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At 27°C temperature, the kinetic energy of an ideal gas is $E_{1}$ . If the temperature is increased to 327°C, then kinetic energy would be

$2 E_{1}$

$\frac{1}{2} E_{1}$

$\sqrt{2} E_{1}$

$\frac{1}{\sqrt{2}} E_{1}$

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Explanation

1. As temperature changes from 27°C (300 K) to 327°C(600 K) i.e. doubled so kinetic energy also

doubled because E ∝ T.

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The average kinetic energy of a gas molecule at $27 ° C$ is $6.21 \times 10^{- 21} J .$ Its average kinetic energy at $227 ° C$ will be

$52.2 \times 10^{- 21} J$

$5.22 \times 10^{- 21} J$

$10.35 \times 10^{- 21} J$

$11.35 \times 10^{- 21} J$

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Explanation

$3 . E \propto T \Rightarrow \frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow \frac{6.21 \times 10^{- 21}}{E_{2}} = \frac{(273 + 27)}{(273 + 227)} = \frac{300}{500} \Rightarrow E_{2} = 10.35 \times 10^{- 21} J$

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The average translational kinetic energy of $O_{2}$ (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of $N_{2}$ (molar mass 28) molecules in eV at the same temperature is

0.0015

0.003

0.048

0.768

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Explanation

3. Average translation $K . E .$ at any temperature $T$ is given by ${(K . E .)}_{av} = \frac{3}{2} kT$ (k = Boltzmann's constant)

This is same for all gases at same temperature.

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The average translational energy and the r.m.s. speed of molecules in a sample of oxygen gas at 300 K are $6.21 \times 10^{- 21} J$ and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)

$12.42 \times 10^{21} J, 968 m / s$

$8.78 \times 10^{21} J, 684 m / s$

$6.21 \times 10^{21} J, 968 m / s$

$12.42 \times 10^{- 21} J, 684 m / s$

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Explanation

4. We know that average translation $K . E .$ of a molecules $= \frac{3}{2} kT$

At average $K . E . = 6.21 \times 10^{- 21} J$

At $600 K,$ average $K . E . = 2 \times (6.21 \times 10^{- 21})$

$= 12.42 \times 10^{- 21} J$

We know that $v_{rms} = \sqrt{\frac{3 kT}{m}}$

At $300 K, v_{rms} = 484 m / s$

At $600 K, v_{rms} = \sqrt{2} \times 484 = 684 m / s$

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At 0 K which of the following properties of a gas will be zero

Kinetic energy

Potential energy

Vibrational energy

Density

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Explanation

1. At $T = 0 K, v_{rms} = 0$

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A gas mixture consists of molecules of type 1, 2 and 3, with molar masses $m_{1} > m_{2} > m_{3} . V_{rms}$ and $\bar{K}$ are the r.m.s. speed and average kinetic energy of the gases. Which of the following is true

${(V_{rms})}_{1} < {(V_{rms})}_{2} < {(V_{rms})}_{3} and {(\bar{K})}_{1} = {(\bar{K})}_{2} = ({\bar{K}}_{3})$

${(V_{rms})}_{1} = {(V_{rms})}_{2} = {(V_{rms})}_{3} and {(\bar{K})}_{1} = {(\bar{K})}_{2} > {(\bar{K})}_{3}$

${(V_{rms})}_{1} > {(V_{rms})}_{2} > {(V_{rms})}_{3} and {(\bar{K})}_{1} < {(\bar{K})}_{2} > ({\bar{K}}_{3})$

${(V_{rms})}_{1} > {(V_{rms})}_{2} > {(V_{rms})}_{3} and {(\bar{K})}_{1} < {(\bar{K})}_{2} < {(\bar{K})}_{3}$

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Explanation

1. $V_{rms} \frac{1}{\sqrt{M}} \Rightarrow {(V_{rms})}_{1} < {(V_{rms})}_{2} < {(V_{rms})}_{3}$ also in mixture temperature of each gas will be same, hence

kinetic energy also remains same.

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Two ideal gases at absolute temperature $T_{1}$ and $T_{2}$ are mixed. There is no loss of energy. The masses of the molecules are $m_{1}$ and $m_{2}$ and the number of molecules in the gases are $n_{1}$ and $n_{2}$ respectively. The temperature of mixture will be

$\frac{T_{1} + T_{2}}{2} energy$

$\frac{T_{1} + T_{2}}{n_{1} n_{2}}$

$\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

$(T_{1} + T_{2})$

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Explanation

3. According to law of conservation of energy

$E = E_{1} + E_{2} \frac{f}{2} (n_{1} + n_{2}) kT = \frac{f}{2} n_{1} {kT}_{1} + \frac{f}{2} n_{2} {kT}_{2} \Rightarrow T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

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The molecules of an ideal gas at a certain temperature have

Only potential energy

Only kinetic energy

Potential and kinetic energy both

None of the above

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Explanation

2. There is no inter molecular force in ideal gas therefore P.E. = 0 and molecules have only kinetic energy.

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Mean kinetic energy per degree of freedom of gas molecules is

$\frac{3}{2} kT$

$kT$

$\frac{1}{2} kT$

$\frac{3}{2} RT$

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Explanation

3. According to law of equipartition of energy, kinetic energy per degree of freedom of a gas molecule is $\frac{1}{2} kT .$

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The temperature at which the average translational kinetic energy of a molecule is equal to the energy gained by an electron in accelerating from rest through a potential difference of 1 volt is

$4.6 \times 10^{3} K$

$11.6 \times 10^{3} K$

$23.2 \times 10^{3} K$

$7.7 \times 10^{3} K$

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Explanation

$4 . \frac{3}{2} kT = 1 eV \Rightarrow T = \frac{2}{3} \frac{eV}{k} = \frac{\frac{2}{3} \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23}} = 7.7 \times 10^{3} K$

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.