Question

Calculate the root mean square velocity of oxygen molecules at 25c

Answer 1

V$_{rms}$ = $\sqrt{\frac{3RT}{M} }$

V$_{rms}$ =  $\sqrt{\frac{3*0.0821*298}{32} }$

V$_{rms}$ =  $\sqrt{\frac{73.4}{32} }$

V$_{rms}$ =√2.3

Answer 2

Answer:

The root mean square velocity of oxygen molecules at 25°C is 1.51m/s.

Explanation:

The root mean square velocity of gaseous molecules is calculated as,

$V_r_m_s=\sqrt{\frac{3RT}{M} }$        (1)

Where,

$V_r_m_s$=root mean square velocity

R=universal gas constant=0.0821 atm-L/mol-K

T=temperature in kelvin

M=molar mass of gaseous molecules

From the question, we are having,

The gas is oxygen so the molar mass (M)=32

The gas molecules are at temperature(T)=25°C+273K=298K

Now by solving equation (1) we get;

$V_r_m_s=\sqrt{\frac{3\times 0.0821\times 298}{32} }$

$V_r_m_s=1.51m/s$

Hence, the root mean square velocity of oxygen molecules at 25°C is 1.51m/s.

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