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Kinetic Theory of Gases

139319 Specific heat of a gas undergoing adiabatic change is

1 Zero

2 Infinite

3 Positive

4 Negative

AP EAMCET-23.08.2021

Kinetic Theory of Gases

139323 For an ideal gas, if the ratio of Molar specific heats $γ = 1.4$ , then the specific heat at constant pressure $C_{P}$ , specific heat at constant volume $C_{V}$ and corresponding molecule are respectively

1

\frac{9}{2} R, \frac{7}{2} R

, polyatomic

2

\frac{7}{2} R, \frac{5}{2} R

, non - rigid diatomic

3

\frac{7}{2} R, \frac{5}{2} R

, rigid diatomic

4

\frac{5}{2} R, \frac{3}{2} R

, monoatomic

Explanation:

C Ratio of molar specific heat $γ = 1.4$
$\frac{C_{P}}{C_{V}} = γ = 1.4$
$\frac{C_{P}}{C_{V}} = \frac{14}{10} = \frac{7}{5} (∵ \frac{C_{P}}{C_{V}} = \frac{7}{5}$ so, diatomic rigid $)$
$C_{V} = \frac{f}{2} R$
So, $C_{V} = \frac{5}{2} R (∵ γ = \frac{7}{5} = 1 + \frac{2}{f} \Rightarrow f = 5)$
$C_{P} = (1 + \frac{n}{2}) R$
$C_{P} = \frac{7}{2} R$
And molecule are rigid diatomic.

MHT-CET 2020

Kinetic Theory of Gases

139324 For a gas, $\frac{R}{C_{v}} = 0.4$ where ' $R$ ' is universal gas constant and $C_{V}$ is the molar specific heat at constant volume. The gas is made up of molecule which are

1 Monoatomic

2 rigid diatomic

3 non-rigid diatomic

4 polyatomic

Explanation:

B Given that,
$\frac{R}{C_{V}} = 0.4$
$R = 0.4 C_{V}$
$∵ C_{P} - C_{V} = R$
So, $C_{P} - C_{V} = 0.4 C_{V}$
$C_{P} = 1.4 C_{V}$
$\frac{C_{P}}{C_{V}} = 1.4 = \frac{7}{5}$
Ratio of $\frac{C_{P}}{C_{V}} = \frac{7}{5}$ , Therefore, gas is made up of rigid diatomic molecule.

MHT-CET 2020

Kinetic Theory of Gases

139325 The molar specific heats of an ideal gas at constant pressure and constant volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $γ = \frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{p}$ is equal to

1

\frac{γ - 1}{R}

2

\frac{(γ - 1)^{2}}{R}

3

\frac{γ - 1}{γ R}

4

\frac{γ R}{γ - 1}

Explanation:

D Given that,
$\frac{C_{P}}{C_{V}} = γ$
We know $C_{P} - C_{V} = R$
From eq (i),
$\frac{C_{P} - C_{V}}{C_{V}} = γ - 1$
$\frac{R}{C_{V}} = γ - 1 (∵ C_{P} - C_{V} = R)$
$\frac{γ R}{C_{P}} = γ - 1 (∵ C_{V} = \frac{C_{P}}{γ})$
$C_{P} = \frac{γ R}{γ - 1}$

MHT-CET 2020

Kinetic Theory of Gases

139319 Specific heat of a gas undergoing adiabatic change is

1 Zero

2 Infinite

3 Positive

4 Negative

AP EAMCET-23.08.2021

Kinetic Theory of Gases

139323 For an ideal gas, if the ratio of Molar specific heats $γ = 1.4$ , then the specific heat at constant pressure $C_{P}$ , specific heat at constant volume $C_{V}$ and corresponding molecule are respectively

1

\frac{9}{2} R, \frac{7}{2} R

, polyatomic

2

\frac{7}{2} R, \frac{5}{2} R

, non - rigid diatomic

3

\frac{7}{2} R, \frac{5}{2} R

, rigid diatomic

4

\frac{5}{2} R, \frac{3}{2} R

, monoatomic

Explanation:

C Ratio of molar specific heat $γ = 1.4$
$\frac{C_{P}}{C_{V}} = γ = 1.4$
$\frac{C_{P}}{C_{V}} = \frac{14}{10} = \frac{7}{5} (∵ \frac{C_{P}}{C_{V}} = \frac{7}{5}$ so, diatomic rigid $)$
$C_{V} = \frac{f}{2} R$
So, $C_{V} = \frac{5}{2} R (∵ γ = \frac{7}{5} = 1 + \frac{2}{f} \Rightarrow f = 5)$
$C_{P} = (1 + \frac{n}{2}) R$
$C_{P} = \frac{7}{2} R$
And molecule are rigid diatomic.

MHT-CET 2020

Kinetic Theory of Gases

139324 For a gas, $\frac{R}{C_{v}} = 0.4$ where ' $R$ ' is universal gas constant and $C_{V}$ is the molar specific heat at constant volume. The gas is made up of molecule which are

1 Monoatomic

2 rigid diatomic

3 non-rigid diatomic

4 polyatomic

Explanation:

B Given that,
$\frac{R}{C_{V}} = 0.4$
$R = 0.4 C_{V}$
$∵ C_{P} - C_{V} = R$
So, $C_{P} - C_{V} = 0.4 C_{V}$
$C_{P} = 1.4 C_{V}$
$\frac{C_{P}}{C_{V}} = 1.4 = \frac{7}{5}$
Ratio of $\frac{C_{P}}{C_{V}} = \frac{7}{5}$ , Therefore, gas is made up of rigid diatomic molecule.

MHT-CET 2020

Kinetic Theory of Gases

139325 The molar specific heats of an ideal gas at constant pressure and constant volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $γ = \frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{p}$ is equal to

1

\frac{γ - 1}{R}

2

\frac{(γ - 1)^{2}}{R}

3

\frac{γ - 1}{γ R}

4

\frac{γ R}{γ - 1}

Explanation:

D Given that,
$\frac{C_{P}}{C_{V}} = γ$
We know $C_{P} - C_{V} = R$
From eq (i),
$\frac{C_{P} - C_{V}}{C_{V}} = γ - 1$
$\frac{R}{C_{V}} = γ - 1 (∵ C_{P} - C_{V} = R)$
$\frac{γ R}{C_{P}} = γ - 1 (∵ C_{V} = \frac{C_{P}}{γ})$
$C_{P} = \frac{γ R}{γ - 1}$

MHT-CET 2020

Kinetic Theory of Gases

139319 Specific heat of a gas undergoing adiabatic change is

1 Zero

2 Infinite

3 Positive

4 Negative

AP EAMCET-23.08.2021

Kinetic Theory of Gases

139323 For an ideal gas, if the ratio of Molar specific heats $γ = 1.4$ , then the specific heat at constant pressure $C_{P}$ , specific heat at constant volume $C_{V}$ and corresponding molecule are respectively

1

\frac{9}{2} R, \frac{7}{2} R

, polyatomic

2

\frac{7}{2} R, \frac{5}{2} R

, non - rigid diatomic

3

\frac{7}{2} R, \frac{5}{2} R

, rigid diatomic

4

\frac{5}{2} R, \frac{3}{2} R

, monoatomic

Explanation:

C Ratio of molar specific heat $γ = 1.4$
$\frac{C_{P}}{C_{V}} = γ = 1.4$
$\frac{C_{P}}{C_{V}} = \frac{14}{10} = \frac{7}{5} (∵ \frac{C_{P}}{C_{V}} = \frac{7}{5}$ so, diatomic rigid $)$
$C_{V} = \frac{f}{2} R$
So, $C_{V} = \frac{5}{2} R (∵ γ = \frac{7}{5} = 1 + \frac{2}{f} \Rightarrow f = 5)$
$C_{P} = (1 + \frac{n}{2}) R$
$C_{P} = \frac{7}{2} R$
And molecule are rigid diatomic.

MHT-CET 2020

Kinetic Theory of Gases

139324 For a gas, $\frac{R}{C_{v}} = 0.4$ where ' $R$ ' is universal gas constant and $C_{V}$ is the molar specific heat at constant volume. The gas is made up of molecule which are

1 Monoatomic

2 rigid diatomic

3 non-rigid diatomic

4 polyatomic

Explanation:

B Given that,
$\frac{R}{C_{V}} = 0.4$
$R = 0.4 C_{V}$
$∵ C_{P} - C_{V} = R$
So, $C_{P} - C_{V} = 0.4 C_{V}$
$C_{P} = 1.4 C_{V}$
$\frac{C_{P}}{C_{V}} = 1.4 = \frac{7}{5}$
Ratio of $\frac{C_{P}}{C_{V}} = \frac{7}{5}$ , Therefore, gas is made up of rigid diatomic molecule.

MHT-CET 2020

Kinetic Theory of Gases

139325 The molar specific heats of an ideal gas at constant pressure and constant volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $γ = \frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{p}$ is equal to

1

\frac{γ - 1}{R}

2

\frac{(γ - 1)^{2}}{R}

3

\frac{γ - 1}{γ R}

4

\frac{γ R}{γ - 1}

Explanation:

D Given that,
$\frac{C_{P}}{C_{V}} = γ$
We know $C_{P} - C_{V} = R$
From eq (i),
$\frac{C_{P} - C_{V}}{C_{V}} = γ - 1$
$\frac{R}{C_{V}} = γ - 1 (∵ C_{P} - C_{V} = R)$
$\frac{γ R}{C_{P}} = γ - 1 (∵ C_{V} = \frac{C_{P}}{γ})$
$C_{P} = \frac{γ R}{γ - 1}$

MHT-CET 2020

Kinetic Theory of Gases

139319 Specific heat of a gas undergoing adiabatic change is

1 Zero

2 Infinite

3 Positive

4 Negative

AP EAMCET-23.08.2021

Kinetic Theory of Gases

139323 For an ideal gas, if the ratio of Molar specific heats $γ = 1.4$ , then the specific heat at constant pressure $C_{P}$ , specific heat at constant volume $C_{V}$ and corresponding molecule are respectively

1

\frac{9}{2} R, \frac{7}{2} R

, polyatomic

2

\frac{7}{2} R, \frac{5}{2} R

, non - rigid diatomic

3

\frac{7}{2} R, \frac{5}{2} R

, rigid diatomic

4

\frac{5}{2} R, \frac{3}{2} R

, monoatomic

Explanation:

C Ratio of molar specific heat $γ = 1.4$
$\frac{C_{P}}{C_{V}} = γ = 1.4$
$\frac{C_{P}}{C_{V}} = \frac{14}{10} = \frac{7}{5} (∵ \frac{C_{P}}{C_{V}} = \frac{7}{5}$ so, diatomic rigid $)$
$C_{V} = \frac{f}{2} R$
So, $C_{V} = \frac{5}{2} R (∵ γ = \frac{7}{5} = 1 + \frac{2}{f} \Rightarrow f = 5)$
$C_{P} = (1 + \frac{n}{2}) R$
$C_{P} = \frac{7}{2} R$
And molecule are rigid diatomic.

MHT-CET 2020

Kinetic Theory of Gases

139324 For a gas, $\frac{R}{C_{v}} = 0.4$ where ' $R$ ' is universal gas constant and $C_{V}$ is the molar specific heat at constant volume. The gas is made up of molecule which are

1 Monoatomic

2 rigid diatomic

3 non-rigid diatomic

4 polyatomic

Explanation:

B Given that,
$\frac{R}{C_{V}} = 0.4$
$R = 0.4 C_{V}$
$∵ C_{P} - C_{V} = R$
So, $C_{P} - C_{V} = 0.4 C_{V}$
$C_{P} = 1.4 C_{V}$
$\frac{C_{P}}{C_{V}} = 1.4 = \frac{7}{5}$
Ratio of $\frac{C_{P}}{C_{V}} = \frac{7}{5}$ , Therefore, gas is made up of rigid diatomic molecule.

MHT-CET 2020

Kinetic Theory of Gases

139325 The molar specific heats of an ideal gas at constant pressure and constant volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $γ = \frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{p}$ is equal to

1

\frac{γ - 1}{R}

2

\frac{(γ - 1)^{2}}{R}

3

\frac{γ - 1}{γ R}

4

\frac{γ R}{γ - 1}

Explanation:

D Given that,
$\frac{C_{P}}{C_{V}} = γ$
We know $C_{P} - C_{V} = R$
From eq (i),
$\frac{C_{P} - C_{V}}{C_{V}} = γ - 1$
$\frac{R}{C_{V}} = γ - 1 (∵ C_{P} - C_{V} = R)$
$\frac{γ R}{C_{P}} = γ - 1 (∵ C_{V} = \frac{C_{P}}{γ})$
$C_{P} = \frac{γ R}{γ - 1}$

MHT-CET 2020

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