Question

Calculate the root mean square speed of oxygen molecules having kinetic energy of 8.368 kJ /mol. At what temperature would the molecules have this value of root mean square speed. The temperature is​

Answer 1

Given : kinetic energy of oxygen molecules is 8.368 kJ/mol.

To find : the root mean square speed of oxygen molecules and temperature at which kinetic energy of the gas is 8.368 kJ/mol.

solution : kinetic energy of gaseous molecule is given by, K.E = 3/2 nRT

⇒8.368 × 10³ J/mol = 3/2 × 1 × 8.314 J/mol/K × T

⇒T = (8368 × 2)/(3 × 8.314) = 670 K

Therefore the required temperature is 670K

now root mean square speed of oxygen gas is given by, v = √{3RT/M}

Here M is molecular weight i.e., M = 32 × 10¯³ kg/mol [ for oxygen gas ]

T = 670 K, R = 8.314 J/mol/K

so, v = √{3 × 8.314 × 670/32 × 10¯³}

= √{522.223 × 10³}

= √(522223)

= 722.65 ≈ 723 m/s

Therefore the root mean square speed of oxygen molecules is 723 m/s.

Answer 2

Answer:

Given : kinetic energy of oxygen molecules is 8.368 kJ/mol.

To find : the root mean square speed of oxygen molecules and temperature at which kinetic energy of the gas is 8.368 kJ/mol.

solution : kinetic energy of gaseous molecule is given by, K.E = 3/2 nRT

⇒8.368 × 10³ J/mol = 3/2 × 1 × 8.314 J/mol/K × T

⇒T = (8368 × 2)/(3 × 8.314) = 670 K

Therefore the required temperature is 670K

now root mean square speed of oxygen gas is given by, v = √{3RT/M}

Here M is molecular weight i.e., M = 32 × 10¯³ kg/mol [ for oxygen gas ]

T = 670 K, R = 8.314 J/mol/K

so, v = √{3 × 8.314 × 670/32 × 10¯³}

= √{522.223 × 10³}

= √(522223)

= 722.65 ≈ 723 m/s

Therefore the root mean square speed of oxygen molecules is 723 m/s.

Explanation: