If the average translational kinetic energy of a molecule in a gas is equal to the kinetic energy of an electron accelerating from rest through 10 V, then the temperature of the gas molecule is (Bolt…

Question

If the average translational kinetic energy of a molecule in a gas is equal to the kinetic energy of an electron accelerating from rest through 10 V, then the temperature of the gas molecule is (Boltzmann constant =1.38 10^-23 JK^-1 )

Accepted Answer

Correct answer: D. (d) According to question, the translation KE of a molecule of gas = 32 kt According to the question, 32 kt= eV ⇒ 32 kt=10e ⇒T= 20e3K_B= 20 1.6 10^-193 1.38 10^-23=77.3 10^3 K

$If the average translational kinetic energy of a molecule in a gas is equal to the kinetic energy of an electron accelerating from rest through 10 V, then the temperature of the gas molecule is (Boltzmann constant = 1.38 \times 10^{- 23} JK^{- 1})$

Solution

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