QBManager

Kinetic Theory of Gases

138938 A monoatomic gas $(γ = \frac{5}{3})$ is suddenly compressed to $1 / 8^{th}$ of the original volume adiabatically, then the pressure of the gas will become how many times larger than pressure originally?

1 32

2

40 / 3

3 8

4 same as before

Explanation:

A Given that,
For monoatomic gas,
$γ = \frac{5}{3}$
As we know that, for adiabatic process
${PV}^{γ} = constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{8 V_{1}}{V_{1}})}^{5 / 3} = {(2^{3})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = 2^{5} = 32$
$P_{2} = 32 P_{1}$

J and K CET- 2000

Kinetic Theory of Gases

138940 An ideal gas undergoes a process that has the net effect of doubling its temperature and doubling its pressure. If $V$ was the initial volume of the gas, the final volume is

1

V

2

2 V

3

4 V

4

\frac{V}{4}

Explanation:

A For an ideal gas equation,
$PV = nRT$
$∴ \frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{P_{1} V_{1}}{2 P_{1} V_{2}} = \frac{T_{1}}{2 T_{1}} [∵ P_{2} = 2 P_{1}, T_{2} = 2 T_{1}]$
$\frac{V_{1}}{2 V_{2}} = \frac{T_{1}}{2 T_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{1}{1} \Rightarrow V_{2} = V_{1} = V (∵ V_{1} = V)$

J and K CET- 1997

Kinetic Theory of Gases

138943 The volume of a gas at $20^{\circ} C$ is $100 {cm}^{3}$ at 1 atmospheric pressure. If it is heated to $100^{\circ} C$ , its volume becomes $125 {cm}^{3}$ at the same pressure, then coefficient of gas at constant pressure is

1

3.6 \times 10^{- 3} /^{\circ} C

2

3.6 \times 10^{- 4} /^{\circ} C

3

2.5 \times 10^{- 3} /^{\circ} C

4

1.25 \times 10^{- 3} /^{\circ} C

Explanation:

A Given,
Initial volume of a gas $(V_{i}) = 100 {cm}^{3}$
Initial temperature of a gas $(T_{i}) = 20^{\circ} C$
Final volume of a gas $(V_{f}) = 125 {cm}^{3}$
Final temperature of gas $(T_{f}) = 100^{\circ} C$
We know that,
$V_{f} = V_{i} (1 + γ Δ T)$
$125 = 100 [1 + γ \times (100 - 20)]$
$1 + γ (80) = \frac{125}{100}$
$1 + 80 \times γ = \frac{5}{4}$
$80 \times γ = \frac{5}{4} - 1$
$80 \times γ = \frac{1}{4}$
$γ = \frac{1}{80 \times 4}$
$γ = 3.125 \times 10^{- 3} /^{\circ} C$
$γ \approx 3.6 \times 10^{- 3} /^{\circ} C$

UP CPMT-2001

Kinetic Theory of Gases

138945 1 mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$ , then final temperature of the gas will be

1

(T - 4) K

2

(T + 4) K

3

(T - 2.4) K

4

(T + 2.4) K

Explanation:

A Given,
Initial temperature $= TK$
Adiabatic work $= 6 R$ Joule
$\frac{C_{P}}{C_{V}} = γ = 5 / 3$
We know that,
$Work done = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$= \frac{{nRT}_{1} - {nRT}_{2}}{(γ - 1)}$
$[∵ PV = nRT Ideal gas equation]$
$6 R = \frac{nR [T_{1} - T_{2}]}{γ - 1}$
$6 = \frac{T - T_{2}}{5 / 3 - 1}$
$6 = \frac{3}{2} [T - T_{2}]$
$4 = T - T_{2}$
$∴ T_{2} = (T - 4) K$

JIPMER-2010

Kinetic Theory of Gases

138938 A monoatomic gas $(γ = \frac{5}{3})$ is suddenly compressed to $1 / 8^{th}$ of the original volume adiabatically, then the pressure of the gas will become how many times larger than pressure originally?

1 32

2

40 / 3

3 8

4 same as before

Explanation:

A Given that,
For monoatomic gas,
$γ = \frac{5}{3}$
As we know that, for adiabatic process
${PV}^{γ} = constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{8 V_{1}}{V_{1}})}^{5 / 3} = {(2^{3})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = 2^{5} = 32$
$P_{2} = 32 P_{1}$

J and K CET- 2000

Kinetic Theory of Gases

138940 An ideal gas undergoes a process that has the net effect of doubling its temperature and doubling its pressure. If $V$ was the initial volume of the gas, the final volume is

1

V

2

2 V

3

4 V

4

\frac{V}{4}

Explanation:

A For an ideal gas equation,
$PV = nRT$
$∴ \frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{P_{1} V_{1}}{2 P_{1} V_{2}} = \frac{T_{1}}{2 T_{1}} [∵ P_{2} = 2 P_{1}, T_{2} = 2 T_{1}]$
$\frac{V_{1}}{2 V_{2}} = \frac{T_{1}}{2 T_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{1}{1} \Rightarrow V_{2} = V_{1} = V (∵ V_{1} = V)$

J and K CET- 1997

Kinetic Theory of Gases

138943 The volume of a gas at $20^{\circ} C$ is $100 {cm}^{3}$ at 1 atmospheric pressure. If it is heated to $100^{\circ} C$ , its volume becomes $125 {cm}^{3}$ at the same pressure, then coefficient of gas at constant pressure is

1

3.6 \times 10^{- 3} /^{\circ} C

2

3.6 \times 10^{- 4} /^{\circ} C

3

2.5 \times 10^{- 3} /^{\circ} C

4

1.25 \times 10^{- 3} /^{\circ} C

Explanation:

A Given,
Initial volume of a gas $(V_{i}) = 100 {cm}^{3}$
Initial temperature of a gas $(T_{i}) = 20^{\circ} C$
Final volume of a gas $(V_{f}) = 125 {cm}^{3}$
Final temperature of gas $(T_{f}) = 100^{\circ} C$
We know that,
$V_{f} = V_{i} (1 + γ Δ T)$
$125 = 100 [1 + γ \times (100 - 20)]$
$1 + γ (80) = \frac{125}{100}$
$1 + 80 \times γ = \frac{5}{4}$
$80 \times γ = \frac{5}{4} - 1$
$80 \times γ = \frac{1}{4}$
$γ = \frac{1}{80 \times 4}$
$γ = 3.125 \times 10^{- 3} /^{\circ} C$
$γ \approx 3.6 \times 10^{- 3} /^{\circ} C$

UP CPMT-2001

Kinetic Theory of Gases

138945 1 mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$ , then final temperature of the gas will be

1

(T - 4) K

2

(T + 4) K

3

(T - 2.4) K

4

(T + 2.4) K

Explanation:

A Given,
Initial temperature $= TK$
Adiabatic work $= 6 R$ Joule
$\frac{C_{P}}{C_{V}} = γ = 5 / 3$
We know that,
$Work done = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$= \frac{{nRT}_{1} - {nRT}_{2}}{(γ - 1)}$
$[∵ PV = nRT Ideal gas equation]$
$6 R = \frac{nR [T_{1} - T_{2}]}{γ - 1}$
$6 = \frac{T - T_{2}}{5 / 3 - 1}$
$6 = \frac{3}{2} [T - T_{2}]$
$4 = T - T_{2}$
$∴ T_{2} = (T - 4) K$

JIPMER-2010

Kinetic Theory of Gases

138938 A monoatomic gas $(γ = \frac{5}{3})$ is suddenly compressed to $1 / 8^{th}$ of the original volume adiabatically, then the pressure of the gas will become how many times larger than pressure originally?

1 32

2

40 / 3

3 8

4 same as before

Explanation:

A Given that,
For monoatomic gas,
$γ = \frac{5}{3}$
As we know that, for adiabatic process
${PV}^{γ} = constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{8 V_{1}}{V_{1}})}^{5 / 3} = {(2^{3})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = 2^{5} = 32$
$P_{2} = 32 P_{1}$

J and K CET- 2000

Kinetic Theory of Gases

138940 An ideal gas undergoes a process that has the net effect of doubling its temperature and doubling its pressure. If $V$ was the initial volume of the gas, the final volume is

1

V

2

2 V

3

4 V

4

\frac{V}{4}

Explanation:

A For an ideal gas equation,
$PV = nRT$
$∴ \frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{P_{1} V_{1}}{2 P_{1} V_{2}} = \frac{T_{1}}{2 T_{1}} [∵ P_{2} = 2 P_{1}, T_{2} = 2 T_{1}]$
$\frac{V_{1}}{2 V_{2}} = \frac{T_{1}}{2 T_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{1}{1} \Rightarrow V_{2} = V_{1} = V (∵ V_{1} = V)$

J and K CET- 1997

Kinetic Theory of Gases

138943 The volume of a gas at $20^{\circ} C$ is $100 {cm}^{3}$ at 1 atmospheric pressure. If it is heated to $100^{\circ} C$ , its volume becomes $125 {cm}^{3}$ at the same pressure, then coefficient of gas at constant pressure is

1

3.6 \times 10^{- 3} /^{\circ} C

2

3.6 \times 10^{- 4} /^{\circ} C

3

2.5 \times 10^{- 3} /^{\circ} C

4

1.25 \times 10^{- 3} /^{\circ} C

Explanation:

A Given,
Initial volume of a gas $(V_{i}) = 100 {cm}^{3}$
Initial temperature of a gas $(T_{i}) = 20^{\circ} C$
Final volume of a gas $(V_{f}) = 125 {cm}^{3}$
Final temperature of gas $(T_{f}) = 100^{\circ} C$
We know that,
$V_{f} = V_{i} (1 + γ Δ T)$
$125 = 100 [1 + γ \times (100 - 20)]$
$1 + γ (80) = \frac{125}{100}$
$1 + 80 \times γ = \frac{5}{4}$
$80 \times γ = \frac{5}{4} - 1$
$80 \times γ = \frac{1}{4}$
$γ = \frac{1}{80 \times 4}$
$γ = 3.125 \times 10^{- 3} /^{\circ} C$
$γ \approx 3.6 \times 10^{- 3} /^{\circ} C$

UP CPMT-2001

Kinetic Theory of Gases

138945 1 mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$ , then final temperature of the gas will be

1

(T - 4) K

2

(T + 4) K

3

(T - 2.4) K

4

(T + 2.4) K

Explanation:

A Given,
Initial temperature $= TK$
Adiabatic work $= 6 R$ Joule
$\frac{C_{P}}{C_{V}} = γ = 5 / 3$
We know that,
$Work done = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$= \frac{{nRT}_{1} - {nRT}_{2}}{(γ - 1)}$
$[∵ PV = nRT Ideal gas equation]$
$6 R = \frac{nR [T_{1} - T_{2}]}{γ - 1}$
$6 = \frac{T - T_{2}}{5 / 3 - 1}$
$6 = \frac{3}{2} [T - T_{2}]$
$4 = T - T_{2}$
$∴ T_{2} = (T - 4) K$

JIPMER-2010

Kinetic Theory of Gases

138938 A monoatomic gas $(γ = \frac{5}{3})$ is suddenly compressed to $1 / 8^{th}$ of the original volume adiabatically, then the pressure of the gas will become how many times larger than pressure originally?

1 32

2

40 / 3

3 8

4 same as before

Explanation:

A Given that,
For monoatomic gas,
$γ = \frac{5}{3}$
As we know that, for adiabatic process
${PV}^{γ} = constant$
$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ}$
$\frac{P_{2}}{P_{1}} = {(\frac{8 V_{1}}{V_{1}})}^{5 / 3} = {(2^{3})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = 2^{5} = 32$
$P_{2} = 32 P_{1}$

J and K CET- 2000

Kinetic Theory of Gases

138940 An ideal gas undergoes a process that has the net effect of doubling its temperature and doubling its pressure. If $V$ was the initial volume of the gas, the final volume is

1

V

2

2 V

3

4 V

4

\frac{V}{4}

Explanation:

A For an ideal gas equation,
$PV = nRT$
$∴ \frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{P_{1} V_{1}}{2 P_{1} V_{2}} = \frac{T_{1}}{2 T_{1}} [∵ P_{2} = 2 P_{1}, T_{2} = 2 T_{1}]$
$\frac{V_{1}}{2 V_{2}} = \frac{T_{1}}{2 T_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{1}{1} \Rightarrow V_{2} = V_{1} = V (∵ V_{1} = V)$

J and K CET- 1997

Kinetic Theory of Gases

138943 The volume of a gas at $20^{\circ} C$ is $100 {cm}^{3}$ at 1 atmospheric pressure. If it is heated to $100^{\circ} C$ , its volume becomes $125 {cm}^{3}$ at the same pressure, then coefficient of gas at constant pressure is

1

3.6 \times 10^{- 3} /^{\circ} C

2

3.6 \times 10^{- 4} /^{\circ} C

3

2.5 \times 10^{- 3} /^{\circ} C

4

1.25 \times 10^{- 3} /^{\circ} C

Explanation:

A Given,
Initial volume of a gas $(V_{i}) = 100 {cm}^{3}$
Initial temperature of a gas $(T_{i}) = 20^{\circ} C$
Final volume of a gas $(V_{f}) = 125 {cm}^{3}$
Final temperature of gas $(T_{f}) = 100^{\circ} C$
We know that,
$V_{f} = V_{i} (1 + γ Δ T)$
$125 = 100 [1 + γ \times (100 - 20)]$
$1 + γ (80) = \frac{125}{100}$
$1 + 80 \times γ = \frac{5}{4}$
$80 \times γ = \frac{5}{4} - 1$
$80 \times γ = \frac{1}{4}$
$γ = \frac{1}{80 \times 4}$
$γ = 3.125 \times 10^{- 3} /^{\circ} C$
$γ \approx 3.6 \times 10^{- 3} /^{\circ} C$

UP CPMT-2001

Kinetic Theory of Gases

138945 1 mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$ , then final temperature of the gas will be

1

(T - 4) K

2

(T + 4) K

3

(T - 2.4) K

4

(T + 2.4) K

Explanation:

A Given,
Initial temperature $= TK$
Adiabatic work $= 6 R$ Joule
$\frac{C_{P}}{C_{V}} = γ = 5 / 3$
We know that,
$Work done = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
$= \frac{{nRT}_{1} - {nRT}_{2}}{(γ - 1)}$
$[∵ PV = nRT Ideal gas equation]$
$6 R = \frac{nR [T_{1} - T_{2}]}{γ - 1}$
$6 = \frac{T - T_{2}}{5 / 3 - 1}$
$6 = \frac{3}{2} [T - T_{2}]$
$4 = T - T_{2}$
$∴ T_{2} = (T - 4) K$

JIPMER-2010

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