Question

A monatomic 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 K).) A. 350 K B. 300 K C. 280 K D. 320 K

Answer 1

Step-by-step 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

Answer 2

For a monoatomic gas, the average molecular kinetic energy can be written as

`K= (1)/(2)mv^2 = (3)/(2) nRT`
where, in our problem,
`m=0.02 kg` is the mass,
`v=1321 m/s` is the average molecular speed,
`n=5 mol` is the number of moles and
`R=8.31 J/(mol K)`. re-arranging the equation, we can find the temperature T of the gas:

`T= (mv^2)/(3nR)= ((0.02 kg)(1321 m/s)^2)/(3 (5 mol)(8.31 J/(mol k)))=280 K`