Question

Freon-12, CF2Cl2, which has been widely used in air conditioning systems, is considered a threat to the ozone layer in the stratosphere. Calculate the root-mean-square velocity of Freon-12 molecules in the lower stratosphere where the temperature is –65°C.

Answer 1

Answer:

The root mean squared velocity for CF2Cl2 is  $v_{rms}= 207.06 m/s$

Explanation:

From the question we are told that

         The temperature is $T = -65 ^oC = -65+273 = 208K$

Root Mean Square velocity is mathematically represented as

      $v _{rms} = \sqrt{\frac{3RT}{MW} }$

 Where  T is the temperature

              MW is the molecular weight of gas

              R is the gas constant with a value of  $R = 8.314 JK^{-1} mol^{-1}$

For  CF2Cl2 its molecular weight is  0.121 kg/mol

     Substituting values

  $v_{rms} = \sqrt{\frac{3 * 8.314 *208}{0.121} }$

          $v_{rms}= 207.06 m/s$

Answer 2

Answer:

The velocity of freon-12 is 207.0998 m/s

Explanation:

The molar mass of freon-12=120.91g/mol=0.1209kg/mol

The root-mean-square velocity of Freon-12 is:

$v=\sqrt{\frac{3RT}{m} }$

Where

R=8.31J/K mol

T=-65°C=208K

Replacing:

$v=\sqrt{\frac{3*8.31*208}{0.1209} } =207.0998m/s$