Question

g the space in between two glass sheets has a volume of 0.025 m3. If the temperature of the gas is 290 K, what is the rms speed of the gas atoms

Answer 1

Complete Question

Argon gas (a monatomic gas) is sometimes used to insulate a double-pane window for purposes of insulation. In a particular window, the space in between two glass sheets has a volume of $0.025 \ m^3$. If the temperature of the gas is 290 K. what is the rms speed of the gas atoms? The atomic mass of Argon is 39.9 u.(l u= 1.66x10-27 kg) Enter your answer in m/s

Answer:

The value is $V_{rms} = 425.75 \ m/s$

Explanation:

From the question we are told that

     The volume of the space in between the two glass sheet is  $a = 0.025 \ m^3$

     The temperature of the gas is $T = 290 \ K$

       The atomic mass of Argon is  $m = 39.9 \ u = 39.9 * 1.66 *10^{-27} = 6.623 *10^{-26} \ kg$

Generally the root mean square velocity is mathematically represented as

      $V_{rms} = \sqrt{\frac{3 K T}{ m } }$

Here K is the Boltzmann constant  with value $k = 1.38 *10^{-23} \ m^3 \cdot kg \cdot s^{-2} \cdot K^{-1}$

So

     $V_{rms} = \sqrt{\frac{3 * 1.38 *10^{-23} * 290 }{ 6.6234*10^{-26} } }$

=>   $V_{rms} = 425.75 \ m/s$