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  • Please,please help me A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 00C. 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

 A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 00C. 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. (kB is Boltzmann constant)                    

 

  • Option 1)

    0

  • Option 2)

    \frac{273 \: k_{B}}{2Mg}

  • Option 3)

    \frac{546 \: k_{B}}{3Mg}

  • Option 4)

    \frac{819 \: k_{B}}{2Mg}

 

Answers (1)

best_answer

As we have learned

Kinetic energy of a gas due to translation per mole -

E= \frac{3}{2}PV

= \frac{3}{2}RT
 

- wherein

R = Universal gas constant

T = temperature of gas

 

 K.E at temprature T = 3/2 K_b T

T is in kelvin 

T = 273 K 

Kinetic energy = T = 3/2 K_b (273)

= 819 K_b / 2

Lets say this particle rises up to height h , then from energy conservativation  mgh = 819 K_b / 2

 

h = 819 K_B / 2mg

 

 

 

 


Option 1)

0

Option 2)

\frac{273 \: k_{B}}{2Mg}

Option 3)

\frac{546 \: k_{B}}{3Mg}

Option 4)

\frac{819 \: k_{B}}{2Mg}

Posted by

Avinash

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