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PHXI13:KINETIC THEORY

360380 For a gas $\frac{R}{C_{V}} = 0.5$ . This gas is made up of molecules which are

1 Diatomic

2 Polyatomic

3 Monoatomic

4 Mixture of diatomic and polyatomic molecules

Explanation:

$C_{V} = \frac{R}{0.5} = \frac{R}{γ - 1} \Rightarrow γ = 1.5$
The given gas is monoatomic

PHXI13:KINETIC THEORY

360381 $C_{P}$ and $C_{V}$ are specific heats at constant pressure and constant volume respectively. It is observed that
$C_{P} - C_{V} =$ a for helium gas
$C_{P} - C_{V} = b$ for oxygen gas
Such that, $b$ is $n$ times the $a$ . Find out value of $n$ .

1

0.53 a

2

0.13 a

3

0.65 a

4

0.95 a

Explanation:

Let molar heat capacity at constant pressure $= S_{P}$
and molar heat capacity at constant volume $= s_{V}$
$∴ s_{P} - s_{V} = R$
Now, principal specific heat, $C = \frac{s}{M}$
$∴ C_{P} - C_{V} = \frac{R}{M}$
$∴$ For $H e, a = \frac{R}{4};$ for $O_{2}, b = \frac{R}{32}$
$∴ \frac{b}{a} = \frac{R}{32} \times \frac{4}{R} = \frac{1}{8}$
$\Rightarrow b = 0.125 a = 0.13 a$
$(R o u n d i n g o f f t o t w o d i g i t s)$

PHXI13:KINETIC THEORY

360382 A mixture of $n_{1}$ moles of monatomic gas and $n_{2}$ moles of diatomic gas has $\frac{C_{P}}{C_{V}} = γ = \frac{17}{11}$ . Then

1

n_{1} = 2 n_{2}

2

7 n_{1} = n_{2}

3

n_{1} = 4 n_{2}

4

3 n_{1} = 5 n_{2}

Explanation:

$γ = \frac{n_{1} C_{R_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}} γ = 1.5$ (Given)
$\begin{aligned} ∴ C_{P_{1}} = \frac{5 R}{2}; C_{P_{2}} = \frac{7 R}{2} \\ C_{V_{1}} = \frac{3 R}{2}; C_{V_{2}} = \frac{5 R}{2}; then n_{1} = 2 n_{2} \end{aligned}$

PHXI13:KINETIC THEORY

360383 Two moles of oxygen is mixed with eight mole of helium. The effective specific heat of the mixture at constant volume is

1

1.7 R

2

1.4 R

3

1.3 R

4

1.9 R

Explanation:

For mixture of gases, let specific heat be $C_{V}$
$C_{V} = \frac{n_{1} {(C_{V})}_{1} + n_{2} {(C_{V})}_{2}}{n_{1} + n_{2}}$
where ${(C_{V})}_{1} = \frac{5 R}{2}, {(C_{V})}_{2} = \frac{3 R}{2}$
$= \frac{2 \times \frac{5 R}{2} + 8 \times \frac{3 R}{2}}{2 + 8} = \frac{17 R}{10} = 1.7 R$

PHXI13:KINETIC THEORY

360384 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and gas constant $(R)$

1

C_{V} = R

2

C_{V} = \frac{R}{2}

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

$\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f} \Rightarrow γ = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} - C_{V} = R \Rightarrow \frac{4}{3} C_{V} - C_{V} = R$
$\Rightarrow \frac{1}{3} C_{V} = R \Rightarrow C_{V} = 3 R$

PHXI13:KINETIC THEORY

360380 For a gas $\frac{R}{C_{V}} = 0.5$ . This gas is made up of molecules which are

1 Diatomic

2 Polyatomic

3 Monoatomic

4 Mixture of diatomic and polyatomic molecules

Explanation:

$C_{V} = \frac{R}{0.5} = \frac{R}{γ - 1} \Rightarrow γ = 1.5$
The given gas is monoatomic

PHXI13:KINETIC THEORY

360381 $C_{P}$ and $C_{V}$ are specific heats at constant pressure and constant volume respectively. It is observed that
$C_{P} - C_{V} =$ a for helium gas
$C_{P} - C_{V} = b$ for oxygen gas
Such that, $b$ is $n$ times the $a$ . Find out value of $n$ .

1

0.53 a

2

0.13 a

3

0.65 a

4

0.95 a

Explanation:

Let molar heat capacity at constant pressure $= S_{P}$
and molar heat capacity at constant volume $= s_{V}$
$∴ s_{P} - s_{V} = R$
Now, principal specific heat, $C = \frac{s}{M}$
$∴ C_{P} - C_{V} = \frac{R}{M}$
$∴$ For $H e, a = \frac{R}{4};$ for $O_{2}, b = \frac{R}{32}$
$∴ \frac{b}{a} = \frac{R}{32} \times \frac{4}{R} = \frac{1}{8}$
$\Rightarrow b = 0.125 a = 0.13 a$
$(R o u n d i n g o f f t o t w o d i g i t s)$

PHXI13:KINETIC THEORY

360382 A mixture of $n_{1}$ moles of monatomic gas and $n_{2}$ moles of diatomic gas has $\frac{C_{P}}{C_{V}} = γ = \frac{17}{11}$ . Then

1

n_{1} = 2 n_{2}

2

7 n_{1} = n_{2}

3

n_{1} = 4 n_{2}

4

3 n_{1} = 5 n_{2}

Explanation:

$γ = \frac{n_{1} C_{R_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}} γ = 1.5$ (Given)
$\begin{aligned} ∴ C_{P_{1}} = \frac{5 R}{2}; C_{P_{2}} = \frac{7 R}{2} \\ C_{V_{1}} = \frac{3 R}{2}; C_{V_{2}} = \frac{5 R}{2}; then n_{1} = 2 n_{2} \end{aligned}$

PHXI13:KINETIC THEORY

360383 Two moles of oxygen is mixed with eight mole of helium. The effective specific heat of the mixture at constant volume is

1

1.7 R

2

1.4 R

3

1.3 R

4

1.9 R

Explanation:

For mixture of gases, let specific heat be $C_{V}$
$C_{V} = \frac{n_{1} {(C_{V})}_{1} + n_{2} {(C_{V})}_{2}}{n_{1} + n_{2}}$
where ${(C_{V})}_{1} = \frac{5 R}{2}, {(C_{V})}_{2} = \frac{3 R}{2}$
$= \frac{2 \times \frac{5 R}{2} + 8 \times \frac{3 R}{2}}{2 + 8} = \frac{17 R}{10} = 1.7 R$

PHXI13:KINETIC THEORY

360384 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and gas constant $(R)$

1

C_{V} = R

2

C_{V} = \frac{R}{2}

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

$\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f} \Rightarrow γ = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} - C_{V} = R \Rightarrow \frac{4}{3} C_{V} - C_{V} = R$
$\Rightarrow \frac{1}{3} C_{V} = R \Rightarrow C_{V} = 3 R$

PHXI13:KINETIC THEORY

360380 For a gas $\frac{R}{C_{V}} = 0.5$ . This gas is made up of molecules which are

1 Diatomic

2 Polyatomic

3 Monoatomic

4 Mixture of diatomic and polyatomic molecules

Explanation:

$C_{V} = \frac{R}{0.5} = \frac{R}{γ - 1} \Rightarrow γ = 1.5$
The given gas is monoatomic

PHXI13:KINETIC THEORY

360381 $C_{P}$ and $C_{V}$ are specific heats at constant pressure and constant volume respectively. It is observed that
$C_{P} - C_{V} =$ a for helium gas
$C_{P} - C_{V} = b$ for oxygen gas
Such that, $b$ is $n$ times the $a$ . Find out value of $n$ .

1

0.53 a

2

0.13 a

3

0.65 a

4

0.95 a

Explanation:

Let molar heat capacity at constant pressure $= S_{P}$
and molar heat capacity at constant volume $= s_{V}$
$∴ s_{P} - s_{V} = R$
Now, principal specific heat, $C = \frac{s}{M}$
$∴ C_{P} - C_{V} = \frac{R}{M}$
$∴$ For $H e, a = \frac{R}{4};$ for $O_{2}, b = \frac{R}{32}$
$∴ \frac{b}{a} = \frac{R}{32} \times \frac{4}{R} = \frac{1}{8}$
$\Rightarrow b = 0.125 a = 0.13 a$
$(R o u n d i n g o f f t o t w o d i g i t s)$

PHXI13:KINETIC THEORY

360382 A mixture of $n_{1}$ moles of monatomic gas and $n_{2}$ moles of diatomic gas has $\frac{C_{P}}{C_{V}} = γ = \frac{17}{11}$ . Then

1

n_{1} = 2 n_{2}

2

7 n_{1} = n_{2}

3

n_{1} = 4 n_{2}

4

3 n_{1} = 5 n_{2}

Explanation:

$γ = \frac{n_{1} C_{R_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}} γ = 1.5$ (Given)
$\begin{aligned} ∴ C_{P_{1}} = \frac{5 R}{2}; C_{P_{2}} = \frac{7 R}{2} \\ C_{V_{1}} = \frac{3 R}{2}; C_{V_{2}} = \frac{5 R}{2}; then n_{1} = 2 n_{2} \end{aligned}$

PHXI13:KINETIC THEORY

360383 Two moles of oxygen is mixed with eight mole of helium. The effective specific heat of the mixture at constant volume is

1

1.7 R

2

1.4 R

3

1.3 R

4

1.9 R

Explanation:

For mixture of gases, let specific heat be $C_{V}$
$C_{V} = \frac{n_{1} {(C_{V})}_{1} + n_{2} {(C_{V})}_{2}}{n_{1} + n_{2}}$
where ${(C_{V})}_{1} = \frac{5 R}{2}, {(C_{V})}_{2} = \frac{3 R}{2}$
$= \frac{2 \times \frac{5 R}{2} + 8 \times \frac{3 R}{2}}{2 + 8} = \frac{17 R}{10} = 1.7 R$

PHXI13:KINETIC THEORY

360384 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and gas constant $(R)$

1

C_{V} = R

2

C_{V} = \frac{R}{2}

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

$\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f} \Rightarrow γ = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} - C_{V} = R \Rightarrow \frac{4}{3} C_{V} - C_{V} = R$
$\Rightarrow \frac{1}{3} C_{V} = R \Rightarrow C_{V} = 3 R$

PHXI13:KINETIC THEORY

360380 For a gas $\frac{R}{C_{V}} = 0.5$ . This gas is made up of molecules which are

1 Diatomic

2 Polyatomic

3 Monoatomic

4 Mixture of diatomic and polyatomic molecules

Explanation:

$C_{V} = \frac{R}{0.5} = \frac{R}{γ - 1} \Rightarrow γ = 1.5$
The given gas is monoatomic

PHXI13:KINETIC THEORY

360381 $C_{P}$ and $C_{V}$ are specific heats at constant pressure and constant volume respectively. It is observed that
$C_{P} - C_{V} =$ a for helium gas
$C_{P} - C_{V} = b$ for oxygen gas
Such that, $b$ is $n$ times the $a$ . Find out value of $n$ .

1

0.53 a

2

0.13 a

3

0.65 a

4

0.95 a

Explanation:

Let molar heat capacity at constant pressure $= S_{P}$
and molar heat capacity at constant volume $= s_{V}$
$∴ s_{P} - s_{V} = R$
Now, principal specific heat, $C = \frac{s}{M}$
$∴ C_{P} - C_{V} = \frac{R}{M}$
$∴$ For $H e, a = \frac{R}{4};$ for $O_{2}, b = \frac{R}{32}$
$∴ \frac{b}{a} = \frac{R}{32} \times \frac{4}{R} = \frac{1}{8}$
$\Rightarrow b = 0.125 a = 0.13 a$
$(R o u n d i n g o f f t o t w o d i g i t s)$

PHXI13:KINETIC THEORY

360382 A mixture of $n_{1}$ moles of monatomic gas and $n_{2}$ moles of diatomic gas has $\frac{C_{P}}{C_{V}} = γ = \frac{17}{11}$ . Then

1

n_{1} = 2 n_{2}

2

7 n_{1} = n_{2}

3

n_{1} = 4 n_{2}

4

3 n_{1} = 5 n_{2}

Explanation:

$γ = \frac{n_{1} C_{R_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}} γ = 1.5$ (Given)
$\begin{aligned} ∴ C_{P_{1}} = \frac{5 R}{2}; C_{P_{2}} = \frac{7 R}{2} \\ C_{V_{1}} = \frac{3 R}{2}; C_{V_{2}} = \frac{5 R}{2}; then n_{1} = 2 n_{2} \end{aligned}$

PHXI13:KINETIC THEORY

360383 Two moles of oxygen is mixed with eight mole of helium. The effective specific heat of the mixture at constant volume is

1

1.7 R

2

1.4 R

3

1.3 R

4

1.9 R

Explanation:

For mixture of gases, let specific heat be $C_{V}$
$C_{V} = \frac{n_{1} {(C_{V})}_{1} + n_{2} {(C_{V})}_{2}}{n_{1} + n_{2}}$
where ${(C_{V})}_{1} = \frac{5 R}{2}, {(C_{V})}_{2} = \frac{3 R}{2}$
$= \frac{2 \times \frac{5 R}{2} + 8 \times \frac{3 R}{2}}{2 + 8} = \frac{17 R}{10} = 1.7 R$

PHXI13:KINETIC THEORY

360384 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and gas constant $(R)$

1

C_{V} = R

2

C_{V} = \frac{R}{2}

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

$\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f} \Rightarrow γ = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} - C_{V} = R \Rightarrow \frac{4}{3} C_{V} - C_{V} = R$
$\Rightarrow \frac{1}{3} C_{V} = R \Rightarrow C_{V} = 3 R$

PHXI13:KINETIC THEORY

360380 For a gas $\frac{R}{C_{V}} = 0.5$ . This gas is made up of molecules which are

1 Diatomic

2 Polyatomic

3 Monoatomic

4 Mixture of diatomic and polyatomic molecules

Explanation:

$C_{V} = \frac{R}{0.5} = \frac{R}{γ - 1} \Rightarrow γ = 1.5$
The given gas is monoatomic

PHXI13:KINETIC THEORY

360381 $C_{P}$ and $C_{V}$ are specific heats at constant pressure and constant volume respectively. It is observed that
$C_{P} - C_{V} =$ a for helium gas
$C_{P} - C_{V} = b$ for oxygen gas
Such that, $b$ is $n$ times the $a$ . Find out value of $n$ .

1

0.53 a

2

0.13 a

3

0.65 a

4

0.95 a

Explanation:

Let molar heat capacity at constant pressure $= S_{P}$
and molar heat capacity at constant volume $= s_{V}$
$∴ s_{P} - s_{V} = R$
Now, principal specific heat, $C = \frac{s}{M}$
$∴ C_{P} - C_{V} = \frac{R}{M}$
$∴$ For $H e, a = \frac{R}{4};$ for $O_{2}, b = \frac{R}{32}$
$∴ \frac{b}{a} = \frac{R}{32} \times \frac{4}{R} = \frac{1}{8}$
$\Rightarrow b = 0.125 a = 0.13 a$
$(R o u n d i n g o f f t o t w o d i g i t s)$

PHXI13:KINETIC THEORY

360382 A mixture of $n_{1}$ moles of monatomic gas and $n_{2}$ moles of diatomic gas has $\frac{C_{P}}{C_{V}} = γ = \frac{17}{11}$ . Then

1

n_{1} = 2 n_{2}

2

7 n_{1} = n_{2}

3

n_{1} = 4 n_{2}

4

3 n_{1} = 5 n_{2}

Explanation:

$γ = \frac{n_{1} C_{R_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}} γ = 1.5$ (Given)
$\begin{aligned} ∴ C_{P_{1}} = \frac{5 R}{2}; C_{P_{2}} = \frac{7 R}{2} \\ C_{V_{1}} = \frac{3 R}{2}; C_{V_{2}} = \frac{5 R}{2}; then n_{1} = 2 n_{2} \end{aligned}$

PHXI13:KINETIC THEORY

360383 Two moles of oxygen is mixed with eight mole of helium. The effective specific heat of the mixture at constant volume is

1

1.7 R

2

1.4 R

3

1.3 R

4

1.9 R

Explanation:

For mixture of gases, let specific heat be $C_{V}$
$C_{V} = \frac{n_{1} {(C_{V})}_{1} + n_{2} {(C_{V})}_{2}}{n_{1} + n_{2}}$
where ${(C_{V})}_{1} = \frac{5 R}{2}, {(C_{V})}_{2} = \frac{3 R}{2}$
$= \frac{2 \times \frac{5 R}{2} + 8 \times \frac{3 R}{2}}{2 + 8} = \frac{17 R}{10} = 1.7 R$

PHXI13:KINETIC THEORY

360384 For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat $(C_{V})$ and gas constant $(R)$

1

C_{V} = R

2

C_{V} = \frac{R}{2}

3

C_{V} = 2 R

4

C_{V} = 3 R

Explanation:

$\frac{C_{P}}{C_{V}} = γ = 1 + \frac{2}{f} \Rightarrow γ = 1 + \frac{2}{6} = \frac{4}{3}$
$C_{P} - C_{V} = R \Rightarrow \frac{4}{3} C_{V} - C_{V} = R$
$\Rightarrow \frac{1}{3} C_{V} = R \Rightarrow C_{V} = 3 R$

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