QBManager

CHXI06:STATES OF MATTER

314061 Refer to the figure given and identify the correct statement.
(In all options, 'a' and 'b' are Van der Waals constant)
supporting img

1 For gas

A, a = 0

and

Z

will linearly depend on pressure

2 Gas

C

is a real gas and we can find '

a

' and '

b

' if intersection data is given

3 All van der Waal gases will behave like gas

C

and give positive slope at high pressure

4 For gas

B, b = 0

and

Z

will linearly depend on pressure

Explanation:

For gas $A, a = 0, Z = 1 + \frac{P b}{R T}$ implies $Z$ varies linearly with pressure.
For gas $B, b = 0, Z = 1 - \frac{a}{V R T}$ . Hence $Z$ does not vary linearly with pressure.
Given the intersection data for gas $C$ , it is possible to find the values of ' $a$ ' and ' $b$ '. All Van der Wall's gases like gas $C$ , gives positive slope at high pressures.

CHXI06:STATES OF MATTER

314062 The compressibility factor for definite amount of van der Waals' gas at $0^{\circ} C$ and 100 atm is found to be 0.5 . Assuming the volume of gas molecules negligible, the van der Waals' constant $a$ for a gas is ____ .

1 1.256

2 12.56

3 125.6

4 1256

Explanation:

$Z = \frac{P V}{n R T} = 0.5$
Now, $[P + \frac{n^{2} a}{V^{2}}] [V - n b] = n R T$
$[P + \frac{n^{2} a}{V^{2}}] [V] = n R T (b$ is negligible $)$
$P V^{2} - n R T V + n^{2} a = 0$
$∴ V = \frac{n R T \pm \sqrt{n^{2} R^{2} T^{2} - 4 n^{2} a \times b}}{2 P}$
Since, $V$ is constant at given $P$ and $T$ , thus, discriminant is 0 .
$∴ n^{2} R^{2} T^{2} = 4 n^{2} a P o r a = \frac{R^{2} T^{2}}{4 P}$
$= \frac{{(0.0821)}^{2} \times {(273)}^{2}}{4 \times 100}$
$= 1.256 L m o l^{- 2} a t m$

CHXI06:STATES OF MATTER

314080 The inversion temperature $(T_{i})$ for a gas is given by

1

a / Rb

2

2 a / Rb

3

\frac{Rb}{a}

4

\frac{2 Rb}{a}

Explanation:

$T_{i} = \frac{2 a}{R b}$

CHXI06:STATES OF MATTER

314066 A gas at $350 K$ and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor $(Z)$ is :

1

Z > 1

and attractive forces are dominant

2

Z > 1

and repulsive forces are dominant

3

Z < 1

and attractive forces are dominant

4

Z < 1

and repulsive forces are dominant

Explanation:

Compressibility factor $(Z) = \frac{V_{r e a l}}{V_{i d e a l}}$
$∵ V_{real} < V_{ideal};$ Hence $Z < 1$
If $Z < 1$ , attractive forces are dominant among gaseous molecules and liquefaction of gas will be easy.

NEET - 2019

CHXI06:STATES OF MATTER

314067 The density of vapour of a substance $(X)$ at 1 $atm$ pressure and $500 K$ is $0.8 kg / m^{3}$ . The vapour effuses through a small hole at a rate of $4 / 5$ times slower than oxygen under the same condition. What is the compressibility factor $(Z)$ of the vapour?

1 0.974

2 1.35

3 1.52

4 1.22

Explanation:

$\frac{r_{X}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{X}}} \Rightarrow 452$
$= \frac{32}{M_{x}} \Rightarrow M_{x} = 50$
$d_{x} = 0.80 kg / m^{3}$
$V_{m} = \frac{1000}{800} \times 50 = 62.5 L$
$Z = \frac{P V_{m}}{R T} = \frac{1 \times 62.5}{0.0821 \times 500} = 1.52$

CHXI06:STATES OF MATTER

314061 Refer to the figure given and identify the correct statement.
(In all options, 'a' and 'b' are Van der Waals constant)
supporting img

1 For gas

A, a = 0

and

Z

will linearly depend on pressure

2 Gas

C

is a real gas and we can find '

a

' and '

b

' if intersection data is given

3 All van der Waal gases will behave like gas

C

and give positive slope at high pressure

4 For gas

B, b = 0

and

Z

will linearly depend on pressure

Explanation:

For gas $A, a = 0, Z = 1 + \frac{P b}{R T}$ implies $Z$ varies linearly with pressure.
For gas $B, b = 0, Z = 1 - \frac{a}{V R T}$ . Hence $Z$ does not vary linearly with pressure.
Given the intersection data for gas $C$ , it is possible to find the values of ' $a$ ' and ' $b$ '. All Van der Wall's gases like gas $C$ , gives positive slope at high pressures.

CHXI06:STATES OF MATTER

314062 The compressibility factor for definite amount of van der Waals' gas at $0^{\circ} C$ and 100 atm is found to be 0.5 . Assuming the volume of gas molecules negligible, the van der Waals' constant $a$ for a gas is ____ .

1 1.256

2 12.56

3 125.6

4 1256

Explanation:

$Z = \frac{P V}{n R T} = 0.5$
Now, $[P + \frac{n^{2} a}{V^{2}}] [V - n b] = n R T$
$[P + \frac{n^{2} a}{V^{2}}] [V] = n R T (b$ is negligible $)$
$P V^{2} - n R T V + n^{2} a = 0$
$∴ V = \frac{n R T \pm \sqrt{n^{2} R^{2} T^{2} - 4 n^{2} a \times b}}{2 P}$
Since, $V$ is constant at given $P$ and $T$ , thus, discriminant is 0 .
$∴ n^{2} R^{2} T^{2} = 4 n^{2} a P o r a = \frac{R^{2} T^{2}}{4 P}$
$= \frac{{(0.0821)}^{2} \times {(273)}^{2}}{4 \times 100}$
$= 1.256 L m o l^{- 2} a t m$

CHXI06:STATES OF MATTER

314080 The inversion temperature $(T_{i})$ for a gas is given by

1

a / Rb

2

2 a / Rb

3

\frac{Rb}{a}

4

\frac{2 Rb}{a}

Explanation:

$T_{i} = \frac{2 a}{R b}$

CHXI06:STATES OF MATTER

314066 A gas at $350 K$ and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor $(Z)$ is :

1

Z > 1

and attractive forces are dominant

2

Z > 1

and repulsive forces are dominant

3

Z < 1

and attractive forces are dominant

4

Z < 1

and repulsive forces are dominant

Explanation:

Compressibility factor $(Z) = \frac{V_{r e a l}}{V_{i d e a l}}$
$∵ V_{real} < V_{ideal};$ Hence $Z < 1$
If $Z < 1$ , attractive forces are dominant among gaseous molecules and liquefaction of gas will be easy.

NEET - 2019

CHXI06:STATES OF MATTER

314067 The density of vapour of a substance $(X)$ at 1 $atm$ pressure and $500 K$ is $0.8 kg / m^{3}$ . The vapour effuses through a small hole at a rate of $4 / 5$ times slower than oxygen under the same condition. What is the compressibility factor $(Z)$ of the vapour?

1 0.974

2 1.35

3 1.52

4 1.22

Explanation:

$\frac{r_{X}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{X}}} \Rightarrow 452$
$= \frac{32}{M_{x}} \Rightarrow M_{x} = 50$
$d_{x} = 0.80 kg / m^{3}$
$V_{m} = \frac{1000}{800} \times 50 = 62.5 L$
$Z = \frac{P V_{m}}{R T} = \frac{1 \times 62.5}{0.0821 \times 500} = 1.52$

CHXI06:STATES OF MATTER

314061 Refer to the figure given and identify the correct statement.
(In all options, 'a' and 'b' are Van der Waals constant)
supporting img

1 For gas

A, a = 0

and

Z

will linearly depend on pressure

2 Gas

C

is a real gas and we can find '

a

' and '

b

' if intersection data is given

3 All van der Waal gases will behave like gas

C

and give positive slope at high pressure

4 For gas

B, b = 0

and

Z

will linearly depend on pressure

Explanation:

For gas $A, a = 0, Z = 1 + \frac{P b}{R T}$ implies $Z$ varies linearly with pressure.
For gas $B, b = 0, Z = 1 - \frac{a}{V R T}$ . Hence $Z$ does not vary linearly with pressure.
Given the intersection data for gas $C$ , it is possible to find the values of ' $a$ ' and ' $b$ '. All Van der Wall's gases like gas $C$ , gives positive slope at high pressures.

CHXI06:STATES OF MATTER

314062 The compressibility factor for definite amount of van der Waals' gas at $0^{\circ} C$ and 100 atm is found to be 0.5 . Assuming the volume of gas molecules negligible, the van der Waals' constant $a$ for a gas is ____ .

1 1.256

2 12.56

3 125.6

4 1256

Explanation:

$Z = \frac{P V}{n R T} = 0.5$
Now, $[P + \frac{n^{2} a}{V^{2}}] [V - n b] = n R T$
$[P + \frac{n^{2} a}{V^{2}}] [V] = n R T (b$ is negligible $)$
$P V^{2} - n R T V + n^{2} a = 0$
$∴ V = \frac{n R T \pm \sqrt{n^{2} R^{2} T^{2} - 4 n^{2} a \times b}}{2 P}$
Since, $V$ is constant at given $P$ and $T$ , thus, discriminant is 0 .
$∴ n^{2} R^{2} T^{2} = 4 n^{2} a P o r a = \frac{R^{2} T^{2}}{4 P}$
$= \frac{{(0.0821)}^{2} \times {(273)}^{2}}{4 \times 100}$
$= 1.256 L m o l^{- 2} a t m$

CHXI06:STATES OF MATTER

314080 The inversion temperature $(T_{i})$ for a gas is given by

1

a / Rb

2

2 a / Rb

3

\frac{Rb}{a}

4

\frac{2 Rb}{a}

Explanation:

$T_{i} = \frac{2 a}{R b}$

CHXI06:STATES OF MATTER

314066 A gas at $350 K$ and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor $(Z)$ is :

1

Z > 1

and attractive forces are dominant

2

Z > 1

and repulsive forces are dominant

3

Z < 1

and attractive forces are dominant

4

Z < 1

and repulsive forces are dominant

Explanation:

Compressibility factor $(Z) = \frac{V_{r e a l}}{V_{i d e a l}}$
$∵ V_{real} < V_{ideal};$ Hence $Z < 1$
If $Z < 1$ , attractive forces are dominant among gaseous molecules and liquefaction of gas will be easy.

NEET - 2019

CHXI06:STATES OF MATTER

314067 The density of vapour of a substance $(X)$ at 1 $atm$ pressure and $500 K$ is $0.8 kg / m^{3}$ . The vapour effuses through a small hole at a rate of $4 / 5$ times slower than oxygen under the same condition. What is the compressibility factor $(Z)$ of the vapour?

1 0.974

2 1.35

3 1.52

4 1.22

Explanation:

$\frac{r_{X}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{X}}} \Rightarrow 452$
$= \frac{32}{M_{x}} \Rightarrow M_{x} = 50$
$d_{x} = 0.80 kg / m^{3}$
$V_{m} = \frac{1000}{800} \times 50 = 62.5 L$
$Z = \frac{P V_{m}}{R T} = \frac{1 \times 62.5}{0.0821 \times 500} = 1.52$

CHXI06:STATES OF MATTER

314061 Refer to the figure given and identify the correct statement.
(In all options, 'a' and 'b' are Van der Waals constant)
supporting img

1 For gas

A, a = 0

and

Z

will linearly depend on pressure

2 Gas

C

is a real gas and we can find '

a

' and '

b

' if intersection data is given

3 All van der Waal gases will behave like gas

C

and give positive slope at high pressure

4 For gas

B, b = 0

and

Z

will linearly depend on pressure

Explanation:

For gas $A, a = 0, Z = 1 + \frac{P b}{R T}$ implies $Z$ varies linearly with pressure.
For gas $B, b = 0, Z = 1 - \frac{a}{V R T}$ . Hence $Z$ does not vary linearly with pressure.
Given the intersection data for gas $C$ , it is possible to find the values of ' $a$ ' and ' $b$ '. All Van der Wall's gases like gas $C$ , gives positive slope at high pressures.

CHXI06:STATES OF MATTER

314062 The compressibility factor for definite amount of van der Waals' gas at $0^{\circ} C$ and 100 atm is found to be 0.5 . Assuming the volume of gas molecules negligible, the van der Waals' constant $a$ for a gas is ____ .

1 1.256

2 12.56

3 125.6

4 1256

Explanation:

$Z = \frac{P V}{n R T} = 0.5$
Now, $[P + \frac{n^{2} a}{V^{2}}] [V - n b] = n R T$
$[P + \frac{n^{2} a}{V^{2}}] [V] = n R T (b$ is negligible $)$
$P V^{2} - n R T V + n^{2} a = 0$
$∴ V = \frac{n R T \pm \sqrt{n^{2} R^{2} T^{2} - 4 n^{2} a \times b}}{2 P}$
Since, $V$ is constant at given $P$ and $T$ , thus, discriminant is 0 .
$∴ n^{2} R^{2} T^{2} = 4 n^{2} a P o r a = \frac{R^{2} T^{2}}{4 P}$
$= \frac{{(0.0821)}^{2} \times {(273)}^{2}}{4 \times 100}$
$= 1.256 L m o l^{- 2} a t m$

CHXI06:STATES OF MATTER

314080 The inversion temperature $(T_{i})$ for a gas is given by

1

a / Rb

2

2 a / Rb

3

\frac{Rb}{a}

4

\frac{2 Rb}{a}

Explanation:

$T_{i} = \frac{2 a}{R b}$

CHXI06:STATES OF MATTER

314066 A gas at $350 K$ and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor $(Z)$ is :

1

Z > 1

and attractive forces are dominant

2

Z > 1

and repulsive forces are dominant

3

Z < 1

and attractive forces are dominant

4

Z < 1

and repulsive forces are dominant

Explanation:

Compressibility factor $(Z) = \frac{V_{r e a l}}{V_{i d e a l}}$
$∵ V_{real} < V_{ideal};$ Hence $Z < 1$
If $Z < 1$ , attractive forces are dominant among gaseous molecules and liquefaction of gas will be easy.

NEET - 2019

CHXI06:STATES OF MATTER

314067 The density of vapour of a substance $(X)$ at 1 $atm$ pressure and $500 K$ is $0.8 kg / m^{3}$ . The vapour effuses through a small hole at a rate of $4 / 5$ times slower than oxygen under the same condition. What is the compressibility factor $(Z)$ of the vapour?

1 0.974

2 1.35

3 1.52

4 1.22

Explanation:

$\frac{r_{X}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{X}}} \Rightarrow 452$
$= \frac{32}{M_{x}} \Rightarrow M_{x} = 50$
$d_{x} = 0.80 kg / m^{3}$
$V_{m} = \frac{1000}{800} \times 50 = 62.5 L$
$Z = \frac{P V_{m}}{R T} = \frac{1 \times 62.5}{0.0821 \times 500} = 1.52$

CHXI06:STATES OF MATTER

314061 Refer to the figure given and identify the correct statement.
(In all options, 'a' and 'b' are Van der Waals constant)
supporting img

1 For gas

A, a = 0

and

Z

will linearly depend on pressure

2 Gas

C

is a real gas and we can find '

a

' and '

b

' if intersection data is given

3 All van der Waal gases will behave like gas

C

and give positive slope at high pressure

4 For gas

B, b = 0

and

Z

will linearly depend on pressure

Explanation:

For gas $A, a = 0, Z = 1 + \frac{P b}{R T}$ implies $Z$ varies linearly with pressure.
For gas $B, b = 0, Z = 1 - \frac{a}{V R T}$ . Hence $Z$ does not vary linearly with pressure.
Given the intersection data for gas $C$ , it is possible to find the values of ' $a$ ' and ' $b$ '. All Van der Wall's gases like gas $C$ , gives positive slope at high pressures.

CHXI06:STATES OF MATTER

314062 The compressibility factor for definite amount of van der Waals' gas at $0^{\circ} C$ and 100 atm is found to be 0.5 . Assuming the volume of gas molecules negligible, the van der Waals' constant $a$ for a gas is ____ .

1 1.256

2 12.56

3 125.6

4 1256

Explanation:

$Z = \frac{P V}{n R T} = 0.5$
Now, $[P + \frac{n^{2} a}{V^{2}}] [V - n b] = n R T$
$[P + \frac{n^{2} a}{V^{2}}] [V] = n R T (b$ is negligible $)$
$P V^{2} - n R T V + n^{2} a = 0$
$∴ V = \frac{n R T \pm \sqrt{n^{2} R^{2} T^{2} - 4 n^{2} a \times b}}{2 P}$
Since, $V$ is constant at given $P$ and $T$ , thus, discriminant is 0 .
$∴ n^{2} R^{2} T^{2} = 4 n^{2} a P o r a = \frac{R^{2} T^{2}}{4 P}$
$= \frac{{(0.0821)}^{2} \times {(273)}^{2}}{4 \times 100}$
$= 1.256 L m o l^{- 2} a t m$

CHXI06:STATES OF MATTER

314080 The inversion temperature $(T_{i})$ for a gas is given by

1

a / Rb

2

2 a / Rb

3

\frac{Rb}{a}

4

\frac{2 Rb}{a}

Explanation:

$T_{i} = \frac{2 a}{R b}$

CHXI06:STATES OF MATTER

314066 A gas at $350 K$ and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor $(Z)$ is :

1

Z > 1

and attractive forces are dominant

2

Z > 1

and repulsive forces are dominant

3

Z < 1

and attractive forces are dominant

4

Z < 1

and repulsive forces are dominant

Explanation:

Compressibility factor $(Z) = \frac{V_{r e a l}}{V_{i d e a l}}$
$∵ V_{real} < V_{ideal};$ Hence $Z < 1$
If $Z < 1$ , attractive forces are dominant among gaseous molecules and liquefaction of gas will be easy.

NEET - 2019

CHXI06:STATES OF MATTER

314067 The density of vapour of a substance $(X)$ at 1 $atm$ pressure and $500 K$ is $0.8 kg / m^{3}$ . The vapour effuses through a small hole at a rate of $4 / 5$ times slower than oxygen under the same condition. What is the compressibility factor $(Z)$ of the vapour?

1 0.974

2 1.35

3 1.52

4 1.22

Explanation:

$\frac{r_{X}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{X}}} \Rightarrow 452$
$= \frac{32}{M_{x}} \Rightarrow M_{x} = 50$
$d_{x} = 0.80 kg / m^{3}$
$V_{m} = \frac{1000}{800} \times 50 = 62.5 L$
$Z = \frac{P V_{m}}{R T} = \frac{1 \times 62.5}{0.0821 \times 500} = 1.52$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: