Question

What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas)

at 308 K? Use the equation: -mv2
2
For m, use 0.02000 kg. Remember that R = 8.31 J/(mol-K).
= ER
3
2
nRT
A. 1540 m/s
B. 876 m/s
C. 87.6 m/s
O D. 15,400 m/s

Answer 1

I think it is 887m/s I hope this helps if not I’m really sry

Answer 2

v = 876 m/s

Step-by-step 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,

`(1)/(2)mv^2=(3)/(2)nRT\\\\v=\sqrt{(3nRT)/(m)} \\\\v=\sqrt{(3* 2* 8.314* 308)/(0.02)} \\\\v=876.47\ m/s`

So, the correct option is (B).