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Kinetic Theory of Gases

138941 Which of the following is not an assumption of the kinetic theory of gases?

1 The molecules travel in straight paths until they undergo collision with other molecules

2 Molecules of the gas are small hard spheres, occupying negligible volume compared with the total volume of the gas

3 The molecules do not undergo any collisions at all

4 The molecules undergo elastic collisions only

J and K-CET-2012

Kinetic Theory of Gases

138947 Relation between pressure (p) and energy (E) of a gas is

1

p = \frac{2}{3} E

2

p = \frac{1}{3} E

3

p = \frac{3}{2} E

4

p = 3 E

Explanation:

A We know that,
Kinetic energy of gas (K.E) $= \frac{1}{2} M V_{rms}^{2}$
RMS velocity of gas molecules $(V_{rms}) = \sqrt{\frac{3 RT}{M}}$ then,
$K . E = \frac{1}{2} M V_{rms}^{2}$
$= \frac{1}{2} M \times \frac{3 RT}{M}$
$K . E = \frac{3}{2} RT$
$K . E = \frac{3}{2} PV$
Hence, $P = \frac{2 K . E}{3 V}$
So, pressure is ${[\frac{2}{3}]}^{rd}$ kinetic energy per unit volume of gas.

AIPMT-1991

Kinetic Theory of Gases

138948 A diatomic gas initially at $18^{\circ} C$ is compressed adiabatically to one-eight of its original volume. The temperature after compression will be

1

18^{\circ} C

2

668.4 K

3

{395.4}^{\circ} C

4

144^{\circ} C

Explanation:

B Given,
Initial temperature $T_{1} = 18^{\circ} C = 291 K$
Initial volume $V_{i} = V$
Final volume $V_{f} = \frac{V}{8}$
Specific heat ratio for diatomic $(γ) = \frac{7}{5}$
We know that, For adiabatically compression process,
${PV}^{γ} = Constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
Ideal gas equation,
$PV = nRT$
From equation (i) and (ii)
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
$T_{2} = 291 (8)^{\frac{7}{5} - 1}$
$T_{2} = 291 (8)^{2 / 5}$
$T_{2} = 668.54 K$
$T_{2} ◻ 668.4 K$

AIPMT-1996

Kinetic Theory of Gases

138949 The equation of state for $5 g$ of oxygen at a pressure $p$ and temperature $T$ , when occupying a volume $V$ , will be

1

pV = (\frac{5}{32}) RT

2

pV = 5 RT

3

pV = (\frac{5}{2}) RT

4

pV = (\frac{5}{16}) RT

Explanation:

A Given, mass (m) $= 5 g$
Molecular weight of $O_{2} = 32$
From ideal gas equation
$PV = nRT$
$PV = (\frac{m}{M}) RT$
$PV = (\frac{5}{32}) RT$

DPMT 2000

Kinetic Theory of Gases

138950 The gas in vessel is subjected to a pressure of $20 atm$ at a temperature $27^{\circ} C$ . The pressure of the gas in a vessel after one-half of the gas is released from the vessel and the temperature of the remainder is raised by $50^{\circ} C$ is

1

8.5 atm

2

11.7 atm

3

17 atm

4

10.8 atm

Explanation:

B Ideal gas equation
$PV = \frac{m}{M} RT$
Initial pressure $(P_{1}) = 20 atm$
Initial temperature $(T_{1}) = 27 + 273 = 300 K$
Now, $20 \times V = \frac{m}{M} \times R \times 300$
Final Temperature $(T_{2}) = 300 + 50 = 350 K$
Final mass $(m^{'}) = m / 2$
$∴ P^{'} \times V = \frac{m / 2}{M} \times R \times 350$
Divide eqn (ii) by (i)
$\frac{P^{'}}{20} = \frac{350}{2 \times 300}$
$P^{'} = 11.7 atm$

AP EMCET(Medical)-2011

Kinetic Theory of Gases

138941 Which of the following is not an assumption of the kinetic theory of gases?

1 The molecules travel in straight paths until they undergo collision with other molecules

2 Molecules of the gas are small hard spheres, occupying negligible volume compared with the total volume of the gas

3 The molecules do not undergo any collisions at all

4 The molecules undergo elastic collisions only

J and K-CET-2012

Kinetic Theory of Gases

138947 Relation between pressure (p) and energy (E) of a gas is

1

p = \frac{2}{3} E

2

p = \frac{1}{3} E

3

p = \frac{3}{2} E

4

p = 3 E

Explanation:

A We know that,
Kinetic energy of gas (K.E) $= \frac{1}{2} M V_{rms}^{2}$
RMS velocity of gas molecules $(V_{rms}) = \sqrt{\frac{3 RT}{M}}$ then,
$K . E = \frac{1}{2} M V_{rms}^{2}$
$= \frac{1}{2} M \times \frac{3 RT}{M}$
$K . E = \frac{3}{2} RT$
$K . E = \frac{3}{2} PV$
Hence, $P = \frac{2 K . E}{3 V}$
So, pressure is ${[\frac{2}{3}]}^{rd}$ kinetic energy per unit volume of gas.

AIPMT-1991

Kinetic Theory of Gases

138948 A diatomic gas initially at $18^{\circ} C$ is compressed adiabatically to one-eight of its original volume. The temperature after compression will be

1

18^{\circ} C

2

668.4 K

3

{395.4}^{\circ} C

4

144^{\circ} C

Explanation:

B Given,
Initial temperature $T_{1} = 18^{\circ} C = 291 K$
Initial volume $V_{i} = V$
Final volume $V_{f} = \frac{V}{8}$
Specific heat ratio for diatomic $(γ) = \frac{7}{5}$
We know that, For adiabatically compression process,
${PV}^{γ} = Constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
Ideal gas equation,
$PV = nRT$
From equation (i) and (ii)
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
$T_{2} = 291 (8)^{\frac{7}{5} - 1}$
$T_{2} = 291 (8)^{2 / 5}$
$T_{2} = 668.54 K$
$T_{2} ◻ 668.4 K$

AIPMT-1996

Kinetic Theory of Gases

138949 The equation of state for $5 g$ of oxygen at a pressure $p$ and temperature $T$ , when occupying a volume $V$ , will be

1

pV = (\frac{5}{32}) RT

2

pV = 5 RT

3

pV = (\frac{5}{2}) RT

4

pV = (\frac{5}{16}) RT

Explanation:

A Given, mass (m) $= 5 g$
Molecular weight of $O_{2} = 32$
From ideal gas equation
$PV = nRT$
$PV = (\frac{m}{M}) RT$
$PV = (\frac{5}{32}) RT$

DPMT 2000

Kinetic Theory of Gases

138950 The gas in vessel is subjected to a pressure of $20 atm$ at a temperature $27^{\circ} C$ . The pressure of the gas in a vessel after one-half of the gas is released from the vessel and the temperature of the remainder is raised by $50^{\circ} C$ is

1

8.5 atm

2

11.7 atm

3

17 atm

4

10.8 atm

Explanation:

B Ideal gas equation
$PV = \frac{m}{M} RT$
Initial pressure $(P_{1}) = 20 atm$
Initial temperature $(T_{1}) = 27 + 273 = 300 K$
Now, $20 \times V = \frac{m}{M} \times R \times 300$
Final Temperature $(T_{2}) = 300 + 50 = 350 K$
Final mass $(m^{'}) = m / 2$
$∴ P^{'} \times V = \frac{m / 2}{M} \times R \times 350$
Divide eqn (ii) by (i)
$\frac{P^{'}}{20} = \frac{350}{2 \times 300}$
$P^{'} = 11.7 atm$

AP EMCET(Medical)-2011

Kinetic Theory of Gases

138941 Which of the following is not an assumption of the kinetic theory of gases?

1 The molecules travel in straight paths until they undergo collision with other molecules

2 Molecules of the gas are small hard spheres, occupying negligible volume compared with the total volume of the gas

3 The molecules do not undergo any collisions at all

4 The molecules undergo elastic collisions only

J and K-CET-2012

Kinetic Theory of Gases

138947 Relation between pressure (p) and energy (E) of a gas is

1

p = \frac{2}{3} E

2

p = \frac{1}{3} E

3

p = \frac{3}{2} E

4

p = 3 E

Explanation:

A We know that,
Kinetic energy of gas (K.E) $= \frac{1}{2} M V_{rms}^{2}$
RMS velocity of gas molecules $(V_{rms}) = \sqrt{\frac{3 RT}{M}}$ then,
$K . E = \frac{1}{2} M V_{rms}^{2}$
$= \frac{1}{2} M \times \frac{3 RT}{M}$
$K . E = \frac{3}{2} RT$
$K . E = \frac{3}{2} PV$
Hence, $P = \frac{2 K . E}{3 V}$
So, pressure is ${[\frac{2}{3}]}^{rd}$ kinetic energy per unit volume of gas.

AIPMT-1991

Kinetic Theory of Gases

138948 A diatomic gas initially at $18^{\circ} C$ is compressed adiabatically to one-eight of its original volume. The temperature after compression will be

1

18^{\circ} C

2

668.4 K

3

{395.4}^{\circ} C

4

144^{\circ} C

Explanation:

B Given,
Initial temperature $T_{1} = 18^{\circ} C = 291 K$
Initial volume $V_{i} = V$
Final volume $V_{f} = \frac{V}{8}$
Specific heat ratio for diatomic $(γ) = \frac{7}{5}$
We know that, For adiabatically compression process,
${PV}^{γ} = Constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
Ideal gas equation,
$PV = nRT$
From equation (i) and (ii)
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
$T_{2} = 291 (8)^{\frac{7}{5} - 1}$
$T_{2} = 291 (8)^{2 / 5}$
$T_{2} = 668.54 K$
$T_{2} ◻ 668.4 K$

AIPMT-1996

Kinetic Theory of Gases

138949 The equation of state for $5 g$ of oxygen at a pressure $p$ and temperature $T$ , when occupying a volume $V$ , will be

1

pV = (\frac{5}{32}) RT

2

pV = 5 RT

3

pV = (\frac{5}{2}) RT

4

pV = (\frac{5}{16}) RT

Explanation:

A Given, mass (m) $= 5 g$
Molecular weight of $O_{2} = 32$
From ideal gas equation
$PV = nRT$
$PV = (\frac{m}{M}) RT$
$PV = (\frac{5}{32}) RT$

DPMT 2000

Kinetic Theory of Gases

138950 The gas in vessel is subjected to a pressure of $20 atm$ at a temperature $27^{\circ} C$ . The pressure of the gas in a vessel after one-half of the gas is released from the vessel and the temperature of the remainder is raised by $50^{\circ} C$ is

1

8.5 atm

2

11.7 atm

3

17 atm

4

10.8 atm

Explanation:

B Ideal gas equation
$PV = \frac{m}{M} RT$
Initial pressure $(P_{1}) = 20 atm$
Initial temperature $(T_{1}) = 27 + 273 = 300 K$
Now, $20 \times V = \frac{m}{M} \times R \times 300$
Final Temperature $(T_{2}) = 300 + 50 = 350 K$
Final mass $(m^{'}) = m / 2$
$∴ P^{'} \times V = \frac{m / 2}{M} \times R \times 350$
Divide eqn (ii) by (i)
$\frac{P^{'}}{20} = \frac{350}{2 \times 300}$
$P^{'} = 11.7 atm$

AP EMCET(Medical)-2011

Kinetic Theory of Gases

138941 Which of the following is not an assumption of the kinetic theory of gases?

1 The molecules travel in straight paths until they undergo collision with other molecules

2 Molecules of the gas are small hard spheres, occupying negligible volume compared with the total volume of the gas

3 The molecules do not undergo any collisions at all

4 The molecules undergo elastic collisions only

J and K-CET-2012

Kinetic Theory of Gases

138947 Relation between pressure (p) and energy (E) of a gas is

1

p = \frac{2}{3} E

2

p = \frac{1}{3} E

3

p = \frac{3}{2} E

4

p = 3 E

Explanation:

A We know that,
Kinetic energy of gas (K.E) $= \frac{1}{2} M V_{rms}^{2}$
RMS velocity of gas molecules $(V_{rms}) = \sqrt{\frac{3 RT}{M}}$ then,
$K . E = \frac{1}{2} M V_{rms}^{2}$
$= \frac{1}{2} M \times \frac{3 RT}{M}$
$K . E = \frac{3}{2} RT$
$K . E = \frac{3}{2} PV$
Hence, $P = \frac{2 K . E}{3 V}$
So, pressure is ${[\frac{2}{3}]}^{rd}$ kinetic energy per unit volume of gas.

AIPMT-1991

Kinetic Theory of Gases

138948 A diatomic gas initially at $18^{\circ} C$ is compressed adiabatically to one-eight of its original volume. The temperature after compression will be

1

18^{\circ} C

2

668.4 K

3

{395.4}^{\circ} C

4

144^{\circ} C

Explanation:

B Given,
Initial temperature $T_{1} = 18^{\circ} C = 291 K$
Initial volume $V_{i} = V$
Final volume $V_{f} = \frac{V}{8}$
Specific heat ratio for diatomic $(γ) = \frac{7}{5}$
We know that, For adiabatically compression process,
${PV}^{γ} = Constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
Ideal gas equation,
$PV = nRT$
From equation (i) and (ii)
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
$T_{2} = 291 (8)^{\frac{7}{5} - 1}$
$T_{2} = 291 (8)^{2 / 5}$
$T_{2} = 668.54 K$
$T_{2} ◻ 668.4 K$

AIPMT-1996

Kinetic Theory of Gases

138949 The equation of state for $5 g$ of oxygen at a pressure $p$ and temperature $T$ , when occupying a volume $V$ , will be

1

pV = (\frac{5}{32}) RT

2

pV = 5 RT

3

pV = (\frac{5}{2}) RT

4

pV = (\frac{5}{16}) RT

Explanation:

A Given, mass (m) $= 5 g$
Molecular weight of $O_{2} = 32$
From ideal gas equation
$PV = nRT$
$PV = (\frac{m}{M}) RT$
$PV = (\frac{5}{32}) RT$

DPMT 2000

Kinetic Theory of Gases

138950 The gas in vessel is subjected to a pressure of $20 atm$ at a temperature $27^{\circ} C$ . The pressure of the gas in a vessel after one-half of the gas is released from the vessel and the temperature of the remainder is raised by $50^{\circ} C$ is

1

8.5 atm

2

11.7 atm

3

17 atm

4

10.8 atm

Explanation:

B Ideal gas equation
$PV = \frac{m}{M} RT$
Initial pressure $(P_{1}) = 20 atm$
Initial temperature $(T_{1}) = 27 + 273 = 300 K$
Now, $20 \times V = \frac{m}{M} \times R \times 300$
Final Temperature $(T_{2}) = 300 + 50 = 350 K$
Final mass $(m^{'}) = m / 2$
$∴ P^{'} \times V = \frac{m / 2}{M} \times R \times 350$
Divide eqn (ii) by (i)
$\frac{P^{'}}{20} = \frac{350}{2 \times 300}$
$P^{'} = 11.7 atm$

AP EMCET(Medical)-2011

Kinetic Theory of Gases

138941 Which of the following is not an assumption of the kinetic theory of gases?

1 The molecules travel in straight paths until they undergo collision with other molecules

2 Molecules of the gas are small hard spheres, occupying negligible volume compared with the total volume of the gas

3 The molecules do not undergo any collisions at all

4 The molecules undergo elastic collisions only

J and K-CET-2012

Kinetic Theory of Gases

138947 Relation between pressure (p) and energy (E) of a gas is

1

p = \frac{2}{3} E

2

p = \frac{1}{3} E

3

p = \frac{3}{2} E

4

p = 3 E

Explanation:

A We know that,
Kinetic energy of gas (K.E) $= \frac{1}{2} M V_{rms}^{2}$
RMS velocity of gas molecules $(V_{rms}) = \sqrt{\frac{3 RT}{M}}$ then,
$K . E = \frac{1}{2} M V_{rms}^{2}$
$= \frac{1}{2} M \times \frac{3 RT}{M}$
$K . E = \frac{3}{2} RT$
$K . E = \frac{3}{2} PV$
Hence, $P = \frac{2 K . E}{3 V}$
So, pressure is ${[\frac{2}{3}]}^{rd}$ kinetic energy per unit volume of gas.

AIPMT-1991

Kinetic Theory of Gases

138948 A diatomic gas initially at $18^{\circ} C$ is compressed adiabatically to one-eight of its original volume. The temperature after compression will be

1

18^{\circ} C

2

668.4 K

3

{395.4}^{\circ} C

4

144^{\circ} C

Explanation:

B Given,
Initial temperature $T_{1} = 18^{\circ} C = 291 K$
Initial volume $V_{i} = V$
Final volume $V_{f} = \frac{V}{8}$
Specific heat ratio for diatomic $(γ) = \frac{7}{5}$
We know that, For adiabatically compression process,
${PV}^{γ} = Constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
Ideal gas equation,
$PV = nRT$
From equation (i) and (ii)
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
$T_{2} = 291 (8)^{\frac{7}{5} - 1}$
$T_{2} = 291 (8)^{2 / 5}$
$T_{2} = 668.54 K$
$T_{2} ◻ 668.4 K$

AIPMT-1996

Kinetic Theory of Gases

138949 The equation of state for $5 g$ of oxygen at a pressure $p$ and temperature $T$ , when occupying a volume $V$ , will be

1

pV = (\frac{5}{32}) RT

2

pV = 5 RT

3

pV = (\frac{5}{2}) RT

4

pV = (\frac{5}{16}) RT

Explanation:

A Given, mass (m) $= 5 g$
Molecular weight of $O_{2} = 32$
From ideal gas equation
$PV = nRT$
$PV = (\frac{m}{M}) RT$
$PV = (\frac{5}{32}) RT$

DPMT 2000

Kinetic Theory of Gases

138950 The gas in vessel is subjected to a pressure of $20 atm$ at a temperature $27^{\circ} C$ . The pressure of the gas in a vessel after one-half of the gas is released from the vessel and the temperature of the remainder is raised by $50^{\circ} C$ is

1

8.5 atm

2

11.7 atm

3

17 atm

4

10.8 atm

Explanation:

B Ideal gas equation
$PV = \frac{m}{M} RT$
Initial pressure $(P_{1}) = 20 atm$
Initial temperature $(T_{1}) = 27 + 273 = 300 K$
Now, $20 \times V = \frac{m}{M} \times R \times 300$
Final Temperature $(T_{2}) = 300 + 50 = 350 K$
Final mass $(m^{'}) = m / 2$
$∴ P^{'} \times V = \frac{m / 2}{M} \times R \times 350$
Divide eqn (ii) by (i)
$\frac{P^{'}}{20} = \frac{350}{2 \times 300}$
$P^{'} = 11.7 atm$

AP EMCET(Medical)-2011

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