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

The velocity of freon-12 is 207.0998 m/s

Step-by-step explanation:

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

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

`v=\sqrt{(3RT)/(m) }`

Where

R=8.31J/K mol

T=-65°C=208K

Replacing:

`v=\sqrt{(3*8.31*208)/(0.1209) } =207.0998m/s`

Answer 2

The root mean squared velocity for CF2Cl2 is
`v_(rms)= 207.06 m/s`

Step-by-step 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{(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{(3 * 8.314 *208)/(0.121) }`

`v_(rms)= 207.06 m/s`