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

the answer is a. at the equator

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.