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PHXI13:KINETIC THEORY

360290 The ratio of translational and rational kinetic energy at 100 $K$ temperature is 3:2. Find the internal energy of one mole gas at this temperature

1

2075 J

2

3120 J

3

5128 J

4

7126 J

Explanation:

we use $\frac{K_{T}}{K_{R}} = \frac{\frac{3}{2} n R T}{\frac{(f - 3)}{2} n R T} = \frac{3}{2} \Rightarrow \frac{3}{f - 3} = \frac{3}{2}$
$\Rightarrow 6 = 3 f - 9 \Rightarrow f = 5$
Now $U = \frac{f}{2} n R T = \frac{5}{2} \times 1 \times 8.3 \times 100 = 2075 J$

PHXI13:KINETIC THEORY

360291 The amount of heat energy required to raise the temperature of $8 g$ of Helium at $N T P$ from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{4} N_{a} k_{B} (\frac{T_{2}}{T_{1}})

2

\frac{3}{8} N_{a} k_{B} (T_{2} - T_{1})

3

\frac{3}{2} N_{a} k_{B} (T_{2} - T_{i})

4

\frac{3}{4} N_{a} k_{B} (T_{2} - T_{1})

Explanation:

Number of moles in $\lg H e = \frac{1}{4}$
Amount of heat energy required to raise its temperature from $T_{1} K$ to $T_{2} K$
$= μ C_{V} Δ T$
$(\frac{1}{4}) \times (\frac{3}{2} R) \times (T_{2} - T_{1})$
$= \frac{3}{8} k_{B} N_{A} (T_{2} - T_{1})$

PHXI13:KINETIC THEORY

360292 A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from $300 K$ to $600 K$ , the average K.E. of helium atoms is

1 Halved

2 Increased by factor

\sqrt{2}

3 Doubled

4 Unchanged

Explanation:

Kinetic energy $\propto$ Temperature. Hence if temperature is doubles, kinetic energy will also be doubled.

PHXI13:KINETIC THEORY

360293 Energy of 10 non rigid diatomic molecules at temperature $T$ is

1

35 R T

2

70 K_{B} T

3

\frac{7}{2} R T

4

35 K_{B} T

Explanation:

For non rigid diatomic molecule, degree of freedom is given by, $f = 7$
Energy of one molecule $= \frac{f}{2} K_{B} T$
Energy of 10 molecules
$= 10 \times \frac{f}{2} K_{B} T = 5 \times 7 K_{B} T = 35 K_{B} T$
So, correct option is (4).

JEE - 2024

PHXI13:KINETIC THEORY

360290 The ratio of translational and rational kinetic energy at 100 $K$ temperature is 3:2. Find the internal energy of one mole gas at this temperature

1

2075 J

2

3120 J

3

5128 J

4

7126 J

Explanation:

we use $\frac{K_{T}}{K_{R}} = \frac{\frac{3}{2} n R T}{\frac{(f - 3)}{2} n R T} = \frac{3}{2} \Rightarrow \frac{3}{f - 3} = \frac{3}{2}$
$\Rightarrow 6 = 3 f - 9 \Rightarrow f = 5$
Now $U = \frac{f}{2} n R T = \frac{5}{2} \times 1 \times 8.3 \times 100 = 2075 J$

PHXI13:KINETIC THEORY

360291 The amount of heat energy required to raise the temperature of $8 g$ of Helium at $N T P$ from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{4} N_{a} k_{B} (\frac{T_{2}}{T_{1}})

2

\frac{3}{8} N_{a} k_{B} (T_{2} - T_{1})

3

\frac{3}{2} N_{a} k_{B} (T_{2} - T_{i})

4

\frac{3}{4} N_{a} k_{B} (T_{2} - T_{1})

Explanation:

Number of moles in $\lg H e = \frac{1}{4}$
Amount of heat energy required to raise its temperature from $T_{1} K$ to $T_{2} K$
$= μ C_{V} Δ T$
$(\frac{1}{4}) \times (\frac{3}{2} R) \times (T_{2} - T_{1})$
$= \frac{3}{8} k_{B} N_{A} (T_{2} - T_{1})$

PHXI13:KINETIC THEORY

360292 A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from $300 K$ to $600 K$ , the average K.E. of helium atoms is

1 Halved

2 Increased by factor

\sqrt{2}

3 Doubled

4 Unchanged

Explanation:

Kinetic energy $\propto$ Temperature. Hence if temperature is doubles, kinetic energy will also be doubled.

PHXI13:KINETIC THEORY

360293 Energy of 10 non rigid diatomic molecules at temperature $T$ is

1

35 R T

2

70 K_{B} T

3

\frac{7}{2} R T

4

35 K_{B} T

Explanation:

For non rigid diatomic molecule, degree of freedom is given by, $f = 7$
Energy of one molecule $= \frac{f}{2} K_{B} T$
Energy of 10 molecules
$= 10 \times \frac{f}{2} K_{B} T = 5 \times 7 K_{B} T = 35 K_{B} T$
So, correct option is (4).

JEE - 2024

PHXI13:KINETIC THEORY

360290 The ratio of translational and rational kinetic energy at 100 $K$ temperature is 3:2. Find the internal energy of one mole gas at this temperature

1

2075 J

2

3120 J

3

5128 J

4

7126 J

Explanation:

we use $\frac{K_{T}}{K_{R}} = \frac{\frac{3}{2} n R T}{\frac{(f - 3)}{2} n R T} = \frac{3}{2} \Rightarrow \frac{3}{f - 3} = \frac{3}{2}$
$\Rightarrow 6 = 3 f - 9 \Rightarrow f = 5$
Now $U = \frac{f}{2} n R T = \frac{5}{2} \times 1 \times 8.3 \times 100 = 2075 J$

PHXI13:KINETIC THEORY

360291 The amount of heat energy required to raise the temperature of $8 g$ of Helium at $N T P$ from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{4} N_{a} k_{B} (\frac{T_{2}}{T_{1}})

2

\frac{3}{8} N_{a} k_{B} (T_{2} - T_{1})

3

\frac{3}{2} N_{a} k_{B} (T_{2} - T_{i})

4

\frac{3}{4} N_{a} k_{B} (T_{2} - T_{1})

Explanation:

Number of moles in $\lg H e = \frac{1}{4}$
Amount of heat energy required to raise its temperature from $T_{1} K$ to $T_{2} K$
$= μ C_{V} Δ T$
$(\frac{1}{4}) \times (\frac{3}{2} R) \times (T_{2} - T_{1})$
$= \frac{3}{8} k_{B} N_{A} (T_{2} - T_{1})$

PHXI13:KINETIC THEORY

360292 A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from $300 K$ to $600 K$ , the average K.E. of helium atoms is

1 Halved

2 Increased by factor

\sqrt{2}

3 Doubled

4 Unchanged

Explanation:

Kinetic energy $\propto$ Temperature. Hence if temperature is doubles, kinetic energy will also be doubled.

PHXI13:KINETIC THEORY

360293 Energy of 10 non rigid diatomic molecules at temperature $T$ is

1

35 R T

2

70 K_{B} T

3

\frac{7}{2} R T

4

35 K_{B} T

Explanation:

For non rigid diatomic molecule, degree of freedom is given by, $f = 7$
Energy of one molecule $= \frac{f}{2} K_{B} T$
Energy of 10 molecules
$= 10 \times \frac{f}{2} K_{B} T = 5 \times 7 K_{B} T = 35 K_{B} T$
So, correct option is (4).

JEE - 2024

PHXI13:KINETIC THEORY

360290 The ratio of translational and rational kinetic energy at 100 $K$ temperature is 3:2. Find the internal energy of one mole gas at this temperature

1

2075 J

2

3120 J

3

5128 J

4

7126 J

Explanation:

we use $\frac{K_{T}}{K_{R}} = \frac{\frac{3}{2} n R T}{\frac{(f - 3)}{2} n R T} = \frac{3}{2} \Rightarrow \frac{3}{f - 3} = \frac{3}{2}$
$\Rightarrow 6 = 3 f - 9 \Rightarrow f = 5$
Now $U = \frac{f}{2} n R T = \frac{5}{2} \times 1 \times 8.3 \times 100 = 2075 J$

PHXI13:KINETIC THEORY

360291 The amount of heat energy required to raise the temperature of $8 g$ of Helium at $N T P$ from $T_{1} K$ to $T_{2} K$ is

1

\frac{3}{4} N_{a} k_{B} (\frac{T_{2}}{T_{1}})

2

\frac{3}{8} N_{a} k_{B} (T_{2} - T_{1})

3

\frac{3}{2} N_{a} k_{B} (T_{2} - T_{i})

4

\frac{3}{4} N_{a} k_{B} (T_{2} - T_{1})

Explanation:

Number of moles in $\lg H e = \frac{1}{4}$
Amount of heat energy required to raise its temperature from $T_{1} K$ to $T_{2} K$
$= μ C_{V} Δ T$
$(\frac{1}{4}) \times (\frac{3}{2} R) \times (T_{2} - T_{1})$
$= \frac{3}{8} k_{B} N_{A} (T_{2} - T_{1})$

PHXI13:KINETIC THEORY

360292 A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from $300 K$ to $600 K$ , the average K.E. of helium atoms is

1 Halved

2 Increased by factor

\sqrt{2}

3 Doubled

4 Unchanged

Explanation:

Kinetic energy $\propto$ Temperature. Hence if temperature is doubles, kinetic energy will also be doubled.

PHXI13:KINETIC THEORY

360293 Energy of 10 non rigid diatomic molecules at temperature $T$ is

1

35 R T

2

70 K_{B} T

3

\frac{7}{2} R T

4

35 K_{B} T

Explanation:

For non rigid diatomic molecule, degree of freedom is given by, $f = 7$
Energy of one molecule $= \frac{f}{2} K_{B} T$
Energy of 10 molecules
$= 10 \times \frac{f}{2} K_{B} T = 5 \times 7 K_{B} T = 35 K_{B} T$
So, correct option is (4).

JEE - 2024