = \sqrt{\frac{8 R T}{π M}}; U_{r m s} = \sqrt{\frac{3 R T}{M}}$
Average velocity of "A" at temperature $T, U_{A V} (A) = \sqrt{\frac{8 R T}{π M_{A}}}$
RMS velocity of "B" at temperature
$T, U_{r m s} (B) = \sqrt{\frac{3 R T}{M_{B}}}$
$∴ \sqrt{\frac{8 R T}{π M_{A}}} = \sqrt{\frac{3 R T}{M_{B}}}$
on solving, $M_{A} = \frac{56}{66} M_{B}$
Let the temperature of $B$ be maintained at $T$ and that of $A$ be changed to $T_{2}$
Now, rms of both A and B should be equal
$\sqrt{\frac{3 R T_{2}}{M_{A}}} = \sqrt{\frac{3 R T}{M_{B}}}$
$\frac{T_{2}}{M_{A}} = \frac{T}{M_{B}}$
$M_{A} = \frac{56}{66} M_{B}$
$\frac{66 T_{2}}{56 M_{B}} = \frac{T}{M_{B}}$
$66 T_{2} = 56 T$
$T_{2} = \frac{56 T}{66}; T_{2} < T$

CHXI06:STATES OF MATTER

314309 Which of the following statement is incorrect?

1 RMS velocity depends on the molecular weight of the gas.

2 R.M.S. velocity depends on temperature of the gas.

3 R.M.S. velocity depends on the number of molecules present in unit volume.

4 R.M.S velocity is greater than average velocity and most probable velocity.

CHXI06:STATES OF MATTER

314310 At what temperature will the average speed of ${CH}_{4}$ molecules have the same value as $O_{2}$ gas at $300 K$ ?

1

1200 K

2

150 K

3

600 K

4

300 K

Explanation:

Given ,
Average speed of ${CH}_{4}$ molecules = Average speed of $O_{2}$ molecules
${(\sqrt{\frac{8 R T}{π M}})}_{C H_{4}} = {(\sqrt{\frac{8 R T}{π M}})}_{O_{2}}$
${(\frac{T}{M})}_{C H_{4}} = {(\frac{T}{M})}_{O_{2}}$
${(\frac{T}{16})}_{C H_{4}} = {(\frac{300}{32})}_{O_{2}}$
$T_{C H_{4}} = 150 K$

CHXI06:STATES OF MATTER

314311 At what temperature will average speed of the molecule of the second member of the series $C_{n} H_{2 n}$ be the same of ${Cl}_{2}$ at $627^{\circ} C$ ?

1

259.4 K

2

400 K

3

532.4 K

4 None of these

Explanation:

Second member of
$C_{n} H_{2 n}$ series $= C_{3} H_{6} = 42$
${(U_{A V})}_{C_{3} H_{6}} = {(U_{A V})}_{C_{2}}$
$\sqrt{\frac{8 R T_{1}}{π M_{1}}} = \sqrt{\frac{8 R T_{2}}{π M_{2}}}$
$\frac{900}{71} = \frac{T_{2}}{42}$
$T_{2} = 532.4 K$

CHXI06:STATES OF MATTER

314308 For two gases, $A$ and $B$ with molecular masses $M_{A}$ , and $M_{B}$ , it is observed that at a certain temperature $T$ the mean velocity of $A$ is equal to the root mean square velocity of $B$ . Thus, the mean velocity of $A$ can be made equal to the mean velocity of $B$ , if

1 A is at temperature T, and B at

T_{2}, T_{2} > T

2 A is lowered to a temperature

T_{2} < T

while

B

is at

T

3 Both A and B are raised to a higher temperature.

4 Both A and B are placed at lower temperature

Explanation:

$U_{A V} = \sqrt{\frac{8 R T}{π M}}; U_{r m s} = \sqrt{\frac{3 R T}{M}}$
Average velocity of "A" at temperature $T, U_{A V} (A) = \sqrt{\frac{8 R T}{π M_{A}}}$
RMS velocity of "B" at temperature
$T, U_{r m s} (B) = \sqrt{\frac{3 R T}{M_{B}}}$
$∴ \sqrt{\frac{8 R T}{π M_{A}}} = \sqrt{\frac{3 R T}{M_{B}}}$
on solving, $M_{A} = \frac{56}{66} M_{B}$
Let the temperature of $B$ be maintained at $T$ and that of $A$ be changed to $T_{2}$
Now, rms of both A and B should be equal
$\sqrt{\frac{3 R T_{2}}{M_{A}}} = \sqrt{\frac{3 R T}{M_{B}}}$
$\frac{T_{2}}{M_{A}} = \frac{T}{M_{B}}$
$M_{A} = \frac{56}{66} M_{B}$
$\frac{66 T_{2}}{56 M_{B}} = \frac{T}{M_{B}}$
$66 T_{2} = 56 T$
$T_{2} = \frac{56 T}{66}; T_{2} < T$

CHXI06:STATES OF MATTER

314309 Which of the following statement is incorrect?

1 RMS velocity depends on the molecular weight of the gas.

2 R.M.S. velocity depends on temperature of the gas.

3 R.M.S. velocity depends on the number of molecules present in unit volume.

4 R.M.S velocity is greater than average velocity and most probable velocity.

CHXI06:STATES OF MATTER

314310 At what temperature will the average speed of ${CH}_{4}$ molecules have the same value as $O_{2}$ gas at $300 K$ ?

1

1200 K

2

150 K

3

600 K

4

300 K

Explanation:

Given ,
Average speed of ${CH}_{4}$ molecules = Average speed of $O_{2}$ molecules
${(\sqrt{\frac{8 R T}{π M}})}_{C H_{4}} = {(\sqrt{\frac{8 R T}{π M}})}_{O_{2}}$
${(\frac{T}{M})}_{C H_{4}} = {(\frac{T}{M})}_{O_{2}}$
${(\frac{T}{16})}_{C H_{4}} = {(\frac{300}{32})}_{O_{2}}$
$T_{C H_{4}} = 150 K$

CHXI06:STATES OF MATTER

314311 At what temperature will average speed of the molecule of the second member of the series $C_{n} H_{2 n}$ be the same of ${Cl}_{2}$ at $627^{\circ} C$ ?

1

259.4 K

2

400 K

3

532.4 K

4 None of these

Explanation:

Second member of
$C_{n} H_{2 n}$ series $= C_{3} H_{6} = 42$
${(U_{A V})}_{C_{3} H_{6}} = {(U_{A V})}_{C_{2}}$
$\sqrt{\frac{8 R T_{1}}{π M_{1}}} = \sqrt{\frac{8 R T_{2}}{π M_{2}}}$
$\frac{900}{71} = \frac{T_{2}}{42}$
$T_{2} = 532.4 K$

CHXI06:STATES OF MATTER

314308 For two gases, $A$ and $B$ with molecular masses $M_{A}$ , and $M_{B}$ , it is observed that at a certain temperature $T$ the mean velocity of $A$ is equal to the root mean square velocity of $B$ . Thus, the mean velocity of $A$ can be made equal to the mean velocity of $B$ , if

1 A is at temperature T, and B at

T_{2}, T_{2} > T

2 A is lowered to a temperature

T_{2} < T

while

B

is at

T

3 Both A and B are raised to a higher temperature.

4 Both A and B are placed at lower temperature

Explanation:

$U_{A V} = \sqrt{\frac{8 R T}{π M}}; U_{r m s} = \sqrt{\frac{3 R T}{M}}$
Average velocity of "A" at temperature $T, U_{A V} (A) = \sqrt{\frac{8 R T}{π M_{A}}}$
RMS velocity of "B" at temperature
$T, U_{r m s} (B) = \sqrt{\frac{3 R T}{M_{B}}}$
$∴ \sqrt{\frac{8 R T}{π M_{A}}} = \sqrt{\frac{3 R T}{M_{B}}}$
on solving, $M_{A} = \frac{56}{66} M_{B}$
Let the temperature of $B$ be maintained at $T$ and that of $A$ be changed to $T_{2}$
Now, rms of both A and B should be equal
$\sqrt{\frac{3 R T_{2}}{M_{A}}} = \sqrt{\frac{3 R T}{M_{B}}}$
$\frac{T_{2}}{M_{A}} = \frac{T}{M_{B}}$
$M_{A} = \frac{56}{66} M_{B}$
$\frac{66 T_{2}}{56 M_{B}} = \frac{T}{M_{B}}$
$66 T_{2} = 56 T$
$T_{2} = \frac{56 T}{66}; T_{2} < T$

CHXI06:STATES OF MATTER

314309 Which of the following statement is incorrect?

1 RMS velocity depends on the molecular weight of the gas.

2 R.M.S. velocity depends on temperature of the gas.

3 R.M.S. velocity depends on the number of molecules present in unit volume.

4 R.M.S velocity is greater than average velocity and most probable velocity.

CHXI06:STATES OF MATTER

314310 At what temperature will the average speed of ${CH}_{4}$ molecules have the same value as $O_{2}$ gas at $300 K$ ?

1

1200 K

2

150 K

3

600 K

4

300 K

Explanation:

Given ,
Average speed of ${CH}_{4}$ molecules = Average speed of $O_{2}$ molecules
${(\sqrt{\frac{8 R T}{π M}})}_{C H_{4}} = {(\sqrt{\frac{8 R T}{π M}})}_{O_{2}}$
${(\frac{T}{M})}_{C H_{4}} = {(\frac{T}{M})}_{O_{2}}$
${(\frac{T}{16})}_{C H_{4}} = {(\frac{300}{32})}_{O_{2}}$
$T_{C H_{4}} = 150 K$

CHXI06:STATES OF MATTER

314311 At what temperature will average speed of the molecule of the second member of the series $C_{n} H_{2 n}$ be the same of ${Cl}_{2}$ at $627^{\circ} C$ ?

1

259.4 K

2

400 K

3

532.4 K

4 None of these

Explanation:

Second member of
$C_{n} H_{2 n}$ series $= C_{3} H_{6} = 42$
${(U_{A V})}_{C_{3} H_{6}} = {(U_{A V})}_{C_{2}}$
$\sqrt{\frac{8 R T_{1}}{π M_{1}}} = \sqrt{\frac{8 R T_{2}}{π M_{2}}}$
$\frac{900}{71} = \frac{T_{2}}{42}$
$T_{2} = 532.4 K$