JEE Main 2014 · 19 Apr · Shift 1 · Physics Q13

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PhysicsSingle correct+4 / -1Question 13 of 90

A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 0°C. If it were to go up straight without colliding with any other molecules, how high it would rise ? Assume that the height attained is much less than radius of the earth. ( $k_B$ is Boltzmann constant)

\dfrac{273k_B}{2Mg}

\dfrac{546k_B}{3Mg}

\dfrac{819k_B}{2Mg}

Answer and worked solution

Official answer: D

At 0°C the mean kinetic energy is $\frac32k_B(273)$ . Setting it equal to $Mgh$ :
$h=\frac{3\times273\,k_B}{2Mg}=\frac{819k_B}{2Mg}.$

Topics:Kinetic interpretation of temperature

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 13