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The ratio of mean kinetic energy of hydrogen and nitrogen at temperature 300 K and 450 K respectively is
$ E \alpha T $ $ \therefore { E_1 \over E_2 } = { T_1 \over T_2 } $
Pressure of an ideal gas is increased by keeping temperature constant what is the effect on kinetic energy of molecules.
Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constant whether pressure is increased or decreased.
A sealed container with negligible co-efficient of volumetric expansion contains helium (a monoatomic gas) when it is heated from 200 K to 600 K, the average K.E. of helium atom is
Kinetic energy is directly proportional to temperature. Hence if temperature is doubled, kinetic energy will also be doubled.
The mean kinetic energy of a gas at 300 K is 100J. mean energy of the gas at 450 K is equal to
$ Average kinetic energy \alpha Temperature$ $ \Rightarrow { E_1 \over E_2 } = { T_1 \over T_2 } $
At what temperature is the kinetic energy of a gas molecule double that of its value at $ 27 ^\circ C$
$ E \alpha T $ $ \therefore { E_1 \over E_2 } = { T_1 \over T_2 } $
The average kinetic energy of a gas molecule at $ 27 ^\circ C$ is $ 6.21 \times 10^{–21} J $ . Its average kinetic energy at $ 227 ^\circ C$ will be
$ E \alpha T $ $ \therefore { E_1 \over E_2 } = { T_1 \over T_2 } $
The average translational energy and rms speed of molecules in 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)
Average translational K.E. of a molecule is = $ { 3 \over 2 } k_B T $ At 300 K, 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 $ \nu_{rms} = \sqrt { 3k_B T \over m } $ At 300 K, $ \nu_{rms} = 484 ms^{–1} $ At 600 K, $ \nu_{rms } = \sqrt 2 \times 484 = 684 ms^{-1} $
The average translational kinetic energy of $O_2$ (molar mass 32) molecules at a particular temperature is 0.068 eV. The translational kinetic energy of $N_2$ (molar mass 28) molecules in eV at the same temperature is
Average translational K.E. of a molecule is = $ { 3 \over 2 } k_B T $ (Where, kB = Boltzmann's constant ) This is same, for all gases at same temperature.
At O K which of the follwing properties of a gas will be zero
At 0 K Kinetic energy is zero.
The kinetic energy of one mole gas at 300 K temperatue is E. At 400 K temperature kinetic energy E' . The value of E'/E is
$ E = {3 \over 2 } RT \Rightarrow E \alpha T \Rightarrow {E' \over E } = { T' \over T } $
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