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CHXI06:STATES OF MATTER

314329 In the following graph, the dotted line at stage D represents
supporting img

1

\sqrt{\frac{2 R T}{M}}

2

\sqrt{\frac{3 R T}{M}}

3

\sqrt{\frac{8 R T}{π M}}

4

\sqrt{\frac{π R T}{8 M}}

Explanation:

The stage ' $B$ ' shows the most probable velocity, $\sqrt{\frac{2 R T}{M}}$ .

CHXI06:STATES OF MATTER

314330 The average speed at temperature $T^{\circ} C$ of $C H_{4} (g)$ is $\sqrt{\frac{28}{88}} \times 10^{3} {ms}^{- 1}$ . What is the value of $T$ ?

1

24055^{\circ} C

2

- {32.45}^{\circ} C

3

3000^{\circ} C

4

- {24.055}^{\circ} C

Explanation:

$U_{A V} = \sqrt{\frac{8 R T}{π M}} = \sqrt{\frac{28}{88}} = \sqrt{\frac{7}{22}}$
$\frac{8 \times 8.314 \times T}{π \times 16 \times 10^{- 3}} = \frac{7}{22} \times 10^{6}$
$T = \frac{1000 \times 2}{8.314} = 240.55 K$
$T^{\circ} C = 240.55 - 273 = - {32.45}^{\circ} C$

CHXI06:STATES OF MATTER

314331 The most probable velocity of molecules of an imaginary ideal gas at $27^{\circ} C$ in $0.3 m / s$ . The average velocity at $927^{\circ} C$ would be

1

0.3 m / s

2

0.677 m / s

3

0.9 m / s

4

1.2 m / s

Explanation:

$U_{a v} = U_{M P} \times 2 .257 = 0 .3 \times 2 .257 = 0 .677 m / s .$

CHXI06:STATES OF MATTER

314332 At what temperature would the average velocity of $CO$ be equal to the of ${CO}_{2}$ ?

1

(x + 273) \times \frac{7}{11}

2

(x + 273) \times \frac{11}{7}

3

x \times \frac{7}{11}

4

x \times \frac{11}{17}

Explanation:

Average velocity $= \sqrt{\frac{8 R T}{π M}}$
For equal average velocities of $CO$ and ${CO}_{2}$ ,
${\sqrt{(\frac{8 R T}{π M})}}_{C O} = {\sqrt{(\frac{8 R T}{π M})}}_{C O_{2}}$
$\Rightarrow {(\frac{T}{M})}_{C O} = {(\frac{T}{M})}_{C O_{2}}$
$\Rightarrow \frac{T_{C O}}{28} = \frac{(x + 273)}{44}$
$\Rightarrow T_{C O} = \frac{28}{44} (x + 273) = \frac{7}{11} (x + 273)$

CHXI06:STATES OF MATTER

314329 In the following graph, the dotted line at stage D represents
supporting img

1

\sqrt{\frac{2 R T}{M}}

2

\sqrt{\frac{3 R T}{M}}

3

\sqrt{\frac{8 R T}{π M}}

4

\sqrt{\frac{π R T}{8 M}}

Explanation:

The stage ' $B$ ' shows the most probable velocity, $\sqrt{\frac{2 R T}{M}}$ .

CHXI06:STATES OF MATTER

314330 The average speed at temperature $T^{\circ} C$ of $C H_{4} (g)$ is $\sqrt{\frac{28}{88}} \times 10^{3} {ms}^{- 1}$ . What is the value of $T$ ?

1

24055^{\circ} C

2

- {32.45}^{\circ} C

3

3000^{\circ} C

4

- {24.055}^{\circ} C

Explanation:

$U_{A V} = \sqrt{\frac{8 R T}{π M}} = \sqrt{\frac{28}{88}} = \sqrt{\frac{7}{22}}$
$\frac{8 \times 8.314 \times T}{π \times 16 \times 10^{- 3}} = \frac{7}{22} \times 10^{6}$
$T = \frac{1000 \times 2}{8.314} = 240.55 K$
$T^{\circ} C = 240.55 - 273 = - {32.45}^{\circ} C$

CHXI06:STATES OF MATTER

314331 The most probable velocity of molecules of an imaginary ideal gas at $27^{\circ} C$ in $0.3 m / s$ . The average velocity at $927^{\circ} C$ would be

1

0.3 m / s

2

0.677 m / s

3

0.9 m / s

4

1.2 m / s

Explanation:

$U_{a v} = U_{M P} \times 2 .257 = 0 .3 \times 2 .257 = 0 .677 m / s .$

CHXI06:STATES OF MATTER

314332 At what temperature would the average velocity of $CO$ be equal to the of ${CO}_{2}$ ?

1

(x + 273) \times \frac{7}{11}

2

(x + 273) \times \frac{11}{7}

3

x \times \frac{7}{11}

4

x \times \frac{11}{17}

Explanation:

Average velocity $= \sqrt{\frac{8 R T}{π M}}$
For equal average velocities of $CO$ and ${CO}_{2}$ ,
${\sqrt{(\frac{8 R T}{π M})}}_{C O} = {\sqrt{(\frac{8 R T}{π M})}}_{C O_{2}}$
$\Rightarrow {(\frac{T}{M})}_{C O} = {(\frac{T}{M})}_{C O_{2}}$
$\Rightarrow \frac{T_{C O}}{28} = \frac{(x + 273)}{44}$
$\Rightarrow T_{C O} = \frac{28}{44} (x + 273) = \frac{7}{11} (x + 273)$

CHXI06:STATES OF MATTER

314329 In the following graph, the dotted line at stage D represents
supporting img

1

\sqrt{\frac{2 R T}{M}}

2

\sqrt{\frac{3 R T}{M}}

3

\sqrt{\frac{8 R T}{π M}}

4

\sqrt{\frac{π R T}{8 M}}

Explanation:

The stage ' $B$ ' shows the most probable velocity, $\sqrt{\frac{2 R T}{M}}$ .

CHXI06:STATES OF MATTER

314330 The average speed at temperature $T^{\circ} C$ of $C H_{4} (g)$ is $\sqrt{\frac{28}{88}} \times 10^{3} {ms}^{- 1}$ . What is the value of $T$ ?

1

24055^{\circ} C

2

- {32.45}^{\circ} C

3

3000^{\circ} C

4

- {24.055}^{\circ} C

Explanation:

$U_{A V} = \sqrt{\frac{8 R T}{π M}} = \sqrt{\frac{28}{88}} = \sqrt{\frac{7}{22}}$
$\frac{8 \times 8.314 \times T}{π \times 16 \times 10^{- 3}} = \frac{7}{22} \times 10^{6}$
$T = \frac{1000 \times 2}{8.314} = 240.55 K$
$T^{\circ} C = 240.55 - 273 = - {32.45}^{\circ} C$

CHXI06:STATES OF MATTER

314331 The most probable velocity of molecules of an imaginary ideal gas at $27^{\circ} C$ in $0.3 m / s$ . The average velocity at $927^{\circ} C$ would be

1

0.3 m / s

2

0.677 m / s

3

0.9 m / s

4

1.2 m / s

Explanation:

$U_{a v} = U_{M P} \times 2 .257 = 0 .3 \times 2 .257 = 0 .677 m / s .$

CHXI06:STATES OF MATTER

314332 At what temperature would the average velocity of $CO$ be equal to the of ${CO}_{2}$ ?

1

(x + 273) \times \frac{7}{11}

2

(x + 273) \times \frac{11}{7}

3

x \times \frac{7}{11}

4

x \times \frac{11}{17}

Explanation:

Average velocity $= \sqrt{\frac{8 R T}{π M}}$
For equal average velocities of $CO$ and ${CO}_{2}$ ,
${\sqrt{(\frac{8 R T}{π M})}}_{C O} = {\sqrt{(\frac{8 R T}{π M})}}_{C O_{2}}$
$\Rightarrow {(\frac{T}{M})}_{C O} = {(\frac{T}{M})}_{C O_{2}}$
$\Rightarrow \frac{T_{C O}}{28} = \frac{(x + 273)}{44}$
$\Rightarrow T_{C O} = \frac{28}{44} (x + 273) = \frac{7}{11} (x + 273)$

CHXI06:STATES OF MATTER

314329 In the following graph, the dotted line at stage D represents
supporting img

1

\sqrt{\frac{2 R T}{M}}

2

\sqrt{\frac{3 R T}{M}}

3

\sqrt{\frac{8 R T}{π M}}

4

\sqrt{\frac{π R T}{8 M}}

Explanation:

The stage ' $B$ ' shows the most probable velocity, $\sqrt{\frac{2 R T}{M}}$ .

CHXI06:STATES OF MATTER

314330 The average speed at temperature $T^{\circ} C$ of $C H_{4} (g)$ is $\sqrt{\frac{28}{88}} \times 10^{3} {ms}^{- 1}$ . What is the value of $T$ ?

1

24055^{\circ} C

2

- {32.45}^{\circ} C

3

3000^{\circ} C

4

- {24.055}^{\circ} C

Explanation:

$U_{A V} = \sqrt{\frac{8 R T}{π M}} = \sqrt{\frac{28}{88}} = \sqrt{\frac{7}{22}}$
$\frac{8 \times 8.314 \times T}{π \times 16 \times 10^{- 3}} = \frac{7}{22} \times 10^{6}$
$T = \frac{1000 \times 2}{8.314} = 240.55 K$
$T^{\circ} C = 240.55 - 273 = - {32.45}^{\circ} C$

CHXI06:STATES OF MATTER

314331 The most probable velocity of molecules of an imaginary ideal gas at $27^{\circ} C$ in $0.3 m / s$ . The average velocity at $927^{\circ} C$ would be

1

0.3 m / s

2

0.677 m / s

3

0.9 m / s

4

1.2 m / s

Explanation:

$U_{a v} = U_{M P} \times 2 .257 = 0 .3 \times 2 .257 = 0 .677 m / s .$

CHXI06:STATES OF MATTER

314332 At what temperature would the average velocity of $CO$ be equal to the of ${CO}_{2}$ ?

1

(x + 273) \times \frac{7}{11}

2

(x + 273) \times \frac{11}{7}

3

x \times \frac{7}{11}

4

x \times \frac{11}{17}

Explanation:

Average velocity $= \sqrt{\frac{8 R T}{π M}}$
For equal average velocities of $CO$ and ${CO}_{2}$ ,
${\sqrt{(\frac{8 R T}{π M})}}_{C O} = {\sqrt{(\frac{8 R T}{π M})}}_{C O_{2}}$
$\Rightarrow {(\frac{T}{M})}_{C O} = {(\frac{T}{M})}_{C O_{2}}$
$\Rightarrow \frac{T_{C O}}{28} = \frac{(x + 273)}{44}$
$\Rightarrow T_{C O} = \frac{28}{44} (x + 273) = \frac{7}{11} (x + 273)$