MCQ Detailed View

The kinetic energy (KE) of gas molecules is directly related to their speed. According to the kinetic theory of gases, the average kinetic energy of a molecule is given by the formula:

KE=12mv2KE = \frac{1}{2}mv^2KE=21mv2

Where:

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  m = mass of the molecule

  
  


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  v = speed of the molecule

  
  

From this equation, we see that kinetic energy is proportional to the square of the speed of the particles. This means if the speed of gas molecules changes, the kinetic energy changes with the square of that factor.

In this question, the speed of the gas molecules is increased by twice (v → 2v) due to heating. Substituting this into the formula:

KE′=12m(2v)2=12m(4v2)=4×12mv2=4 KEKE' = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times \frac{1}{2}mv^2 = 4 \, KEKE′=21m(2v)2=21m(4v2)=4×21mv2=4KE

Thus, the kinetic energy increases by four times when the speed of gas molecules doubles.

The other options are incorrect:

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  Two times: This would only be correct if KE were directly proportional to speed, but it is proportional to the square of speed.

  
  


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  Decrease: Heating always increases molecular energy, not decreases it.

  
  

This principle is important in physical chemistry for understanding the relationship between temperature, molecular motion, and kinetic energy. As temperature increases, gas molecules move faster, causing their energy and pressure to rise. This relationship explains many gas laws and the behavior of gases under thermal changes.