Answer:
v = 876 m/s
Explanation:
It is given that,
Number of mol of Neon is 2 mol
Temperature, T = 308 K
Mass, m = 0.02 kg
Value of R - 8.31 J/mol-K
We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,
[tex]\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s[/tex]
So, the correct option is (B).