QBManager

Kinetic Theory of Gases

139054 Two identical containers each of volume $V_{0}$ are joined by a small pipe. The containers contain identical gases at temperature $T_{0}$ and pressure $P_{0}$. One container is heated to temperature $2 T_{0}$ while maintaining the other at the same temperature. The common pressure of gas is $P$ and $n$ is the number of moles of gas in container at temperature $2 \mathrm{~T}_{0}$

1 $\mathrm{P}=2 \mathrm{P}_{0}$

2 $\mathrm{P}=\frac{4}{3} \mathrm{P}_{0}$

3 $\mathrm{n}=\frac{2}{3} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

4 $\mathrm{n}=\frac{3}{2} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

Manipal UGET-2014

Kinetic Theory of Gases

139056 The internal energy of an ideal gas depends on which of the following factors?

1 Pressure only

2 Volume only

3 Temperature only

4 Pressure, volume and temperature

SCRA-2010

Kinetic Theory of Gases

139057 An ideal gas with $\gamma=\frac{5}{3}$ is originally of volume
$V_{0}$ and pressure $P_{0}$. If it expands adiabatically to final volume $V_{1}$. What is the work done by the gas in this process?

1 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{2}$

2 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{3}$

3 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{3}$

4 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{2}$

SCRA-2009

Kinetic Theory of Gases

139058 One mole of an ideal gas is taken from $A$ to $B$, from $B$ to $C$ and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure $A$ is $P_{0}$, volume is $V_{0}$, Then, the internal energy :

1 at $\mathrm{A}$ and $\mathrm{B}$ are equal

2 at $\mathrm{A}$ is more than at $\mathrm{B}$

3 at $C$ is less than at $B$

4 at $\mathrm{B}$ is more than at $\mathrm{A}$

Karnataka CET-2012

Kinetic Theory of Gases

139059 The average pressure of an ideal gas is

1 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}^{2}$

2 $\mathrm{P}=\frac{1}{2} \mathrm{mnv}_{\mathrm{av}}$

3 $\mathrm{P}=\frac{1}{4} \mathrm{mnv}_{\mathrm{av}}^{2}$

4 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}$
Where symbols have their usual meanings.

Explanation:

A
Now, consider a cubical box, contains ideal gas. The molecule of ideal gas have motion in different direction (x, $y, z$ direction).
Force on the wall due to velocity in $\mathrm{x}$-direction $\left(\mathrm{v}_{\mathrm{x}}\right)$
Change in the momentum in $x$ direction
$\Delta \mathrm{p}_{\mathrm{x}} =\mathrm{mv}_{\mathrm{x}}-\left(-\mathrm{mv}_{\mathrm{x}}\right)$
$\Delta \mathrm{p}_{\mathrm{x}} =2 \mathrm{mv}_{\mathrm{x}}$
After the collision, the molecule travels a distance of $2 l$ before colliding again with wall 1 .
Thus, time taken by
$\text { Time }=\frac{\text { Distance travelled }}{\text { Velocity }}=\frac{2 l}{\mathrm{v}_{\mathrm{x}}}$
These continuous collisions carry a force,
$\text { Force }=\frac{\text { Change in momentum }}{\text { Change in time }}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$
Putting value of equation (i) and (ii) in equation (iii), we get-
$\text { Force }(\mathrm{F})=\frac{2 \mathrm{mv}_{\mathrm{x}}}{2 l / \mathrm{v}_{\mathrm{x}}}$
$\mathrm{F}=\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l}$
These continuous collision also exert pressure on the wall,
$\text { Pressure } =\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{mv}_{\mathrm{x}}^{2} / l}{(l)^{2}}$
$\mathrm{P} =\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l^{3}}$
Thus pressure exerted by on wall 1 by the collision of $\mathrm{N}$ number of gas molecule is given by-
$\mathrm{P}=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{v}_{\mathrm{x}_{1}}^{2}+\mathrm{v}_{\mathrm{x}_{2}}^{2}+\mathrm{v}_{\mathrm{x}_{3}}^{2}+\ldots .+\mathrm{v}_{\mathrm{x}_{\mathrm{N}}}^{2}\right)$
$=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{~N} \overline{\mathrm{v}}_{\mathrm{x}}^{2}\right)$
$\mathrm{P} =\frac{\mathrm{mN} \overline{\mathrm{v}}_{\mathrm{x}}^{2}}{\mathrm{v}}$
On extending the above equation to three dimension,
$\mathrm{v}^{2}=\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{\mathrm{z}}^{2}$
$\overline{\mathrm{v}}^{2}=\overline{\mathrm{v}}_{\mathrm{x}}^{2}+\overline{\mathrm{v}}_{\mathrm{y}}^{2}+\overline{\mathrm{v}}_{\mathrm{z}}^{2}$
For a large number of gas molecule,
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\overline{\mathrm{v}}_{\mathrm{y}}^{2}=\overline{\mathrm{v}}_{\mathrm{z}}^{2}=\overline{\mathrm{v}}^{2}$
$\overline{\mathrm{v}}^{2}=3 \overline{\mathrm{v}}_{\mathrm{x}}^{2}$
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\frac{1}{3} \overline{\mathrm{v}}^{2}$
Substituting equation (vii) in (vi), we get-
$\mathrm{P}=\frac{\mathrm{Nm}\left(\frac{1}{3} \overline{\mathrm{v}}^{2}\right)}{\mathrm{v}}$
$\mathrm{PV}=\frac{1}{3} \mathrm{Nm} \overline{\mathrm{v}}^{2}$

J and K-CET-2013

Kinetic Theory of Gases

139054 Two identical containers each of volume $V_{0}$ are joined by a small pipe. The containers contain identical gases at temperature $T_{0}$ and pressure $P_{0}$. One container is heated to temperature $2 T_{0}$ while maintaining the other at the same temperature. The common pressure of gas is $P$ and $n$ is the number of moles of gas in container at temperature $2 \mathrm{~T}_{0}$

1 $\mathrm{P}=2 \mathrm{P}_{0}$

2 $\mathrm{P}=\frac{4}{3} \mathrm{P}_{0}$

3 $\mathrm{n}=\frac{2}{3} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

4 $\mathrm{n}=\frac{3}{2} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

Manipal UGET-2014

Kinetic Theory of Gases

139056 The internal energy of an ideal gas depends on which of the following factors?

1 Pressure only

2 Volume only

3 Temperature only

4 Pressure, volume and temperature

SCRA-2010

Kinetic Theory of Gases

139057 An ideal gas with $\gamma=\frac{5}{3}$ is originally of volume
$V_{0}$ and pressure $P_{0}$. If it expands adiabatically to final volume $V_{1}$. What is the work done by the gas in this process?

1 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{2}$

2 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{3}$

3 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{3}$

4 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{2}$

SCRA-2009

Kinetic Theory of Gases

139058 One mole of an ideal gas is taken from $A$ to $B$, from $B$ to $C$ and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure $A$ is $P_{0}$, volume is $V_{0}$, Then, the internal energy :

1 at $\mathrm{A}$ and $\mathrm{B}$ are equal

2 at $\mathrm{A}$ is more than at $\mathrm{B}$

3 at $C$ is less than at $B$

4 at $\mathrm{B}$ is more than at $\mathrm{A}$

Karnataka CET-2012

Kinetic Theory of Gases

139059 The average pressure of an ideal gas is

1 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}^{2}$

2 $\mathrm{P}=\frac{1}{2} \mathrm{mnv}_{\mathrm{av}}$

3 $\mathrm{P}=\frac{1}{4} \mathrm{mnv}_{\mathrm{av}}^{2}$

4 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}$
Where symbols have their usual meanings.

Explanation:

A
Now, consider a cubical box, contains ideal gas. The molecule of ideal gas have motion in different direction (x, $y, z$ direction).
Force on the wall due to velocity in $\mathrm{x}$-direction $\left(\mathrm{v}_{\mathrm{x}}\right)$
Change in the momentum in $x$ direction
$\Delta \mathrm{p}_{\mathrm{x}} =\mathrm{mv}_{\mathrm{x}}-\left(-\mathrm{mv}_{\mathrm{x}}\right)$
$\Delta \mathrm{p}_{\mathrm{x}} =2 \mathrm{mv}_{\mathrm{x}}$
After the collision, the molecule travels a distance of $2 l$ before colliding again with wall 1 .
Thus, time taken by
$\text { Time }=\frac{\text { Distance travelled }}{\text { Velocity }}=\frac{2 l}{\mathrm{v}_{\mathrm{x}}}$
These continuous collisions carry a force,
$\text { Force }=\frac{\text { Change in momentum }}{\text { Change in time }}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$
Putting value of equation (i) and (ii) in equation (iii), we get-
$\text { Force }(\mathrm{F})=\frac{2 \mathrm{mv}_{\mathrm{x}}}{2 l / \mathrm{v}_{\mathrm{x}}}$
$\mathrm{F}=\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l}$
These continuous collision also exert pressure on the wall,
$\text { Pressure } =\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{mv}_{\mathrm{x}}^{2} / l}{(l)^{2}}$
$\mathrm{P} =\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l^{3}}$
Thus pressure exerted by on wall 1 by the collision of $\mathrm{N}$ number of gas molecule is given by-
$\mathrm{P}=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{v}_{\mathrm{x}_{1}}^{2}+\mathrm{v}_{\mathrm{x}_{2}}^{2}+\mathrm{v}_{\mathrm{x}_{3}}^{2}+\ldots .+\mathrm{v}_{\mathrm{x}_{\mathrm{N}}}^{2}\right)$
$=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{~N} \overline{\mathrm{v}}_{\mathrm{x}}^{2}\right)$
$\mathrm{P} =\frac{\mathrm{mN} \overline{\mathrm{v}}_{\mathrm{x}}^{2}}{\mathrm{v}}$
On extending the above equation to three dimension,
$\mathrm{v}^{2}=\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{\mathrm{z}}^{2}$
$\overline{\mathrm{v}}^{2}=\overline{\mathrm{v}}_{\mathrm{x}}^{2}+\overline{\mathrm{v}}_{\mathrm{y}}^{2}+\overline{\mathrm{v}}_{\mathrm{z}}^{2}$
For a large number of gas molecule,
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\overline{\mathrm{v}}_{\mathrm{y}}^{2}=\overline{\mathrm{v}}_{\mathrm{z}}^{2}=\overline{\mathrm{v}}^{2}$
$\overline{\mathrm{v}}^{2}=3 \overline{\mathrm{v}}_{\mathrm{x}}^{2}$
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\frac{1}{3} \overline{\mathrm{v}}^{2}$
Substituting equation (vii) in (vi), we get-
$\mathrm{P}=\frac{\mathrm{Nm}\left(\frac{1}{3} \overline{\mathrm{v}}^{2}\right)}{\mathrm{v}}$
$\mathrm{PV}=\frac{1}{3} \mathrm{Nm} \overline{\mathrm{v}}^{2}$

J and K-CET-2013

Kinetic Theory of Gases

139054 Two identical containers each of volume $V_{0}$ are joined by a small pipe. The containers contain identical gases at temperature $T_{0}$ and pressure $P_{0}$. One container is heated to temperature $2 T_{0}$ while maintaining the other at the same temperature. The common pressure of gas is $P$ and $n$ is the number of moles of gas in container at temperature $2 \mathrm{~T}_{0}$

1 $\mathrm{P}=2 \mathrm{P}_{0}$

2 $\mathrm{P}=\frac{4}{3} \mathrm{P}_{0}$

3 $\mathrm{n}=\frac{2}{3} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

4 $\mathrm{n}=\frac{3}{2} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

Manipal UGET-2014

Kinetic Theory of Gases

139056 The internal energy of an ideal gas depends on which of the following factors?

1 Pressure only

2 Volume only

3 Temperature only

4 Pressure, volume and temperature

SCRA-2010

Kinetic Theory of Gases

139057 An ideal gas with $\gamma=\frac{5}{3}$ is originally of volume
$V_{0}$ and pressure $P_{0}$. If it expands adiabatically to final volume $V_{1}$. What is the work done by the gas in this process?

1 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{2}$

2 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{3}$

3 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{3}$

4 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{2}$

SCRA-2009

Kinetic Theory of Gases

139058 One mole of an ideal gas is taken from $A$ to $B$, from $B$ to $C$ and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure $A$ is $P_{0}$, volume is $V_{0}$, Then, the internal energy :

1 at $\mathrm{A}$ and $\mathrm{B}$ are equal

2 at $\mathrm{A}$ is more than at $\mathrm{B}$

3 at $C$ is less than at $B$

4 at $\mathrm{B}$ is more than at $\mathrm{A}$

Karnataka CET-2012

Kinetic Theory of Gases

139059 The average pressure of an ideal gas is

1 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}^{2}$

2 $\mathrm{P}=\frac{1}{2} \mathrm{mnv}_{\mathrm{av}}$

3 $\mathrm{P}=\frac{1}{4} \mathrm{mnv}_{\mathrm{av}}^{2}$

4 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}$
Where symbols have their usual meanings.

Explanation:

A
Now, consider a cubical box, contains ideal gas. The molecule of ideal gas have motion in different direction (x, $y, z$ direction).
Force on the wall due to velocity in $\mathrm{x}$-direction $\left(\mathrm{v}_{\mathrm{x}}\right)$
Change in the momentum in $x$ direction
$\Delta \mathrm{p}_{\mathrm{x}} =\mathrm{mv}_{\mathrm{x}}-\left(-\mathrm{mv}_{\mathrm{x}}\right)$
$\Delta \mathrm{p}_{\mathrm{x}} =2 \mathrm{mv}_{\mathrm{x}}$
After the collision, the molecule travels a distance of $2 l$ before colliding again with wall 1 .
Thus, time taken by
$\text { Time }=\frac{\text { Distance travelled }}{\text { Velocity }}=\frac{2 l}{\mathrm{v}_{\mathrm{x}}}$
These continuous collisions carry a force,
$\text { Force }=\frac{\text { Change in momentum }}{\text { Change in time }}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$
Putting value of equation (i) and (ii) in equation (iii), we get-
$\text { Force }(\mathrm{F})=\frac{2 \mathrm{mv}_{\mathrm{x}}}{2 l / \mathrm{v}_{\mathrm{x}}}$
$\mathrm{F}=\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l}$
These continuous collision also exert pressure on the wall,
$\text { Pressure } =\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{mv}_{\mathrm{x}}^{2} / l}{(l)^{2}}$
$\mathrm{P} =\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l^{3}}$
Thus pressure exerted by on wall 1 by the collision of $\mathrm{N}$ number of gas molecule is given by-
$\mathrm{P}=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{v}_{\mathrm{x}_{1}}^{2}+\mathrm{v}_{\mathrm{x}_{2}}^{2}+\mathrm{v}_{\mathrm{x}_{3}}^{2}+\ldots .+\mathrm{v}_{\mathrm{x}_{\mathrm{N}}}^{2}\right)$
$=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{~N} \overline{\mathrm{v}}_{\mathrm{x}}^{2}\right)$
$\mathrm{P} =\frac{\mathrm{mN} \overline{\mathrm{v}}_{\mathrm{x}}^{2}}{\mathrm{v}}$
On extending the above equation to three dimension,
$\mathrm{v}^{2}=\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{\mathrm{z}}^{2}$
$\overline{\mathrm{v}}^{2}=\overline{\mathrm{v}}_{\mathrm{x}}^{2}+\overline{\mathrm{v}}_{\mathrm{y}}^{2}+\overline{\mathrm{v}}_{\mathrm{z}}^{2}$
For a large number of gas molecule,
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\overline{\mathrm{v}}_{\mathrm{y}}^{2}=\overline{\mathrm{v}}_{\mathrm{z}}^{2}=\overline{\mathrm{v}}^{2}$
$\overline{\mathrm{v}}^{2}=3 \overline{\mathrm{v}}_{\mathrm{x}}^{2}$
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\frac{1}{3} \overline{\mathrm{v}}^{2}$
Substituting equation (vii) in (vi), we get-
$\mathrm{P}=\frac{\mathrm{Nm}\left(\frac{1}{3} \overline{\mathrm{v}}^{2}\right)}{\mathrm{v}}$
$\mathrm{PV}=\frac{1}{3} \mathrm{Nm} \overline{\mathrm{v}}^{2}$

J and K-CET-2013

Kinetic Theory of Gases

139054 Two identical containers each of volume $V_{0}$ are joined by a small pipe. The containers contain identical gases at temperature $T_{0}$ and pressure $P_{0}$. One container is heated to temperature $2 T_{0}$ while maintaining the other at the same temperature. The common pressure of gas is $P$ and $n$ is the number of moles of gas in container at temperature $2 \mathrm{~T}_{0}$

1 $\mathrm{P}=2 \mathrm{P}_{0}$

2 $\mathrm{P}=\frac{4}{3} \mathrm{P}_{0}$

3 $\mathrm{n}=\frac{2}{3} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

4 $\mathrm{n}=\frac{3}{2} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

Manipal UGET-2014

Kinetic Theory of Gases

139056 The internal energy of an ideal gas depends on which of the following factors?

1 Pressure only

2 Volume only

3 Temperature only

4 Pressure, volume and temperature

SCRA-2010

Kinetic Theory of Gases

139057 An ideal gas with $\gamma=\frac{5}{3}$ is originally of volume
$V_{0}$ and pressure $P_{0}$. If it expands adiabatically to final volume $V_{1}$. What is the work done by the gas in this process?

1 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{2}$

2 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{3}$

3 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{3}$

4 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{2}$

SCRA-2009

Kinetic Theory of Gases

139058 One mole of an ideal gas is taken from $A$ to $B$, from $B$ to $C$ and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure $A$ is $P_{0}$, volume is $V_{0}$, Then, the internal energy :

1 at $\mathrm{A}$ and $\mathrm{B}$ are equal

2 at $\mathrm{A}$ is more than at $\mathrm{B}$

3 at $C$ is less than at $B$

4 at $\mathrm{B}$ is more than at $\mathrm{A}$

Karnataka CET-2012

Kinetic Theory of Gases

139059 The average pressure of an ideal gas is

1 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}^{2}$

2 $\mathrm{P}=\frac{1}{2} \mathrm{mnv}_{\mathrm{av}}$

3 $\mathrm{P}=\frac{1}{4} \mathrm{mnv}_{\mathrm{av}}^{2}$

4 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}$
Where symbols have their usual meanings.

Explanation:

A
Now, consider a cubical box, contains ideal gas. The molecule of ideal gas have motion in different direction (x, $y, z$ direction).
Force on the wall due to velocity in $\mathrm{x}$-direction $\left(\mathrm{v}_{\mathrm{x}}\right)$
Change in the momentum in $x$ direction
$\Delta \mathrm{p}_{\mathrm{x}} =\mathrm{mv}_{\mathrm{x}}-\left(-\mathrm{mv}_{\mathrm{x}}\right)$
$\Delta \mathrm{p}_{\mathrm{x}} =2 \mathrm{mv}_{\mathrm{x}}$
After the collision, the molecule travels a distance of $2 l$ before colliding again with wall 1 .
Thus, time taken by
$\text { Time }=\frac{\text { Distance travelled }}{\text { Velocity }}=\frac{2 l}{\mathrm{v}_{\mathrm{x}}}$
These continuous collisions carry a force,
$\text { Force }=\frac{\text { Change in momentum }}{\text { Change in time }}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$
Putting value of equation (i) and (ii) in equation (iii), we get-
$\text { Force }(\mathrm{F})=\frac{2 \mathrm{mv}_{\mathrm{x}}}{2 l / \mathrm{v}_{\mathrm{x}}}$
$\mathrm{F}=\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l}$
These continuous collision also exert pressure on the wall,
$\text { Pressure } =\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{mv}_{\mathrm{x}}^{2} / l}{(l)^{2}}$
$\mathrm{P} =\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l^{3}}$
Thus pressure exerted by on wall 1 by the collision of $\mathrm{N}$ number of gas molecule is given by-
$\mathrm{P}=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{v}_{\mathrm{x}_{1}}^{2}+\mathrm{v}_{\mathrm{x}_{2}}^{2}+\mathrm{v}_{\mathrm{x}_{3}}^{2}+\ldots .+\mathrm{v}_{\mathrm{x}_{\mathrm{N}}}^{2}\right)$
$=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{~N} \overline{\mathrm{v}}_{\mathrm{x}}^{2}\right)$
$\mathrm{P} =\frac{\mathrm{mN} \overline{\mathrm{v}}_{\mathrm{x}}^{2}}{\mathrm{v}}$
On extending the above equation to three dimension,
$\mathrm{v}^{2}=\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{\mathrm{z}}^{2}$
$\overline{\mathrm{v}}^{2}=\overline{\mathrm{v}}_{\mathrm{x}}^{2}+\overline{\mathrm{v}}_{\mathrm{y}}^{2}+\overline{\mathrm{v}}_{\mathrm{z}}^{2}$
For a large number of gas molecule,
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\overline{\mathrm{v}}_{\mathrm{y}}^{2}=\overline{\mathrm{v}}_{\mathrm{z}}^{2}=\overline{\mathrm{v}}^{2}$
$\overline{\mathrm{v}}^{2}=3 \overline{\mathrm{v}}_{\mathrm{x}}^{2}$
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\frac{1}{3} \overline{\mathrm{v}}^{2}$
Substituting equation (vii) in (vi), we get-
$\mathrm{P}=\frac{\mathrm{Nm}\left(\frac{1}{3} \overline{\mathrm{v}}^{2}\right)}{\mathrm{v}}$
$\mathrm{PV}=\frac{1}{3} \mathrm{Nm} \overline{\mathrm{v}}^{2}$

J and K-CET-2013

Kinetic Theory of Gases

139054 Two identical containers each of volume $V_{0}$ are joined by a small pipe. The containers contain identical gases at temperature $T_{0}$ and pressure $P_{0}$. One container is heated to temperature $2 T_{0}$ while maintaining the other at the same temperature. The common pressure of gas is $P$ and $n$ is the number of moles of gas in container at temperature $2 \mathrm{~T}_{0}$

1 $\mathrm{P}=2 \mathrm{P}_{0}$

2 $\mathrm{P}=\frac{4}{3} \mathrm{P}_{0}$

3 $\mathrm{n}=\frac{2}{3} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

4 $\mathrm{n}=\frac{3}{2} \frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{RT}_{0}}$

Manipal UGET-2014

Kinetic Theory of Gases

139056 The internal energy of an ideal gas depends on which of the following factors?

1 Pressure only

2 Volume only

3 Temperature only

4 Pressure, volume and temperature

SCRA-2010

Kinetic Theory of Gases

139057 An ideal gas with $\gamma=\frac{5}{3}$ is originally of volume
$V_{0}$ and pressure $P_{0}$. If it expands adiabatically to final volume $V_{1}$. What is the work done by the gas in this process?

1 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{2}$

2 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{3}$

3 $\frac{2 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{3}{2}}\right]}{3}$

4 $\frac{3 \mathrm{P}_0 \mathrm{~V}_0\left[1-\left(\frac{\mathrm{V}_0}{\mathrm{~V}_1}\right)^{\frac{2}{3}}\right]}{2}$

SCRA-2009

Kinetic Theory of Gases

139058 One mole of an ideal gas is taken from $A$ to $B$, from $B$ to $C$ and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure $A$ is $P_{0}$, volume is $V_{0}$, Then, the internal energy :

1 at $\mathrm{A}$ and $\mathrm{B}$ are equal

2 at $\mathrm{A}$ is more than at $\mathrm{B}$

3 at $C$ is less than at $B$

4 at $\mathrm{B}$ is more than at $\mathrm{A}$

Karnataka CET-2012

Kinetic Theory of Gases

139059 The average pressure of an ideal gas is

1 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}^{2}$

2 $\mathrm{P}=\frac{1}{2} \mathrm{mnv}_{\mathrm{av}}$

3 $\mathrm{P}=\frac{1}{4} \mathrm{mnv}_{\mathrm{av}}^{2}$

4 $\mathrm{P}=\frac{1}{3} \mathrm{mnv}_{\mathrm{av}}$
Where symbols have their usual meanings.

Explanation:

A
Now, consider a cubical box, contains ideal gas. The molecule of ideal gas have motion in different direction (x, $y, z$ direction).
Force on the wall due to velocity in $\mathrm{x}$-direction $\left(\mathrm{v}_{\mathrm{x}}\right)$
Change in the momentum in $x$ direction
$\Delta \mathrm{p}_{\mathrm{x}} =\mathrm{mv}_{\mathrm{x}}-\left(-\mathrm{mv}_{\mathrm{x}}\right)$
$\Delta \mathrm{p}_{\mathrm{x}} =2 \mathrm{mv}_{\mathrm{x}}$
After the collision, the molecule travels a distance of $2 l$ before colliding again with wall 1 .
Thus, time taken by
$\text { Time }=\frac{\text { Distance travelled }}{\text { Velocity }}=\frac{2 l}{\mathrm{v}_{\mathrm{x}}}$
These continuous collisions carry a force,
$\text { Force }=\frac{\text { Change in momentum }}{\text { Change in time }}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$
Putting value of equation (i) and (ii) in equation (iii), we get-
$\text { Force }(\mathrm{F})=\frac{2 \mathrm{mv}_{\mathrm{x}}}{2 l / \mathrm{v}_{\mathrm{x}}}$
$\mathrm{F}=\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l}$
These continuous collision also exert pressure on the wall,
$\text { Pressure } =\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{mv}_{\mathrm{x}}^{2} / l}{(l)^{2}}$
$\mathrm{P} =\frac{\mathrm{mv}_{\mathrm{x}}^{2}}{l^{3}}$
Thus pressure exerted by on wall 1 by the collision of $\mathrm{N}$ number of gas molecule is given by-
$\mathrm{P}=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{v}_{\mathrm{x}_{1}}^{2}+\mathrm{v}_{\mathrm{x}_{2}}^{2}+\mathrm{v}_{\mathrm{x}_{3}}^{2}+\ldots .+\mathrm{v}_{\mathrm{x}_{\mathrm{N}}}^{2}\right)$
$=\frac{\mathrm{m}}{l^{3}}\left(\mathrm{~N} \overline{\mathrm{v}}_{\mathrm{x}}^{2}\right)$
$\mathrm{P} =\frac{\mathrm{mN} \overline{\mathrm{v}}_{\mathrm{x}}^{2}}{\mathrm{v}}$
On extending the above equation to three dimension,
$\mathrm{v}^{2}=\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{\mathrm{z}}^{2}$
$\overline{\mathrm{v}}^{2}=\overline{\mathrm{v}}_{\mathrm{x}}^{2}+\overline{\mathrm{v}}_{\mathrm{y}}^{2}+\overline{\mathrm{v}}_{\mathrm{z}}^{2}$
For a large number of gas molecule,
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\overline{\mathrm{v}}_{\mathrm{y}}^{2}=\overline{\mathrm{v}}_{\mathrm{z}}^{2}=\overline{\mathrm{v}}^{2}$
$\overline{\mathrm{v}}^{2}=3 \overline{\mathrm{v}}_{\mathrm{x}}^{2}$
$\overline{\mathrm{v}}_{\mathrm{x}}^{2}=\frac{1}{3} \overline{\mathrm{v}}^{2}$
Substituting equation (vii) in (vi), we get-
$\mathrm{P}=\frac{\mathrm{Nm}\left(\frac{1}{3} \overline{\mathrm{v}}^{2}\right)}{\mathrm{v}}$
$\mathrm{PV}=\frac{1}{3} \mathrm{Nm} \overline{\mathrm{v}}^{2}$

J and K-CET-2013