Question

Boltzmann’s constant is 1.38066×10⁻²³ J/K, and the universal gas constant is 8.31451 J/K · mol. If 3 mol of a gas is confined to a 6.1 L vessel at a pressure of 7 atm, what is the average kinetic energy of a gas molecule? Answer in units of J.

Answer 1

`3.59* 10^(-21) J`

Step-by-step explanation:

We are given that

Boltzmann's constant,
`k_B=1.38066* 10^(-23) J/K`

Universal gas constant,R=8.31451 J/K

Number of moles,n=3

Volume ,V=6.1 L=
`6.1* 10^(-3)m^3`

Pressure,P=7 atm=
`7* 101325 Pa`

`PV=nRT`

`T=(PV)/(nR)=(7* 101325* 6.1* 10^(-3))/(3* 8.31451)`

T=173.45 K

Average kinetic energy=
`(3)/(2)k_BT`

Average kinetic energy=
`(3)/(2)(1.38066* 10^(-23)* 173.45`

Average kinetic energy=
`3.59* 10^(-21) J`