QBManager

PHXI13:KINETIC THEORY

360367 One mole of a gas occupies $22.4 l i t$ at $N . T . P .$ Calculate the difference between two molar specific heats of the gas. $J = 4200 J / k c a l$ .

1

1.979 k c a l / k m o l K

2

2.378 k c a l / k m o l K

3

4.569 k c a l / k m o l K

4

3.028 k c a l / k m o l K

Explanation:

$V = 22.4 litre = 22.4 \times 10^{- 3} m^{3},$
$J = 4200 J / k c a l$ by ideal gas equation for one mole of a gas, $R = \frac{P V}{T} = \frac{1.013 \times 10^{5} \times 22.4 \times 10^{- 3}}{273}$
$C_{p} - C_{v} = \frac{R}{J} = \frac{1.013 \times 10^{5} \times 22.4}{273 \times 4200} = 1.979$
$k c a l / k m o l K$

PHXI13:KINETIC THEORY

360368 The molar specific heat at constant pressure of an ideal gas is $\frac{7}{2} R$ . The ratio of specific heat at constant pressure to that at constant volume is

1

\frac{7}{5}

2

\frac{8}{7}

3

\frac{5}{7}

4

\frac{9}{7}

Explanation:

Use Mayer's equation
$\begin{aligned} C_{P} - C_{V} = R \\ C_{V} = \frac{7}{2} R - R = \frac{5}{2} R \\ γ = \frac{C_{P}}{C_{V}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} \Rightarrow γ = \frac{7}{5} \end{aligned}$

PHXI13:KINETIC THEORY

360369 A container contain $0.1 m o l$ of $H_{2}$ and $0.1 m o l$ of $O_{2}$ . If the gases are in thermal equilibrium then

1 Only the average kinetic energy of the molecule of

H_{2}

and

O_{2}

is same.

2 Average speed of the molecule of

H_{2}

and

O_{2}

is same.

3 Only the specific heat at constant pressure of two gases is same.

4 The specific heat at constant pressure and the kinetic energy are same for both the gases.

Explanation:

The specific heat at constant pressure $(C_{P})$ is the amount of heat required to raise the temperature of one gram through $1^{\circ} C$ when the pressure of the gas is kept constant. Again, the mean kinetic energy per molecule (3/2) $k T$ depends only upon temperature. Clearly both the specific heats at constant pressure and mean kinetic energy are depending on the temperature which is again same for the two gases.

PHXI13:KINETIC THEORY

360370 Two moles of an ideal gas with $\frac{C_{P}}{C_{V}} = \frac{5}{3}$ are mixed with 3 moles of another ideal gas with $\frac{C_{P}}{C_{V}} = \frac{4}{3}$ . The value of $\frac{C_{P}}{C_{V}}$ for the mixture is

1 1.50

2 1.42

3 1.47

4 1.45

Explanation:

$γ_{max} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}$
$C_{P} = (\frac{γ}{γ - 1}) R$
$C_{V} = \frac{R}{γ - 1}$
$γ_{m i x} = \frac{n_{1} \frac{γ_{1}}{γ_{1} - 1} + \frac{n_{2} γ_{2}}{γ_{2} - 1}}{\frac{n_{1}}{γ_{1} - 1} + \frac{n_{2}}{γ_{2} - 1}}$
$n_{1} = 2, γ_{1} = \frac{5}{3}, n_{2} = 3, γ_{2} = \frac{4}{3}$
We get
$γ_{m i x} = 1.42$

JEE - 2020

PHXI13:KINETIC THEORY

360371 The ratio of molar specific heats of oxygen is

1 1.4

2 1.67

3 1.33

4 1.28

Explanation:

Ratio of molar specific heats, $γ = \frac{C_{P}}{C_{V}}$ $γ = 1 + \frac{2}{f}$
$f =$ number of degrees of freedom
$f = 5$ for $O_{2}$
So $γ = 1 + \frac{2}{5} = 1.4$

KCET - 2024

PHXI13:KINETIC THEORY

360367 One mole of a gas occupies $22.4 l i t$ at $N . T . P .$ Calculate the difference between two molar specific heats of the gas. $J = 4200 J / k c a l$ .

1

1.979 k c a l / k m o l K

2

2.378 k c a l / k m o l K

3

4.569 k c a l / k m o l K

4

3.028 k c a l / k m o l K

Explanation:

$V = 22.4 litre = 22.4 \times 10^{- 3} m^{3},$
$J = 4200 J / k c a l$ by ideal gas equation for one mole of a gas, $R = \frac{P V}{T} = \frac{1.013 \times 10^{5} \times 22.4 \times 10^{- 3}}{273}$
$C_{p} - C_{v} = \frac{R}{J} = \frac{1.013 \times 10^{5} \times 22.4}{273 \times 4200} = 1.979$
$k c a l / k m o l K$

PHXI13:KINETIC THEORY

360368 The molar specific heat at constant pressure of an ideal gas is $\frac{7}{2} R$ . The ratio of specific heat at constant pressure to that at constant volume is

1

\frac{7}{5}

2

\frac{8}{7}

3

\frac{5}{7}

4

\frac{9}{7}

Explanation:

Use Mayer's equation
$\begin{aligned} C_{P} - C_{V} = R \\ C_{V} = \frac{7}{2} R - R = \frac{5}{2} R \\ γ = \frac{C_{P}}{C_{V}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} \Rightarrow γ = \frac{7}{5} \end{aligned}$

PHXI13:KINETIC THEORY

360369 A container contain $0.1 m o l$ of $H_{2}$ and $0.1 m o l$ of $O_{2}$ . If the gases are in thermal equilibrium then

1 Only the average kinetic energy of the molecule of

H_{2}

and

O_{2}

is same.

2 Average speed of the molecule of

H_{2}

and

O_{2}

is same.

3 Only the specific heat at constant pressure of two gases is same.

4 The specific heat at constant pressure and the kinetic energy are same for both the gases.

Explanation:

The specific heat at constant pressure $(C_{P})$ is the amount of heat required to raise the temperature of one gram through $1^{\circ} C$ when the pressure of the gas is kept constant. Again, the mean kinetic energy per molecule (3/2) $k T$ depends only upon temperature. Clearly both the specific heats at constant pressure and mean kinetic energy are depending on the temperature which is again same for the two gases.

PHXI13:KINETIC THEORY

360370 Two moles of an ideal gas with $\frac{C_{P}}{C_{V}} = \frac{5}{3}$ are mixed with 3 moles of another ideal gas with $\frac{C_{P}}{C_{V}} = \frac{4}{3}$ . The value of $\frac{C_{P}}{C_{V}}$ for the mixture is

1 1.50

2 1.42

3 1.47

4 1.45

Explanation:

$γ_{max} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}$
$C_{P} = (\frac{γ}{γ - 1}) R$
$C_{V} = \frac{R}{γ - 1}$
$γ_{m i x} = \frac{n_{1} \frac{γ_{1}}{γ_{1} - 1} + \frac{n_{2} γ_{2}}{γ_{2} - 1}}{\frac{n_{1}}{γ_{1} - 1} + \frac{n_{2}}{γ_{2} - 1}}$
$n_{1} = 2, γ_{1} = \frac{5}{3}, n_{2} = 3, γ_{2} = \frac{4}{3}$
We get
$γ_{m i x} = 1.42$

JEE - 2020

PHXI13:KINETIC THEORY

360371 The ratio of molar specific heats of oxygen is

1 1.4

2 1.67

3 1.33

4 1.28

Explanation:

Ratio of molar specific heats, $γ = \frac{C_{P}}{C_{V}}$ $γ = 1 + \frac{2}{f}$
$f =$ number of degrees of freedom
$f = 5$ for $O_{2}$
So $γ = 1 + \frac{2}{5} = 1.4$

KCET - 2024

PHXI13:KINETIC THEORY

360367 One mole of a gas occupies $22.4 l i t$ at $N . T . P .$ Calculate the difference between two molar specific heats of the gas. $J = 4200 J / k c a l$ .

1

1.979 k c a l / k m o l K

2

2.378 k c a l / k m o l K

3

4.569 k c a l / k m o l K

4

3.028 k c a l / k m o l K

Explanation:

$V = 22.4 litre = 22.4 \times 10^{- 3} m^{3},$
$J = 4200 J / k c a l$ by ideal gas equation for one mole of a gas, $R = \frac{P V}{T} = \frac{1.013 \times 10^{5} \times 22.4 \times 10^{- 3}}{273}$
$C_{p} - C_{v} = \frac{R}{J} = \frac{1.013 \times 10^{5} \times 22.4}{273 \times 4200} = 1.979$
$k c a l / k m o l K$

PHXI13:KINETIC THEORY

360368 The molar specific heat at constant pressure of an ideal gas is $\frac{7}{2} R$ . The ratio of specific heat at constant pressure to that at constant volume is

1

\frac{7}{5}

2

\frac{8}{7}

3

\frac{5}{7}

4

\frac{9}{7}

Explanation:

Use Mayer's equation
$\begin{aligned} C_{P} - C_{V} = R \\ C_{V} = \frac{7}{2} R - R = \frac{5}{2} R \\ γ = \frac{C_{P}}{C_{V}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} \Rightarrow γ = \frac{7}{5} \end{aligned}$

PHXI13:KINETIC THEORY

360369 A container contain $0.1 m o l$ of $H_{2}$ and $0.1 m o l$ of $O_{2}$ . If the gases are in thermal equilibrium then

1 Only the average kinetic energy of the molecule of

H_{2}

and

O_{2}

is same.

2 Average speed of the molecule of

H_{2}

and

O_{2}

is same.

3 Only the specific heat at constant pressure of two gases is same.

4 The specific heat at constant pressure and the kinetic energy are same for both the gases.

Explanation:

The specific heat at constant pressure $(C_{P})$ is the amount of heat required to raise the temperature of one gram through $1^{\circ} C$ when the pressure of the gas is kept constant. Again, the mean kinetic energy per molecule (3/2) $k T$ depends only upon temperature. Clearly both the specific heats at constant pressure and mean kinetic energy are depending on the temperature which is again same for the two gases.

PHXI13:KINETIC THEORY

360370 Two moles of an ideal gas with $\frac{C_{P}}{C_{V}} = \frac{5}{3}$ are mixed with 3 moles of another ideal gas with $\frac{C_{P}}{C_{V}} = \frac{4}{3}$ . The value of $\frac{C_{P}}{C_{V}}$ for the mixture is

1 1.50

2 1.42

3 1.47

4 1.45

Explanation:

$γ_{max} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}$
$C_{P} = (\frac{γ}{γ - 1}) R$
$C_{V} = \frac{R}{γ - 1}$
$γ_{m i x} = \frac{n_{1} \frac{γ_{1}}{γ_{1} - 1} + \frac{n_{2} γ_{2}}{γ_{2} - 1}}{\frac{n_{1}}{γ_{1} - 1} + \frac{n_{2}}{γ_{2} - 1}}$
$n_{1} = 2, γ_{1} = \frac{5}{3}, n_{2} = 3, γ_{2} = \frac{4}{3}$
We get
$γ_{m i x} = 1.42$

JEE - 2020

PHXI13:KINETIC THEORY

360371 The ratio of molar specific heats of oxygen is

1 1.4

2 1.67

3 1.33

4 1.28

Explanation:

Ratio of molar specific heats, $γ = \frac{C_{P}}{C_{V}}$ $γ = 1 + \frac{2}{f}$
$f =$ number of degrees of freedom
$f = 5$ for $O_{2}$
So $γ = 1 + \frac{2}{5} = 1.4$

KCET - 2024

PHXI13:KINETIC THEORY

360367 One mole of a gas occupies $22.4 l i t$ at $N . T . P .$ Calculate the difference between two molar specific heats of the gas. $J = 4200 J / k c a l$ .

1

1.979 k c a l / k m o l K

2

2.378 k c a l / k m o l K

3

4.569 k c a l / k m o l K

4

3.028 k c a l / k m o l K

Explanation:

$V = 22.4 litre = 22.4 \times 10^{- 3} m^{3},$
$J = 4200 J / k c a l$ by ideal gas equation for one mole of a gas, $R = \frac{P V}{T} = \frac{1.013 \times 10^{5} \times 22.4 \times 10^{- 3}}{273}$
$C_{p} - C_{v} = \frac{R}{J} = \frac{1.013 \times 10^{5} \times 22.4}{273 \times 4200} = 1.979$
$k c a l / k m o l K$

PHXI13:KINETIC THEORY

360368 The molar specific heat at constant pressure of an ideal gas is $\frac{7}{2} R$ . The ratio of specific heat at constant pressure to that at constant volume is

1

\frac{7}{5}

2

\frac{8}{7}

3

\frac{5}{7}

4

\frac{9}{7}

Explanation:

Use Mayer's equation
$\begin{aligned} C_{P} - C_{V} = R \\ C_{V} = \frac{7}{2} R - R = \frac{5}{2} R \\ γ = \frac{C_{P}}{C_{V}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} \Rightarrow γ = \frac{7}{5} \end{aligned}$

PHXI13:KINETIC THEORY

360369 A container contain $0.1 m o l$ of $H_{2}$ and $0.1 m o l$ of $O_{2}$ . If the gases are in thermal equilibrium then

1 Only the average kinetic energy of the molecule of

H_{2}

and

O_{2}

is same.

2 Average speed of the molecule of

H_{2}

and

O_{2}

is same.

3 Only the specific heat at constant pressure of two gases is same.

4 The specific heat at constant pressure and the kinetic energy are same for both the gases.

Explanation:

The specific heat at constant pressure $(C_{P})$ is the amount of heat required to raise the temperature of one gram through $1^{\circ} C$ when the pressure of the gas is kept constant. Again, the mean kinetic energy per molecule (3/2) $k T$ depends only upon temperature. Clearly both the specific heats at constant pressure and mean kinetic energy are depending on the temperature which is again same for the two gases.

PHXI13:KINETIC THEORY

360370 Two moles of an ideal gas with $\frac{C_{P}}{C_{V}} = \frac{5}{3}$ are mixed with 3 moles of another ideal gas with $\frac{C_{P}}{C_{V}} = \frac{4}{3}$ . The value of $\frac{C_{P}}{C_{V}}$ for the mixture is

1 1.50

2 1.42

3 1.47

4 1.45

Explanation:

$γ_{max} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}$
$C_{P} = (\frac{γ}{γ - 1}) R$
$C_{V} = \frac{R}{γ - 1}$
$γ_{m i x} = \frac{n_{1} \frac{γ_{1}}{γ_{1} - 1} + \frac{n_{2} γ_{2}}{γ_{2} - 1}}{\frac{n_{1}}{γ_{1} - 1} + \frac{n_{2}}{γ_{2} - 1}}$
$n_{1} = 2, γ_{1} = \frac{5}{3}, n_{2} = 3, γ_{2} = \frac{4}{3}$
We get
$γ_{m i x} = 1.42$

JEE - 2020

PHXI13:KINETIC THEORY

360371 The ratio of molar specific heats of oxygen is

1 1.4

2 1.67

3 1.33

4 1.28

Explanation:

Ratio of molar specific heats, $γ = \frac{C_{P}}{C_{V}}$ $γ = 1 + \frac{2}{f}$
$f =$ number of degrees of freedom
$f = 5$ for $O_{2}$
So $γ = 1 + \frac{2}{5} = 1.4$

KCET - 2024

PHXI13:KINETIC THEORY

360367 One mole of a gas occupies $22.4 l i t$ at $N . T . P .$ Calculate the difference between two molar specific heats of the gas. $J = 4200 J / k c a l$ .

1

1.979 k c a l / k m o l K

2

2.378 k c a l / k m o l K

3

4.569 k c a l / k m o l K

4

3.028 k c a l / k m o l K

Explanation:

$V = 22.4 litre = 22.4 \times 10^{- 3} m^{3},$
$J = 4200 J / k c a l$ by ideal gas equation for one mole of a gas, $R = \frac{P V}{T} = \frac{1.013 \times 10^{5} \times 22.4 \times 10^{- 3}}{273}$
$C_{p} - C_{v} = \frac{R}{J} = \frac{1.013 \times 10^{5} \times 22.4}{273 \times 4200} = 1.979$
$k c a l / k m o l K$

PHXI13:KINETIC THEORY

360368 The molar specific heat at constant pressure of an ideal gas is $\frac{7}{2} R$ . The ratio of specific heat at constant pressure to that at constant volume is

1

\frac{7}{5}

2

\frac{8}{7}

3

\frac{5}{7}

4

\frac{9}{7}

Explanation:

Use Mayer's equation
$\begin{aligned} C_{P} - C_{V} = R \\ C_{V} = \frac{7}{2} R - R = \frac{5}{2} R \\ γ = \frac{C_{P}}{C_{V}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} \Rightarrow γ = \frac{7}{5} \end{aligned}$

PHXI13:KINETIC THEORY

360369 A container contain $0.1 m o l$ of $H_{2}$ and $0.1 m o l$ of $O_{2}$ . If the gases are in thermal equilibrium then

1 Only the average kinetic energy of the molecule of

H_{2}

and

O_{2}

is same.

2 Average speed of the molecule of

H_{2}

and

O_{2}

is same.

3 Only the specific heat at constant pressure of two gases is same.

4 The specific heat at constant pressure and the kinetic energy are same for both the gases.

Explanation:

The specific heat at constant pressure $(C_{P})$ is the amount of heat required to raise the temperature of one gram through $1^{\circ} C$ when the pressure of the gas is kept constant. Again, the mean kinetic energy per molecule (3/2) $k T$ depends only upon temperature. Clearly both the specific heats at constant pressure and mean kinetic energy are depending on the temperature which is again same for the two gases.

PHXI13:KINETIC THEORY

360370 Two moles of an ideal gas with $\frac{C_{P}}{C_{V}} = \frac{5}{3}$ are mixed with 3 moles of another ideal gas with $\frac{C_{P}}{C_{V}} = \frac{4}{3}$ . The value of $\frac{C_{P}}{C_{V}}$ for the mixture is

1 1.50

2 1.42

3 1.47

4 1.45

Explanation:

$γ_{max} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}$
$C_{P} = (\frac{γ}{γ - 1}) R$
$C_{V} = \frac{R}{γ - 1}$
$γ_{m i x} = \frac{n_{1} \frac{γ_{1}}{γ_{1} - 1} + \frac{n_{2} γ_{2}}{γ_{2} - 1}}{\frac{n_{1}}{γ_{1} - 1} + \frac{n_{2}}{γ_{2} - 1}}$
$n_{1} = 2, γ_{1} = \frac{5}{3}, n_{2} = 3, γ_{2} = \frac{4}{3}$
We get
$γ_{m i x} = 1.42$

JEE - 2020

PHXI13:KINETIC THEORY

360371 The ratio of molar specific heats of oxygen is

1 1.4

2 1.67

3 1.33

4 1.28

Explanation:

Ratio of molar specific heats, $γ = \frac{C_{P}}{C_{V}}$ $γ = 1 + \frac{2}{f}$
$f =$ number of degrees of freedom
$f = 5$ for $O_{2}$
So $γ = 1 + \frac{2}{5} = 1.4$

KCET - 2024

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: