QBManager

Kinetic Theory of Gases

139075 Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} C$ , if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(R = 8.314 J / mol / K)$

1

116.2 J

2

136.2 J

3

166.2 J

4

186.2 J

Explanation:

C Given, Ideal gas equation condition $V \propto T^{2 / 3}$ Then,
$PV = nRT and Δ T = 30^{\circ} C = 303 K$
$PV = RT$
$P = \frac{R T}{V}$
$P = \frac{R T}{T^{2 / 3}}$
$P = R T^{1 / 3}$
We know that,
$dW = PdV$
$dW = PdV$
$V \propto T^{2 / 3}$
$\frac{dV}{dT} = \frac{2}{3} t^{2 / 3 - 1}$
$dV = \frac{2}{3} T^{- 1 / 3} dT$
$dW = PdV$
$= {RT}^{1 / 3} \times \frac{2}{3} T^{- 1 / 3} dT$
$dW = \frac{2}{3} RdT$
$W = \frac{2}{3} RT$
$W = \frac{2}{3} \times 8.314 \times 30$
$W = 20 \times 8.314 = 166.2 J$
$W = 166.2 J$

AP EAMCET -2012

Kinetic Theory of Gases

139077 A vessel contains $32 gm$ of $O_{2}$ at a temperature $T$ . The pressure of the gas is $P$ . An identical vessel containing $4 gm$ of $H_{2}$ at a temperature $2 T$ has a pressure of

1

8 P

2

4 P

3

P

4

\frac{P}{8}

Explanation:

B Given that, an identical container so volume is constant from ideal gas equation
$PV = nRT {∵ V and R are Constant}$
$P \propto nT$
Then,
$\frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$
$\frac{P_{1}}{P_{2}} = \frac{\frac{32}{32} (T)}{\frac{4}{2} (2 T)} = \frac{1}{4} [∵ P_{1} = P]$
$P_{2} = 4 P$

AMU-2009

Kinetic Theory of Gases

139078 A cylinder contains 12 litres of oxygen at $20^{\circ} C$ and $15 atm$ pressure. The temperature of the gas is raised to $35^{\circ} C$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?

1 9

2 11

3 15

4 17

Explanation:

B Given here,
$V_{1} = 12 litres T_{1} = 20^{\circ} C = 293 K$
$P_{1} = 15 (atm)$
$T_{2} = 35^{\circ} C = 308 K, V_{2} = 17 litres, P_{2} = ?$
We know that,
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{15 \times 12}{293} = \frac{P_{2} \times 17}{308}$
$P_{2} = 11 Litre$

AMU-2007

Kinetic Theory of Gases

139079 A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be

1

4 P_{o}

2

2 P_{o}

3

P_{o}

4

\frac{P_{o}}{2}

Explanation:

B Given,
$m_{2} = \frac{m_{1}}{2}, V_{2} = 2 V_{1}$
We know that
$P = \frac{1}{3} \frac{mN}{V} V_{rms}^{2}$
$P \propto m V_{rms}^{2}$
$\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{V_{2}}{V_{1}})}^{2}$
$= \frac{m_{1} / 2}{m_{1}} {(\frac{2 V_{1}}{V_{1}})}^{2} [∵ m_{2} = \frac{m_{1}}{2}]$
$P_{2} = 2 P_{1}$
$P_{2} = 2 P_{o} [∵ P_{1} = P_{o}]$

2004]

Kinetic Theory of Gases

139080 If a given mass of a gas occupies a volume $100 {cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} C$ , what will be its volume at 4 atmospheric pressure the temperature being the same

1

25 {cm}^{3}

2 0

3

50 {cm}^{3}

4

5 {cm}^{3}

Explanation:

D Given
$P_{2} = 4 atm$
$V_{1} = 100 {cm}^{3}$
$V_{2} = ?$
$T = 100^{\circ} C$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} [T_{1} = T_{2} = T]$
$\frac{1 \times 100}{T} = \frac{4 \times V_{2}}{T}$
$25 = V_{2}$
$V_{2} = 25 {cm}^{3}$

AMU-2002

Kinetic Theory of Gases

139075 Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} C$ , if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(R = 8.314 J / mol / K)$

1

116.2 J

2

136.2 J

3

166.2 J

4

186.2 J

Explanation:

C Given, Ideal gas equation condition $V \propto T^{2 / 3}$ Then,
$PV = nRT and Δ T = 30^{\circ} C = 303 K$
$PV = RT$
$P = \frac{R T}{V}$
$P = \frac{R T}{T^{2 / 3}}$
$P = R T^{1 / 3}$
We know that,
$dW = PdV$
$dW = PdV$
$V \propto T^{2 / 3}$
$\frac{dV}{dT} = \frac{2}{3} t^{2 / 3 - 1}$
$dV = \frac{2}{3} T^{- 1 / 3} dT$
$dW = PdV$
$= {RT}^{1 / 3} \times \frac{2}{3} T^{- 1 / 3} dT$
$dW = \frac{2}{3} RdT$
$W = \frac{2}{3} RT$
$W = \frac{2}{3} \times 8.314 \times 30$
$W = 20 \times 8.314 = 166.2 J$
$W = 166.2 J$

AP EAMCET -2012

Kinetic Theory of Gases

139077 A vessel contains $32 gm$ of $O_{2}$ at a temperature $T$ . The pressure of the gas is $P$ . An identical vessel containing $4 gm$ of $H_{2}$ at a temperature $2 T$ has a pressure of

1

8 P

2

4 P

3

P

4

\frac{P}{8}

Explanation:

B Given that, an identical container so volume is constant from ideal gas equation
$PV = nRT {∵ V and R are Constant}$
$P \propto nT$
Then,
$\frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$
$\frac{P_{1}}{P_{2}} = \frac{\frac{32}{32} (T)}{\frac{4}{2} (2 T)} = \frac{1}{4} [∵ P_{1} = P]$
$P_{2} = 4 P$

AMU-2009

Kinetic Theory of Gases

139078 A cylinder contains 12 litres of oxygen at $20^{\circ} C$ and $15 atm$ pressure. The temperature of the gas is raised to $35^{\circ} C$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?

1 9

2 11

3 15

4 17

Explanation:

B Given here,
$V_{1} = 12 litres T_{1} = 20^{\circ} C = 293 K$
$P_{1} = 15 (atm)$
$T_{2} = 35^{\circ} C = 308 K, V_{2} = 17 litres, P_{2} = ?$
We know that,
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{15 \times 12}{293} = \frac{P_{2} \times 17}{308}$
$P_{2} = 11 Litre$

AMU-2007

Kinetic Theory of Gases

139079 A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be

1

4 P_{o}

2

2 P_{o}

3

P_{o}

4

\frac{P_{o}}{2}

Explanation:

B Given,
$m_{2} = \frac{m_{1}}{2}, V_{2} = 2 V_{1}$
We know that
$P = \frac{1}{3} \frac{mN}{V} V_{rms}^{2}$
$P \propto m V_{rms}^{2}$
$\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{V_{2}}{V_{1}})}^{2}$
$= \frac{m_{1} / 2}{m_{1}} {(\frac{2 V_{1}}{V_{1}})}^{2} [∵ m_{2} = \frac{m_{1}}{2}]$
$P_{2} = 2 P_{1}$
$P_{2} = 2 P_{o} [∵ P_{1} = P_{o}]$

2004]

Kinetic Theory of Gases

139080 If a given mass of a gas occupies a volume $100 {cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} C$ , what will be its volume at 4 atmospheric pressure the temperature being the same

1

25 {cm}^{3}

2 0

3

50 {cm}^{3}

4

5 {cm}^{3}

Explanation:

D Given
$P_{2} = 4 atm$
$V_{1} = 100 {cm}^{3}$
$V_{2} = ?$
$T = 100^{\circ} C$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} [T_{1} = T_{2} = T]$
$\frac{1 \times 100}{T} = \frac{4 \times V_{2}}{T}$
$25 = V_{2}$
$V_{2} = 25 {cm}^{3}$

AMU-2002

Kinetic Theory of Gases

139075 Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} C$ , if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(R = 8.314 J / mol / K)$

1

116.2 J

2

136.2 J

3

166.2 J

4

186.2 J

Explanation:

C Given, Ideal gas equation condition $V \propto T^{2 / 3}$ Then,
$PV = nRT and Δ T = 30^{\circ} C = 303 K$
$PV = RT$
$P = \frac{R T}{V}$
$P = \frac{R T}{T^{2 / 3}}$
$P = R T^{1 / 3}$
We know that,
$dW = PdV$
$dW = PdV$
$V \propto T^{2 / 3}$
$\frac{dV}{dT} = \frac{2}{3} t^{2 / 3 - 1}$
$dV = \frac{2}{3} T^{- 1 / 3} dT$
$dW = PdV$
$= {RT}^{1 / 3} \times \frac{2}{3} T^{- 1 / 3} dT$
$dW = \frac{2}{3} RdT$
$W = \frac{2}{3} RT$
$W = \frac{2}{3} \times 8.314 \times 30$
$W = 20 \times 8.314 = 166.2 J$
$W = 166.2 J$

AP EAMCET -2012

Kinetic Theory of Gases

139077 A vessel contains $32 gm$ of $O_{2}$ at a temperature $T$ . The pressure of the gas is $P$ . An identical vessel containing $4 gm$ of $H_{2}$ at a temperature $2 T$ has a pressure of

1

8 P

2

4 P

3

P

4

\frac{P}{8}

Explanation:

B Given that, an identical container so volume is constant from ideal gas equation
$PV = nRT {∵ V and R are Constant}$
$P \propto nT$
Then,
$\frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$
$\frac{P_{1}}{P_{2}} = \frac{\frac{32}{32} (T)}{\frac{4}{2} (2 T)} = \frac{1}{4} [∵ P_{1} = P]$
$P_{2} = 4 P$

AMU-2009

Kinetic Theory of Gases

139078 A cylinder contains 12 litres of oxygen at $20^{\circ} C$ and $15 atm$ pressure. The temperature of the gas is raised to $35^{\circ} C$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?

1 9

2 11

3 15

4 17

Explanation:

B Given here,
$V_{1} = 12 litres T_{1} = 20^{\circ} C = 293 K$
$P_{1} = 15 (atm)$
$T_{2} = 35^{\circ} C = 308 K, V_{2} = 17 litres, P_{2} = ?$
We know that,
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{15 \times 12}{293} = \frac{P_{2} \times 17}{308}$
$P_{2} = 11 Litre$

AMU-2007

Kinetic Theory of Gases

139079 A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be

1

4 P_{o}

2

2 P_{o}

3

P_{o}

4

\frac{P_{o}}{2}

Explanation:

B Given,
$m_{2} = \frac{m_{1}}{2}, V_{2} = 2 V_{1}$
We know that
$P = \frac{1}{3} \frac{mN}{V} V_{rms}^{2}$
$P \propto m V_{rms}^{2}$
$\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{V_{2}}{V_{1}})}^{2}$
$= \frac{m_{1} / 2}{m_{1}} {(\frac{2 V_{1}}{V_{1}})}^{2} [∵ m_{2} = \frac{m_{1}}{2}]$
$P_{2} = 2 P_{1}$
$P_{2} = 2 P_{o} [∵ P_{1} = P_{o}]$

2004]

Kinetic Theory of Gases

139080 If a given mass of a gas occupies a volume $100 {cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} C$ , what will be its volume at 4 atmospheric pressure the temperature being the same

1

25 {cm}^{3}

2 0

3

50 {cm}^{3}

4

5 {cm}^{3}

Explanation:

D Given
$P_{2} = 4 atm$
$V_{1} = 100 {cm}^{3}$
$V_{2} = ?$
$T = 100^{\circ} C$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} [T_{1} = T_{2} = T]$
$\frac{1 \times 100}{T} = \frac{4 \times V_{2}}{T}$
$25 = V_{2}$
$V_{2} = 25 {cm}^{3}$

AMU-2002

Kinetic Theory of Gases

139075 Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} C$ , if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(R = 8.314 J / mol / K)$

1

116.2 J

2

136.2 J

3

166.2 J

4

186.2 J

Explanation:

C Given, Ideal gas equation condition $V \propto T^{2 / 3}$ Then,
$PV = nRT and Δ T = 30^{\circ} C = 303 K$
$PV = RT$
$P = \frac{R T}{V}$
$P = \frac{R T}{T^{2 / 3}}$
$P = R T^{1 / 3}$
We know that,
$dW = PdV$
$dW = PdV$
$V \propto T^{2 / 3}$
$\frac{dV}{dT} = \frac{2}{3} t^{2 / 3 - 1}$
$dV = \frac{2}{3} T^{- 1 / 3} dT$
$dW = PdV$
$= {RT}^{1 / 3} \times \frac{2}{3} T^{- 1 / 3} dT$
$dW = \frac{2}{3} RdT$
$W = \frac{2}{3} RT$
$W = \frac{2}{3} \times 8.314 \times 30$
$W = 20 \times 8.314 = 166.2 J$
$W = 166.2 J$

AP EAMCET -2012

Kinetic Theory of Gases

139077 A vessel contains $32 gm$ of $O_{2}$ at a temperature $T$ . The pressure of the gas is $P$ . An identical vessel containing $4 gm$ of $H_{2}$ at a temperature $2 T$ has a pressure of

1

8 P

2

4 P

3

P

4

\frac{P}{8}

Explanation:

B Given that, an identical container so volume is constant from ideal gas equation
$PV = nRT {∵ V and R are Constant}$
$P \propto nT$
Then,
$\frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$
$\frac{P_{1}}{P_{2}} = \frac{\frac{32}{32} (T)}{\frac{4}{2} (2 T)} = \frac{1}{4} [∵ P_{1} = P]$
$P_{2} = 4 P$

AMU-2009

Kinetic Theory of Gases

139078 A cylinder contains 12 litres of oxygen at $20^{\circ} C$ and $15 atm$ pressure. The temperature of the gas is raised to $35^{\circ} C$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?

1 9

2 11

3 15

4 17

Explanation:

B Given here,
$V_{1} = 12 litres T_{1} = 20^{\circ} C = 293 K$
$P_{1} = 15 (atm)$
$T_{2} = 35^{\circ} C = 308 K, V_{2} = 17 litres, P_{2} = ?$
We know that,
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{15 \times 12}{293} = \frac{P_{2} \times 17}{308}$
$P_{2} = 11 Litre$

AMU-2007

Kinetic Theory of Gases

139079 A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be

1

4 P_{o}

2

2 P_{o}

3

P_{o}

4

\frac{P_{o}}{2}

Explanation:

B Given,
$m_{2} = \frac{m_{1}}{2}, V_{2} = 2 V_{1}$
We know that
$P = \frac{1}{3} \frac{mN}{V} V_{rms}^{2}$
$P \propto m V_{rms}^{2}$
$\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{V_{2}}{V_{1}})}^{2}$
$= \frac{m_{1} / 2}{m_{1}} {(\frac{2 V_{1}}{V_{1}})}^{2} [∵ m_{2} = \frac{m_{1}}{2}]$
$P_{2} = 2 P_{1}$
$P_{2} = 2 P_{o} [∵ P_{1} = P_{o}]$

2004]

Kinetic Theory of Gases

139080 If a given mass of a gas occupies a volume $100 {cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} C$ , what will be its volume at 4 atmospheric pressure the temperature being the same

1

25 {cm}^{3}

2 0

3

50 {cm}^{3}

4

5 {cm}^{3}

Explanation:

D Given
$P_{2} = 4 atm$
$V_{1} = 100 {cm}^{3}$
$V_{2} = ?$
$T = 100^{\circ} C$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} [T_{1} = T_{2} = T]$
$\frac{1 \times 100}{T} = \frac{4 \times V_{2}}{T}$
$25 = V_{2}$
$V_{2} = 25 {cm}^{3}$

AMU-2002

Kinetic Theory of Gases

139075 Work done to increase the temperature of one mole of an ideal gas by $30^{\circ} C$ , if it is expanding under the condition $V \propto T^{2 / 3}$ is, $(R = 8.314 J / mol / K)$

1

116.2 J

2

136.2 J

3

166.2 J

4

186.2 J

Explanation:

C Given, Ideal gas equation condition $V \propto T^{2 / 3}$ Then,
$PV = nRT and Δ T = 30^{\circ} C = 303 K$
$PV = RT$
$P = \frac{R T}{V}$
$P = \frac{R T}{T^{2 / 3}}$
$P = R T^{1 / 3}$
We know that,
$dW = PdV$
$dW = PdV$
$V \propto T^{2 / 3}$
$\frac{dV}{dT} = \frac{2}{3} t^{2 / 3 - 1}$
$dV = \frac{2}{3} T^{- 1 / 3} dT$
$dW = PdV$
$= {RT}^{1 / 3} \times \frac{2}{3} T^{- 1 / 3} dT$
$dW = \frac{2}{3} RdT$
$W = \frac{2}{3} RT$
$W = \frac{2}{3} \times 8.314 \times 30$
$W = 20 \times 8.314 = 166.2 J$
$W = 166.2 J$

AP EAMCET -2012

Kinetic Theory of Gases

139077 A vessel contains $32 gm$ of $O_{2}$ at a temperature $T$ . The pressure of the gas is $P$ . An identical vessel containing $4 gm$ of $H_{2}$ at a temperature $2 T$ has a pressure of

1

8 P

2

4 P

3

P

4

\frac{P}{8}

Explanation:

B Given that, an identical container so volume is constant from ideal gas equation
$PV = nRT {∵ V and R are Constant}$
$P \propto nT$
Then,
$\frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$
$\frac{P_{1}}{P_{2}} = \frac{\frac{32}{32} (T)}{\frac{4}{2} (2 T)} = \frac{1}{4} [∵ P_{1} = P]$
$P_{2} = 4 P$

AMU-2009

Kinetic Theory of Gases

139078 A cylinder contains 12 litres of oxygen at $20^{\circ} C$ and $15 atm$ pressure. The temperature of the gas is raised to $35^{\circ} C$ and its volume increased to 17 litres. What is the final pressure of the gas (in atm)?

1 9

2 11

3 15

4 17

Explanation:

B Given here,
$V_{1} = 12 litres T_{1} = 20^{\circ} C = 293 K$
$P_{1} = 15 (atm)$
$T_{2} = 35^{\circ} C = 308 K, V_{2} = 17 litres, P_{2} = ?$
We know that,
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{15 \times 12}{293} = \frac{P_{2} \times 17}{308}$
$P_{2} = 11 Litre$

AMU-2007

Kinetic Theory of Gases

139079 A gas at pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be

1

4 P_{o}

2

2 P_{o}

3

P_{o}

4

\frac{P_{o}}{2}

Explanation:

B Given,
$m_{2} = \frac{m_{1}}{2}, V_{2} = 2 V_{1}$
We know that
$P = \frac{1}{3} \frac{mN}{V} V_{rms}^{2}$
$P \propto m V_{rms}^{2}$
$\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{V_{2}}{V_{1}})}^{2}$
$= \frac{m_{1} / 2}{m_{1}} {(\frac{2 V_{1}}{V_{1}})}^{2} [∵ m_{2} = \frac{m_{1}}{2}]$
$P_{2} = 2 P_{1}$
$P_{2} = 2 P_{o} [∵ P_{1} = P_{o}]$

2004]

Kinetic Theory of Gases

139080 If a given mass of a gas occupies a volume $100 {cm}^{3}$ at one atmospheric pressure and temperature of $100^{\circ} C$ , what will be its volume at 4 atmospheric pressure the temperature being the same

1

25 {cm}^{3}

2 0

3

50 {cm}^{3}

4

5 {cm}^{3}

Explanation:

D Given
$P_{2} = 4 atm$
$V_{1} = 100 {cm}^{3}$
$V_{2} = ?$
$T = 100^{\circ} C$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} [T_{1} = T_{2} = T]$
$\frac{1 \times 100}{T} = \frac{4 \times V_{2}}{T}$
$25 = V_{2}$
$V_{2} = 25 {cm}^{3}$

AMU-2002

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