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CHXI06:STATES OF MATTER

314202 Consider the following statements.
I. One mole of each gas at standard temperature and pressure will have same volume.
II. At STP, molar volume of an ideal gas or a combination of ideal gases is $22.7 L {mol}^{- 1}$ .
III. The combined gas law is given by
$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$ .
IV. In ideal gas equation, $R$ is called universal gas constant
Choose the correct statement and select the correct option.

1 Both I and II

2 Both II and III

3 Both I and IV

4 I, II, III and IV

CHXI06:STATES OF MATTER

314203 The density of a gas is $1.964 g {dm}^{- 3}$ at $273 K$ and $76 c m H g$ . The gas is

1

{CH}_{4}

2

C_{2} H_{6}

3

{CO}_{2}

4

Xe

Explanation:

Ideal gas equation,
$p V = n R T \Rightarrow p V = \frac{w}{M} R T$
$[∴ n = \frac{w}{M}]$
$\Rightarrow M = \frac{w}{V} \frac{R T}{p} \Rightarrow M = d \frac{R T}{p}$
$H e r e, d = 1.964 g / d m^{3}$
$= 1.964 \times 10^{- 3} g / c c$
$p = 76 c m = 1 a t m$
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$
$o r = 82.1 c c a t m K^{- 1} m o l^{- 1}$
$T = 273 K$
The molecular weight of ${CO}_{2}$ is $44 g {mol}^{- 1}$ .
$∴$ The gas is ${CO}_{2}$ .

CHXI06:STATES OF MATTER

314204 The pressure and temperature of $4 d m^{3}$ of carbon dioxide gas are doubled. Then, volume of carbon dioxide gas would be

1

2 d m^{3}

2

3 d m^{3}

3

4 d m^{3}

4

8 d m^{3}

Explanation:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{p_{1} \times 4}{T_{1}} = \frac{2 p_{1} \times V_{2}}{2 T_{1}}$
$V_{2} = 4 d m^{3}$

CHXI06:STATES OF MATTER

314205 By the ideal gas law, the pressure of $0.60 m o l e N H_{3} g a s i n a 3 .00 L v e s s e l a t 25^{\circ} C i s$

1

48.9 atm

2

4.89 atm

3

0.489 atm

4 489 atm

Explanation:

Given:
$\begin{array}{l} n = 0.60 m o l, V = 3.00 L \\ T = 25^{\circ} C (25 + 273 = 298 K) \\ F r o m i d e a l g a s e q u a t i o n, w e h a v e \\ V = n R T o r p = \frac{n R T}{V} \\ = \frac{0.60 \times 0.0821 \times 298}{3} = 4.89 a t m \end{array}$

CHXI06:STATES OF MATTER

314206 For a given mass of gas, if pressure is reduced to half and the absolute temperature is increased two times, then the volume would be become

1

V / 4

2

4 V

3

2 V^{2}

4

6 V

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{P_{1} V_{1}}{T_{1}} = [(\frac{P_{1}}{2}) (V_{2})] 2 T_{2}$
$V_{2} = 4 V_{1}$

CHXI06:STATES OF MATTER

314202 Consider the following statements.
I. One mole of each gas at standard temperature and pressure will have same volume.
II. At STP, molar volume of an ideal gas or a combination of ideal gases is $22.7 L {mol}^{- 1}$ .
III. The combined gas law is given by
$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$ .
IV. In ideal gas equation, $R$ is called universal gas constant
Choose the correct statement and select the correct option.

1 Both I and II

2 Both II and III

3 Both I and IV

4 I, II, III and IV

CHXI06:STATES OF MATTER

314203 The density of a gas is $1.964 g {dm}^{- 3}$ at $273 K$ and $76 c m H g$ . The gas is

1

{CH}_{4}

2

C_{2} H_{6}

3

{CO}_{2}

4

Xe

Explanation:

Ideal gas equation,
$p V = n R T \Rightarrow p V = \frac{w}{M} R T$
$[∴ n = \frac{w}{M}]$
$\Rightarrow M = \frac{w}{V} \frac{R T}{p} \Rightarrow M = d \frac{R T}{p}$
$H e r e, d = 1.964 g / d m^{3}$
$= 1.964 \times 10^{- 3} g / c c$
$p = 76 c m = 1 a t m$
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$
$o r = 82.1 c c a t m K^{- 1} m o l^{- 1}$
$T = 273 K$
The molecular weight of ${CO}_{2}$ is $44 g {mol}^{- 1}$ .
$∴$ The gas is ${CO}_{2}$ .

CHXI06:STATES OF MATTER

314204 The pressure and temperature of $4 d m^{3}$ of carbon dioxide gas are doubled. Then, volume of carbon dioxide gas would be

1

2 d m^{3}

2

3 d m^{3}

3

4 d m^{3}

4

8 d m^{3}

Explanation:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{p_{1} \times 4}{T_{1}} = \frac{2 p_{1} \times V_{2}}{2 T_{1}}$
$V_{2} = 4 d m^{3}$

CHXI06:STATES OF MATTER

314205 By the ideal gas law, the pressure of $0.60 m o l e N H_{3} g a s i n a 3 .00 L v e s s e l a t 25^{\circ} C i s$

1

48.9 atm

2

4.89 atm

3

0.489 atm

4 489 atm

Explanation:

Given:
$\begin{array}{l} n = 0.60 m o l, V = 3.00 L \\ T = 25^{\circ} C (25 + 273 = 298 K) \\ F r o m i d e a l g a s e q u a t i o n, w e h a v e \\ V = n R T o r p = \frac{n R T}{V} \\ = \frac{0.60 \times 0.0821 \times 298}{3} = 4.89 a t m \end{array}$

CHXI06:STATES OF MATTER

314206 For a given mass of gas, if pressure is reduced to half and the absolute temperature is increased two times, then the volume would be become

1

V / 4

2

4 V

3

2 V^{2}

4

6 V

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{P_{1} V_{1}}{T_{1}} = [(\frac{P_{1}}{2}) (V_{2})] 2 T_{2}$
$V_{2} = 4 V_{1}$

CHXI06:STATES OF MATTER

314202 Consider the following statements.
I. One mole of each gas at standard temperature and pressure will have same volume.
II. At STP, molar volume of an ideal gas or a combination of ideal gases is $22.7 L {mol}^{- 1}$ .
III. The combined gas law is given by
$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$ .
IV. In ideal gas equation, $R$ is called universal gas constant
Choose the correct statement and select the correct option.

1 Both I and II

2 Both II and III

3 Both I and IV

4 I, II, III and IV

CHXI06:STATES OF MATTER

314203 The density of a gas is $1.964 g {dm}^{- 3}$ at $273 K$ and $76 c m H g$ . The gas is

1

{CH}_{4}

2

C_{2} H_{6}

3

{CO}_{2}

4

Xe

Explanation:

Ideal gas equation,
$p V = n R T \Rightarrow p V = \frac{w}{M} R T$
$[∴ n = \frac{w}{M}]$
$\Rightarrow M = \frac{w}{V} \frac{R T}{p} \Rightarrow M = d \frac{R T}{p}$
$H e r e, d = 1.964 g / d m^{3}$
$= 1.964 \times 10^{- 3} g / c c$
$p = 76 c m = 1 a t m$
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$
$o r = 82.1 c c a t m K^{- 1} m o l^{- 1}$
$T = 273 K$
The molecular weight of ${CO}_{2}$ is $44 g {mol}^{- 1}$ .
$∴$ The gas is ${CO}_{2}$ .

CHXI06:STATES OF MATTER

314204 The pressure and temperature of $4 d m^{3}$ of carbon dioxide gas are doubled. Then, volume of carbon dioxide gas would be

1

2 d m^{3}

2

3 d m^{3}

3

4 d m^{3}

4

8 d m^{3}

Explanation:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{p_{1} \times 4}{T_{1}} = \frac{2 p_{1} \times V_{2}}{2 T_{1}}$
$V_{2} = 4 d m^{3}$

CHXI06:STATES OF MATTER

314205 By the ideal gas law, the pressure of $0.60 m o l e N H_{3} g a s i n a 3 .00 L v e s s e l a t 25^{\circ} C i s$

1

48.9 atm

2

4.89 atm

3

0.489 atm

4 489 atm

Explanation:

Given:
$\begin{array}{l} n = 0.60 m o l, V = 3.00 L \\ T = 25^{\circ} C (25 + 273 = 298 K) \\ F r o m i d e a l g a s e q u a t i o n, w e h a v e \\ V = n R T o r p = \frac{n R T}{V} \\ = \frac{0.60 \times 0.0821 \times 298}{3} = 4.89 a t m \end{array}$

CHXI06:STATES OF MATTER

314206 For a given mass of gas, if pressure is reduced to half and the absolute temperature is increased two times, then the volume would be become

1

V / 4

2

4 V

3

2 V^{2}

4

6 V

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{P_{1} V_{1}}{T_{1}} = [(\frac{P_{1}}{2}) (V_{2})] 2 T_{2}$
$V_{2} = 4 V_{1}$

CHXI06:STATES OF MATTER

314202 Consider the following statements.
I. One mole of each gas at standard temperature and pressure will have same volume.
II. At STP, molar volume of an ideal gas or a combination of ideal gases is $22.7 L {mol}^{- 1}$ .
III. The combined gas law is given by
$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$ .
IV. In ideal gas equation, $R$ is called universal gas constant
Choose the correct statement and select the correct option.

1 Both I and II

2 Both II and III

3 Both I and IV

4 I, II, III and IV

CHXI06:STATES OF MATTER

314203 The density of a gas is $1.964 g {dm}^{- 3}$ at $273 K$ and $76 c m H g$ . The gas is

1

{CH}_{4}

2

C_{2} H_{6}

3

{CO}_{2}

4

Xe

Explanation:

Ideal gas equation,
$p V = n R T \Rightarrow p V = \frac{w}{M} R T$
$[∴ n = \frac{w}{M}]$
$\Rightarrow M = \frac{w}{V} \frac{R T}{p} \Rightarrow M = d \frac{R T}{p}$
$H e r e, d = 1.964 g / d m^{3}$
$= 1.964 \times 10^{- 3} g / c c$
$p = 76 c m = 1 a t m$
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$
$o r = 82.1 c c a t m K^{- 1} m o l^{- 1}$
$T = 273 K$
The molecular weight of ${CO}_{2}$ is $44 g {mol}^{- 1}$ .
$∴$ The gas is ${CO}_{2}$ .

CHXI06:STATES OF MATTER

314204 The pressure and temperature of $4 d m^{3}$ of carbon dioxide gas are doubled. Then, volume of carbon dioxide gas would be

1

2 d m^{3}

2

3 d m^{3}

3

4 d m^{3}

4

8 d m^{3}

Explanation:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{p_{1} \times 4}{T_{1}} = \frac{2 p_{1} \times V_{2}}{2 T_{1}}$
$V_{2} = 4 d m^{3}$

CHXI06:STATES OF MATTER

314205 By the ideal gas law, the pressure of $0.60 m o l e N H_{3} g a s i n a 3 .00 L v e s s e l a t 25^{\circ} C i s$

1

48.9 atm

2

4.89 atm

3

0.489 atm

4 489 atm

Explanation:

Given:
$\begin{array}{l} n = 0.60 m o l, V = 3.00 L \\ T = 25^{\circ} C (25 + 273 = 298 K) \\ F r o m i d e a l g a s e q u a t i o n, w e h a v e \\ V = n R T o r p = \frac{n R T}{V} \\ = \frac{0.60 \times 0.0821 \times 298}{3} = 4.89 a t m \end{array}$

CHXI06:STATES OF MATTER

314206 For a given mass of gas, if pressure is reduced to half and the absolute temperature is increased two times, then the volume would be become

1

V / 4

2

4 V

3

2 V^{2}

4

6 V

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{P_{1} V_{1}}{T_{1}} = [(\frac{P_{1}}{2}) (V_{2})] 2 T_{2}$
$V_{2} = 4 V_{1}$

CHXI06:STATES OF MATTER

314202 Consider the following statements.
I. One mole of each gas at standard temperature and pressure will have same volume.
II. At STP, molar volume of an ideal gas or a combination of ideal gases is $22.7 L {mol}^{- 1}$ .
III. The combined gas law is given by
$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$ .
IV. In ideal gas equation, $R$ is called universal gas constant
Choose the correct statement and select the correct option.

1 Both I and II

2 Both II and III

3 Both I and IV

4 I, II, III and IV

CHXI06:STATES OF MATTER

314203 The density of a gas is $1.964 g {dm}^{- 3}$ at $273 K$ and $76 c m H g$ . The gas is

1

{CH}_{4}

2

C_{2} H_{6}

3

{CO}_{2}

4

Xe

Explanation:

Ideal gas equation,
$p V = n R T \Rightarrow p V = \frac{w}{M} R T$
$[∴ n = \frac{w}{M}]$
$\Rightarrow M = \frac{w}{V} \frac{R T}{p} \Rightarrow M = d \frac{R T}{p}$
$H e r e, d = 1.964 g / d m^{3}$
$= 1.964 \times 10^{- 3} g / c c$
$p = 76 c m = 1 a t m$
$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$
$o r = 82.1 c c a t m K^{- 1} m o l^{- 1}$
$T = 273 K$
The molecular weight of ${CO}_{2}$ is $44 g {mol}^{- 1}$ .
$∴$ The gas is ${CO}_{2}$ .

CHXI06:STATES OF MATTER

314204 The pressure and temperature of $4 d m^{3}$ of carbon dioxide gas are doubled. Then, volume of carbon dioxide gas would be

1

2 d m^{3}

2

3 d m^{3}

3

4 d m^{3}

4

8 d m^{3}

Explanation:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$
$\Rightarrow \frac{p_{1} \times 4}{T_{1}} = \frac{2 p_{1} \times V_{2}}{2 T_{1}}$
$V_{2} = 4 d m^{3}$

CHXI06:STATES OF MATTER

314205 By the ideal gas law, the pressure of $0.60 m o l e N H_{3} g a s i n a 3 .00 L v e s s e l a t 25^{\circ} C i s$

1

48.9 atm

2

4.89 atm

3

0.489 atm

4 489 atm

Explanation:

Given:
$\begin{array}{l} n = 0.60 m o l, V = 3.00 L \\ T = 25^{\circ} C (25 + 273 = 298 K) \\ F r o m i d e a l g a s e q u a t i o n, w e h a v e \\ V = n R T o r p = \frac{n R T}{V} \\ = \frac{0.60 \times 0.0821 \times 298}{3} = 4.89 a t m \end{array}$

CHXI06:STATES OF MATTER

314206 For a given mass of gas, if pressure is reduced to half and the absolute temperature is increased two times, then the volume would be become

1

V / 4

2

4 V

3

2 V^{2}

4

6 V

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{P_{1} V_{1}}{T_{1}} = [(\frac{P_{1}}{2}) (V_{2})] 2 T_{2}$
$V_{2} = 4 V_{1}$

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