QBManager

Kinetic Theory of Gases

139037 An experiment is carried on a fixed amount of gas at different temperature and at high pressure such that it deviates from the ideal gas behaviour. The variation of $\frac{p V}{R T}$ with $p$ is shown in the diagram. The correct variation will correspond to

1 Curve

A

2 Curve B

3 Curve

C

4 Curve D

Explanation:

B Here, an experiment is carried on a fixed amount of gas at different temperature and the condition is given at low pressure and high temperature that means its density will be significantly less.
Then we can use ideal gas equation formula-
$∵ PV = nRT$
$n = \frac{PV}{RT}$
Therefore, $\frac{PV}{RT} =$ constant
But when the pressure is increased then the decrease in volume will not take place in same proportion.
Therefore, $\frac{PV}{RT}$ will increase
Thus, we can say that the correct variation will correspond to curve $B$ .

Manipal UGET-2014 UP CPMT-2014

Kinetic Theory of Gases

139038 The equation $(p + \frac{a}{V^{2}}) (V - b) =$ constant. The unit of a is

1 Dyne

\times {cm}^{5}

2 Dyne

\times {cm}^{4}

3 Dyne

/ {cm}^{3}

4 Dyne/cm

^{2}

Explanation:

B According to principal dimensional homognity
$(P + \frac{a}{v^{2}}) (v - b) = constant$
We know-
$P = \frac{F}{A} = Dyne / {cm}^{2}$
$\frac{a}{v^{2}} = \frac{a}{{({cm}^{3})}^{2}} = Dyne / {cm}^{2}$
The units of $a = \frac{Dyne}{(cm)^{2}} \times {({cm}^{3})}^{2}$
$= \frac{Dyne}{(cm)^{2}} \times {cm}^{6}$
$= Dyne \times {cm}^{4}$

Manipal UGET-2014

Kinetic Theory of Gases

139039 Air is filled in a bottle at atmospheric pressure and it is corked at $35^{\circ} C$ . If the cork can come out at 3 atmospheric pressure, then upto what temperature should the bottle be heated in order to remove the cork.

1

{325.5}^{\circ} C

2

851^{\circ} C

3

651^{\circ} C

4 None of these

Explanation:

C Given that,
$P_{1} = 1 atm$
$T_{1} = 35^{\circ} C = (35 + 273 K) = 308 K$
$P_{2} = 3 atm, T_{2} = ?$
From Gay Lusac's law,
Volume $=$ constant
$\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1}{3} = \frac{308}{T_{2}}$
$T_{2} = 308 \times 3$
$= 924 K$
$T_{2} = (924 - 273)^{\circ} C$
$= 651^{\circ} C$

JCECE-2016

Kinetic Theory of Gases

139040 A partition divides a container having insulated walls into two compartments. The same gas fills the two compartments (see figure). The ratio of the number of molecules in compartments I and $II$ is

1

6 : 1

2

1 : 6

3

4 : 1

4

1 : 4

Explanation:

B According to ideal gas equation,
$PV = {nRT}^{2}$
$P_{1} V_{1} = n_{1} {RT}_{1}$
$pV = n_{1} RT$
and
$P_{2} V_{2} = n_{2} {RT}_{2}$
$(2 p) (3 V) = n_{2} RT$
$6 pV = n_{2} RT$
From equation (i) and equation (ii)
$n_{1} RT = \frac{n_{2} RT}{6}$
$\frac{n_{1}}{n_{2}} = \frac{1}{6}$
$n_{1} : n_{2} = 1 : 6$

JCECE-2015

Kinetic Theory of Gases

139041 Air is pumped into an automobile tube upto a pressure of $200 kPa$ in the morning when the air temperature is $22^{\circ} C$ . During the day, temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

212 kPa

2

209 kPa

3

206 kPa

4

200 kPa

Explanation:

B Given that,
$T_{1} = 22^{\circ} C T_{2} = 42^{\circ} C$
$= 22 + 273 = 42 + 273$
$= 295 K = 315 K$
$P_{1} = 200 kPa$
$V_{2} = V_{1} + 0.02 V_{1} = 1.02 V_{1}$
We know that,
$PV = nRT$
Where $\frac{PV}{T} =$ constant
Then, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$∴ \frac{T_{2}}{T_{1}} = \frac{P_{2} V_{2}}{P_{1} V_{1}}$
$\frac{315}{295} = \frac{P_{2}}{200} \times 1.02$
$P_{2} = \frac{315 \times 200}{295 \times 1.02}$
$P_{2} = 209.37 kPa = 209 kPa$

JCECE-2010]

Kinetic Theory of Gases

139037 An experiment is carried on a fixed amount of gas at different temperature and at high pressure such that it deviates from the ideal gas behaviour. The variation of $\frac{p V}{R T}$ with $p$ is shown in the diagram. The correct variation will correspond to

1 Curve

A

2 Curve B

3 Curve

C

4 Curve D

Explanation:

B Here, an experiment is carried on a fixed amount of gas at different temperature and the condition is given at low pressure and high temperature that means its density will be significantly less.
Then we can use ideal gas equation formula-
$∵ PV = nRT$
$n = \frac{PV}{RT}$
Therefore, $\frac{PV}{RT} =$ constant
But when the pressure is increased then the decrease in volume will not take place in same proportion.
Therefore, $\frac{PV}{RT}$ will increase
Thus, we can say that the correct variation will correspond to curve $B$ .

Manipal UGET-2014 UP CPMT-2014

Kinetic Theory of Gases

139038 The equation $(p + \frac{a}{V^{2}}) (V - b) =$ constant. The unit of a is

1 Dyne

\times {cm}^{5}

2 Dyne

\times {cm}^{4}

3 Dyne

/ {cm}^{3}

4 Dyne/cm

^{2}

Explanation:

B According to principal dimensional homognity
$(P + \frac{a}{v^{2}}) (v - b) = constant$
We know-
$P = \frac{F}{A} = Dyne / {cm}^{2}$
$\frac{a}{v^{2}} = \frac{a}{{({cm}^{3})}^{2}} = Dyne / {cm}^{2}$
The units of $a = \frac{Dyne}{(cm)^{2}} \times {({cm}^{3})}^{2}$
$= \frac{Dyne}{(cm)^{2}} \times {cm}^{6}$
$= Dyne \times {cm}^{4}$

Manipal UGET-2014

Kinetic Theory of Gases

139039 Air is filled in a bottle at atmospheric pressure and it is corked at $35^{\circ} C$ . If the cork can come out at 3 atmospheric pressure, then upto what temperature should the bottle be heated in order to remove the cork.

1

{325.5}^{\circ} C

2

851^{\circ} C

3

651^{\circ} C

4 None of these

Explanation:

C Given that,
$P_{1} = 1 atm$
$T_{1} = 35^{\circ} C = (35 + 273 K) = 308 K$
$P_{2} = 3 atm, T_{2} = ?$
From Gay Lusac's law,
Volume $=$ constant
$\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1}{3} = \frac{308}{T_{2}}$
$T_{2} = 308 \times 3$
$= 924 K$
$T_{2} = (924 - 273)^{\circ} C$
$= 651^{\circ} C$

JCECE-2016

Kinetic Theory of Gases

139040 A partition divides a container having insulated walls into two compartments. The same gas fills the two compartments (see figure). The ratio of the number of molecules in compartments I and $II$ is

1

6 : 1

2

1 : 6

3

4 : 1

4

1 : 4

Explanation:

B According to ideal gas equation,
$PV = {nRT}^{2}$
$P_{1} V_{1} = n_{1} {RT}_{1}$
$pV = n_{1} RT$
and
$P_{2} V_{2} = n_{2} {RT}_{2}$
$(2 p) (3 V) = n_{2} RT$
$6 pV = n_{2} RT$
From equation (i) and equation (ii)
$n_{1} RT = \frac{n_{2} RT}{6}$
$\frac{n_{1}}{n_{2}} = \frac{1}{6}$
$n_{1} : n_{2} = 1 : 6$

JCECE-2015

Kinetic Theory of Gases

139041 Air is pumped into an automobile tube upto a pressure of $200 kPa$ in the morning when the air temperature is $22^{\circ} C$ . During the day, temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

212 kPa

2

209 kPa

3

206 kPa

4

200 kPa

Explanation:

B Given that,
$T_{1} = 22^{\circ} C T_{2} = 42^{\circ} C$
$= 22 + 273 = 42 + 273$
$= 295 K = 315 K$
$P_{1} = 200 kPa$
$V_{2} = V_{1} + 0.02 V_{1} = 1.02 V_{1}$
We know that,
$PV = nRT$
Where $\frac{PV}{T} =$ constant
Then, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$∴ \frac{T_{2}}{T_{1}} = \frac{P_{2} V_{2}}{P_{1} V_{1}}$
$\frac{315}{295} = \frac{P_{2}}{200} \times 1.02$
$P_{2} = \frac{315 \times 200}{295 \times 1.02}$
$P_{2} = 209.37 kPa = 209 kPa$

JCECE-2010]

Kinetic Theory of Gases

139037 An experiment is carried on a fixed amount of gas at different temperature and at high pressure such that it deviates from the ideal gas behaviour. The variation of $\frac{p V}{R T}$ with $p$ is shown in the diagram. The correct variation will correspond to

1 Curve

A

2 Curve B

3 Curve

C

4 Curve D

Explanation:

B Here, an experiment is carried on a fixed amount of gas at different temperature and the condition is given at low pressure and high temperature that means its density will be significantly less.
Then we can use ideal gas equation formula-
$∵ PV = nRT$
$n = \frac{PV}{RT}$
Therefore, $\frac{PV}{RT} =$ constant
But when the pressure is increased then the decrease in volume will not take place in same proportion.
Therefore, $\frac{PV}{RT}$ will increase
Thus, we can say that the correct variation will correspond to curve $B$ .

Manipal UGET-2014 UP CPMT-2014

Kinetic Theory of Gases

139038 The equation $(p + \frac{a}{V^{2}}) (V - b) =$ constant. The unit of a is

1 Dyne

\times {cm}^{5}

2 Dyne

\times {cm}^{4}

3 Dyne

/ {cm}^{3}

4 Dyne/cm

^{2}

Explanation:

B According to principal dimensional homognity
$(P + \frac{a}{v^{2}}) (v - b) = constant$
We know-
$P = \frac{F}{A} = Dyne / {cm}^{2}$
$\frac{a}{v^{2}} = \frac{a}{{({cm}^{3})}^{2}} = Dyne / {cm}^{2}$
The units of $a = \frac{Dyne}{(cm)^{2}} \times {({cm}^{3})}^{2}$
$= \frac{Dyne}{(cm)^{2}} \times {cm}^{6}$
$= Dyne \times {cm}^{4}$

Manipal UGET-2014

Kinetic Theory of Gases

139039 Air is filled in a bottle at atmospheric pressure and it is corked at $35^{\circ} C$ . If the cork can come out at 3 atmospheric pressure, then upto what temperature should the bottle be heated in order to remove the cork.

1

{325.5}^{\circ} C

2

851^{\circ} C

3

651^{\circ} C

4 None of these

Explanation:

C Given that,
$P_{1} = 1 atm$
$T_{1} = 35^{\circ} C = (35 + 273 K) = 308 K$
$P_{2} = 3 atm, T_{2} = ?$
From Gay Lusac's law,
Volume $=$ constant
$\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1}{3} = \frac{308}{T_{2}}$
$T_{2} = 308 \times 3$
$= 924 K$
$T_{2} = (924 - 273)^{\circ} C$
$= 651^{\circ} C$

JCECE-2016

Kinetic Theory of Gases

139040 A partition divides a container having insulated walls into two compartments. The same gas fills the two compartments (see figure). The ratio of the number of molecules in compartments I and $II$ is

1

6 : 1

2

1 : 6

3

4 : 1

4

1 : 4

Explanation:

B According to ideal gas equation,
$PV = {nRT}^{2}$
$P_{1} V_{1} = n_{1} {RT}_{1}$
$pV = n_{1} RT$
and
$P_{2} V_{2} = n_{2} {RT}_{2}$
$(2 p) (3 V) = n_{2} RT$
$6 pV = n_{2} RT$
From equation (i) and equation (ii)
$n_{1} RT = \frac{n_{2} RT}{6}$
$\frac{n_{1}}{n_{2}} = \frac{1}{6}$
$n_{1} : n_{2} = 1 : 6$

JCECE-2015

Kinetic Theory of Gases

139041 Air is pumped into an automobile tube upto a pressure of $200 kPa$ in the morning when the air temperature is $22^{\circ} C$ . During the day, temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

212 kPa

2

209 kPa

3

206 kPa

4

200 kPa

Explanation:

B Given that,
$T_{1} = 22^{\circ} C T_{2} = 42^{\circ} C$
$= 22 + 273 = 42 + 273$
$= 295 K = 315 K$
$P_{1} = 200 kPa$
$V_{2} = V_{1} + 0.02 V_{1} = 1.02 V_{1}$
We know that,
$PV = nRT$
Where $\frac{PV}{T} =$ constant
Then, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$∴ \frac{T_{2}}{T_{1}} = \frac{P_{2} V_{2}}{P_{1} V_{1}}$
$\frac{315}{295} = \frac{P_{2}}{200} \times 1.02$
$P_{2} = \frac{315 \times 200}{295 \times 1.02}$
$P_{2} = 209.37 kPa = 209 kPa$

JCECE-2010]

Kinetic Theory of Gases

139037 An experiment is carried on a fixed amount of gas at different temperature and at high pressure such that it deviates from the ideal gas behaviour. The variation of $\frac{p V}{R T}$ with $p$ is shown in the diagram. The correct variation will correspond to

1 Curve

A

2 Curve B

3 Curve

C

4 Curve D

Explanation:

B Here, an experiment is carried on a fixed amount of gas at different temperature and the condition is given at low pressure and high temperature that means its density will be significantly less.
Then we can use ideal gas equation formula-
$∵ PV = nRT$
$n = \frac{PV}{RT}$
Therefore, $\frac{PV}{RT} =$ constant
But when the pressure is increased then the decrease in volume will not take place in same proportion.
Therefore, $\frac{PV}{RT}$ will increase
Thus, we can say that the correct variation will correspond to curve $B$ .

Manipal UGET-2014 UP CPMT-2014

Kinetic Theory of Gases

139038 The equation $(p + \frac{a}{V^{2}}) (V - b) =$ constant. The unit of a is

1 Dyne

\times {cm}^{5}

2 Dyne

\times {cm}^{4}

3 Dyne

/ {cm}^{3}

4 Dyne/cm

^{2}

Explanation:

B According to principal dimensional homognity
$(P + \frac{a}{v^{2}}) (v - b) = constant$
We know-
$P = \frac{F}{A} = Dyne / {cm}^{2}$
$\frac{a}{v^{2}} = \frac{a}{{({cm}^{3})}^{2}} = Dyne / {cm}^{2}$
The units of $a = \frac{Dyne}{(cm)^{2}} \times {({cm}^{3})}^{2}$
$= \frac{Dyne}{(cm)^{2}} \times {cm}^{6}$
$= Dyne \times {cm}^{4}$

Manipal UGET-2014

Kinetic Theory of Gases

139039 Air is filled in a bottle at atmospheric pressure and it is corked at $35^{\circ} C$ . If the cork can come out at 3 atmospheric pressure, then upto what temperature should the bottle be heated in order to remove the cork.

1

{325.5}^{\circ} C

2

851^{\circ} C

3

651^{\circ} C

4 None of these

Explanation:

C Given that,
$P_{1} = 1 atm$
$T_{1} = 35^{\circ} C = (35 + 273 K) = 308 K$
$P_{2} = 3 atm, T_{2} = ?$
From Gay Lusac's law,
Volume $=$ constant
$\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1}{3} = \frac{308}{T_{2}}$
$T_{2} = 308 \times 3$
$= 924 K$
$T_{2} = (924 - 273)^{\circ} C$
$= 651^{\circ} C$

JCECE-2016

Kinetic Theory of Gases

139040 A partition divides a container having insulated walls into two compartments. The same gas fills the two compartments (see figure). The ratio of the number of molecules in compartments I and $II$ is

1

6 : 1

2

1 : 6

3

4 : 1

4

1 : 4

Explanation:

B According to ideal gas equation,
$PV = {nRT}^{2}$
$P_{1} V_{1} = n_{1} {RT}_{1}$
$pV = n_{1} RT$
and
$P_{2} V_{2} = n_{2} {RT}_{2}$
$(2 p) (3 V) = n_{2} RT$
$6 pV = n_{2} RT$
From equation (i) and equation (ii)
$n_{1} RT = \frac{n_{2} RT}{6}$
$\frac{n_{1}}{n_{2}} = \frac{1}{6}$
$n_{1} : n_{2} = 1 : 6$

JCECE-2015

Kinetic Theory of Gases

139041 Air is pumped into an automobile tube upto a pressure of $200 kPa$ in the morning when the air temperature is $22^{\circ} C$ . During the day, temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

212 kPa

2

209 kPa

3

206 kPa

4

200 kPa

Explanation:

B Given that,
$T_{1} = 22^{\circ} C T_{2} = 42^{\circ} C$
$= 22 + 273 = 42 + 273$
$= 295 K = 315 K$
$P_{1} = 200 kPa$
$V_{2} = V_{1} + 0.02 V_{1} = 1.02 V_{1}$
We know that,
$PV = nRT$
Where $\frac{PV}{T} =$ constant
Then, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$∴ \frac{T_{2}}{T_{1}} = \frac{P_{2} V_{2}}{P_{1} V_{1}}$
$\frac{315}{295} = \frac{P_{2}}{200} \times 1.02$
$P_{2} = \frac{315 \times 200}{295 \times 1.02}$
$P_{2} = 209.37 kPa = 209 kPa$

JCECE-2010]

Kinetic Theory of Gases

139037 An experiment is carried on a fixed amount of gas at different temperature and at high pressure such that it deviates from the ideal gas behaviour. The variation of $\frac{p V}{R T}$ with $p$ is shown in the diagram. The correct variation will correspond to

1 Curve

A

2 Curve B

3 Curve

C

4 Curve D

Explanation:

B Here, an experiment is carried on a fixed amount of gas at different temperature and the condition is given at low pressure and high temperature that means its density will be significantly less.
Then we can use ideal gas equation formula-
$∵ PV = nRT$
$n = \frac{PV}{RT}$
Therefore, $\frac{PV}{RT} =$ constant
But when the pressure is increased then the decrease in volume will not take place in same proportion.
Therefore, $\frac{PV}{RT}$ will increase
Thus, we can say that the correct variation will correspond to curve $B$ .

Manipal UGET-2014 UP CPMT-2014

Kinetic Theory of Gases

139038 The equation $(p + \frac{a}{V^{2}}) (V - b) =$ constant. The unit of a is

1 Dyne

\times {cm}^{5}

2 Dyne

\times {cm}^{4}

3 Dyne

/ {cm}^{3}

4 Dyne/cm

^{2}

Explanation:

B According to principal dimensional homognity
$(P + \frac{a}{v^{2}}) (v - b) = constant$
We know-
$P = \frac{F}{A} = Dyne / {cm}^{2}$
$\frac{a}{v^{2}} = \frac{a}{{({cm}^{3})}^{2}} = Dyne / {cm}^{2}$
The units of $a = \frac{Dyne}{(cm)^{2}} \times {({cm}^{3})}^{2}$
$= \frac{Dyne}{(cm)^{2}} \times {cm}^{6}$
$= Dyne \times {cm}^{4}$

Manipal UGET-2014

Kinetic Theory of Gases

139039 Air is filled in a bottle at atmospheric pressure and it is corked at $35^{\circ} C$ . If the cork can come out at 3 atmospheric pressure, then upto what temperature should the bottle be heated in order to remove the cork.

1

{325.5}^{\circ} C

2

851^{\circ} C

3

651^{\circ} C

4 None of these

Explanation:

C Given that,
$P_{1} = 1 atm$
$T_{1} = 35^{\circ} C = (35 + 273 K) = 308 K$
$P_{2} = 3 atm, T_{2} = ?$
From Gay Lusac's law,
Volume $=$ constant
$\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1}{3} = \frac{308}{T_{2}}$
$T_{2} = 308 \times 3$
$= 924 K$
$T_{2} = (924 - 273)^{\circ} C$
$= 651^{\circ} C$

JCECE-2016

Kinetic Theory of Gases

139040 A partition divides a container having insulated walls into two compartments. The same gas fills the two compartments (see figure). The ratio of the number of molecules in compartments I and $II$ is

1

6 : 1

2

1 : 6

3

4 : 1

4

1 : 4

Explanation:

B According to ideal gas equation,
$PV = {nRT}^{2}$
$P_{1} V_{1} = n_{1} {RT}_{1}$
$pV = n_{1} RT$
and
$P_{2} V_{2} = n_{2} {RT}_{2}$
$(2 p) (3 V) = n_{2} RT$
$6 pV = n_{2} RT$
From equation (i) and equation (ii)
$n_{1} RT = \frac{n_{2} RT}{6}$
$\frac{n_{1}}{n_{2}} = \frac{1}{6}$
$n_{1} : n_{2} = 1 : 6$

JCECE-2015

Kinetic Theory of Gases

139041 Air is pumped into an automobile tube upto a pressure of $200 kPa$ in the morning when the air temperature is $22^{\circ} C$ . During the day, temperature rises to $42^{\circ} C$ and the tube expands by $2 %$ . The pressure of the air in the tube at this temperature, will be approximately

1

212 kPa

2

209 kPa

3

206 kPa

4

200 kPa

Explanation:

B Given that,
$T_{1} = 22^{\circ} C T_{2} = 42^{\circ} C$
$= 22 + 273 = 42 + 273$
$= 295 K = 315 K$
$P_{1} = 200 kPa$
$V_{2} = V_{1} + 0.02 V_{1} = 1.02 V_{1}$
We know that,
$PV = nRT$
Where $\frac{PV}{T} =$ constant
Then, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$∴ \frac{T_{2}}{T_{1}} = \frac{P_{2} V_{2}}{P_{1} V_{1}}$
$\frac{315}{295} = \frac{P_{2}}{200} \times 1.02$
$P_{2} = \frac{315 \times 200}{295 \times 1.02}$
$P_{2} = 209.37 kPa = 209 kPa$

JCECE-2010]

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