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CHXI06:STATES OF MATTER

314357 The volume-temperature graphs of a given mass of an ideal gas at constant pressure are shown below:
supporting img

1

p_{1} > p_{3} > p_{2}

2

p_{1} > p_{2} > p_{3}

3

p_{2} > p_{3} > p_{1}

4

p_{2} > p_{1} > p_{3}

Explanation:

We know, $V_{1} < V_{3} < V_{2}$ at 273K(from figure)
$p \propto \frac{1}{V}$
Hence, $p_{1} > p_{3} > p_{2}$

CHXI06:STATES OF MATTER

314358 The temperature of a certain mass of gas was increased from $40^{\circ} C$ to $41^{\circ} C$ at constant pressure. The volume of the gas

1 Will remain constant

2 Will increase by

\frac{1}{273}

of its volume at

273 K

3 Will increase by

\frac{1}{273}

of its volume at

40^{\circ} C

4 Will increase, but the increase in volume cannot be predicted

Explanation:

According to Charle's law,
$\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}$

CHXI06:STATES OF MATTER

314359 The volume of an ideal gas at $27^{\circ} C$ is $30 {cm}^{3}$ . At what temperature (in Celsius) will the volume of the gas be $35 {cm}^{3}$ ?

1 77

2 300

3 30

4 35

Explanation:

$\frac{V_{2}}{V_{1}} = \frac{T_{2}}{T_{1}}$
$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}} = \frac{35 \times 300}{30} = 350 K = 77^{\circ} C$

CHXI06:STATES OF MATTER

314360 A sample of gas occupies 100 mL at $27^{\circ} C$ 0.9 atm pressure. When volume is changed to 60 mL at the same pressure, the temperature of the gas will be ____ $^{\circ} C$ .

1 180

2

-

180

3

-

93

4 93

Explanation:

For the same amount of gas at constant pressure,
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}}$
$T_{2} = \frac{60 \times 300}{100} = 180 K$
$= 180 - 273 = - 93^{\circ} C$

CHXI06:STATES OF MATTER

314357 The volume-temperature graphs of a given mass of an ideal gas at constant pressure are shown below:
supporting img

1

p_{1} > p_{3} > p_{2}

2

p_{1} > p_{2} > p_{3}

3

p_{2} > p_{3} > p_{1}

4

p_{2} > p_{1} > p_{3}

Explanation:

We know, $V_{1} < V_{3} < V_{2}$ at 273K(from figure)
$p \propto \frac{1}{V}$
Hence, $p_{1} > p_{3} > p_{2}$

CHXI06:STATES OF MATTER

314358 The temperature of a certain mass of gas was increased from $40^{\circ} C$ to $41^{\circ} C$ at constant pressure. The volume of the gas

1 Will remain constant

2 Will increase by

\frac{1}{273}

of its volume at

273 K

3 Will increase by

\frac{1}{273}

of its volume at

40^{\circ} C

4 Will increase, but the increase in volume cannot be predicted

Explanation:

According to Charle's law,
$\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}$

CHXI06:STATES OF MATTER

314359 The volume of an ideal gas at $27^{\circ} C$ is $30 {cm}^{3}$ . At what temperature (in Celsius) will the volume of the gas be $35 {cm}^{3}$ ?

1 77

2 300

3 30

4 35

Explanation:

$\frac{V_{2}}{V_{1}} = \frac{T_{2}}{T_{1}}$
$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}} = \frac{35 \times 300}{30} = 350 K = 77^{\circ} C$

CHXI06:STATES OF MATTER

314360 A sample of gas occupies 100 mL at $27^{\circ} C$ 0.9 atm pressure. When volume is changed to 60 mL at the same pressure, the temperature of the gas will be ____ $^{\circ} C$ .

1 180

2

-

180

3

-

93

4 93

Explanation:

For the same amount of gas at constant pressure,
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}}$
$T_{2} = \frac{60 \times 300}{100} = 180 K$
$= 180 - 273 = - 93^{\circ} C$

CHXI06:STATES OF MATTER

314357 The volume-temperature graphs of a given mass of an ideal gas at constant pressure are shown below:
supporting img

1

p_{1} > p_{3} > p_{2}

2

p_{1} > p_{2} > p_{3}

3

p_{2} > p_{3} > p_{1}

4

p_{2} > p_{1} > p_{3}

Explanation:

We know, $V_{1} < V_{3} < V_{2}$ at 273K(from figure)
$p \propto \frac{1}{V}$
Hence, $p_{1} > p_{3} > p_{2}$

CHXI06:STATES OF MATTER

314358 The temperature of a certain mass of gas was increased from $40^{\circ} C$ to $41^{\circ} C$ at constant pressure. The volume of the gas

1 Will remain constant

2 Will increase by

\frac{1}{273}

of its volume at

273 K

3 Will increase by

\frac{1}{273}

of its volume at

40^{\circ} C

4 Will increase, but the increase in volume cannot be predicted

Explanation:

According to Charle's law,
$\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}$

CHXI06:STATES OF MATTER

314359 The volume of an ideal gas at $27^{\circ} C$ is $30 {cm}^{3}$ . At what temperature (in Celsius) will the volume of the gas be $35 {cm}^{3}$ ?

1 77

2 300

3 30

4 35

Explanation:

$\frac{V_{2}}{V_{1}} = \frac{T_{2}}{T_{1}}$
$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}} = \frac{35 \times 300}{30} = 350 K = 77^{\circ} C$

CHXI06:STATES OF MATTER

314360 A sample of gas occupies 100 mL at $27^{\circ} C$ 0.9 atm pressure. When volume is changed to 60 mL at the same pressure, the temperature of the gas will be ____ $^{\circ} C$ .

1 180

2

-

180

3

-

93

4 93

Explanation:

For the same amount of gas at constant pressure,
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}}$
$T_{2} = \frac{60 \times 300}{100} = 180 K$
$= 180 - 273 = - 93^{\circ} C$

CHXI06:STATES OF MATTER

314357 The volume-temperature graphs of a given mass of an ideal gas at constant pressure are shown below:
supporting img

1

p_{1} > p_{3} > p_{2}

2

p_{1} > p_{2} > p_{3}

3

p_{2} > p_{3} > p_{1}

4

p_{2} > p_{1} > p_{3}

Explanation:

We know, $V_{1} < V_{3} < V_{2}$ at 273K(from figure)
$p \propto \frac{1}{V}$
Hence, $p_{1} > p_{3} > p_{2}$

CHXI06:STATES OF MATTER

314358 The temperature of a certain mass of gas was increased from $40^{\circ} C$ to $41^{\circ} C$ at constant pressure. The volume of the gas

1 Will remain constant

2 Will increase by

\frac{1}{273}

of its volume at

273 K

3 Will increase by

\frac{1}{273}

of its volume at

40^{\circ} C

4 Will increase, but the increase in volume cannot be predicted

Explanation:

According to Charle's law,
$\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}$

CHXI06:STATES OF MATTER

314359 The volume of an ideal gas at $27^{\circ} C$ is $30 {cm}^{3}$ . At what temperature (in Celsius) will the volume of the gas be $35 {cm}^{3}$ ?

1 77

2 300

3 30

4 35

Explanation:

$\frac{V_{2}}{V_{1}} = \frac{T_{2}}{T_{1}}$
$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}} = \frac{35 \times 300}{30} = 350 K = 77^{\circ} C$

CHXI06:STATES OF MATTER

314360 A sample of gas occupies 100 mL at $27^{\circ} C$ 0.9 atm pressure. When volume is changed to 60 mL at the same pressure, the temperature of the gas will be ____ $^{\circ} C$ .

1 180

2

-

180

3

-

93

4 93

Explanation:

For the same amount of gas at constant pressure,
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}}$
$T_{2} = \frac{60 \times 300}{100} = 180 K$
$= 180 - 273 = - 93^{\circ} C$

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