What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas) at 308 K? Use the equation: -mv2=nRT

For m, use 0.02000 kg. Remember that R = 8.31 J/(mol-K).

O

A. 876 m/s

O

B. 87.6 m/s

O

C. 1540 m/s

O

D. 15,400 m/s

876m/s

Explanation:

Using the formula

1/2mv² = 3/2nRT

mv² = 3nRT

m is the mass = 0.02kg

v is the average velocity of atom

n is the mole = 2.00atm

T is the temperature = 308K

R is the Boltzmann constant = 8.31 J/(mol-K).

Get the velocity

v² = 3nRT/m

Substitute the given values into the formula

v² = 3×2(8.31)(308)/0.02

v² = 3×5,118.96/0.02

v² = 15356.88/0.02

V = √767844

v = 876.27m/s

Hence the average velocity of the atom us 876m/s

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the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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