Amonatomic gas is measured to have an average speed of 1321 m/s. if the total amount of the gas is 5 mol (which equates to a mass of 0.02 kg), what is the approximate temperature of the gas? (recall that the equation for kinetic energy due to translation in a gas is translational = mv2 = nrt, and r = 8.31 j/(mol a. 350 k b. 300 k c. 280 k d. 320 k

Explanation:

Given that,

Speed v = 1321 m/s

Total amount of gas

n = 5mol

Mass of gas

m = 0.02kg

Gas constant

R = 8.31 J/mol•K

Temperature T =?

For a monoatomic gas, the average kinetic energy can be determined using

½mv² = 3/2 × nRT

½ × 0.02 × 1321² = 3/2 × 5 × 8.31 × T

17450.41 = 62.325 T

Then, T = 17450.41 / 62.325

T = 279.991 K

Then,

T ≈ 280K

The correct option is C

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.