4. What is the rms speed of an nitrogen molecules contained in an `8.5`-`m^3` volume at `2.1\ atm` if the total amount of nitrogen is `1300 mol`?
4. What is the rms speed of an nitrogen molecules contained in an 8.5-m^3 volume at 2.1\ atm if the total amount of nitrogen is 1300 mol?
Hint
`PV = NkT` `v_(rms) = sqrt((3kT)/m)`
Answer
`v_(rms)= 3.9 times 10^2\ m//s`
Show Steps
Given: `V = 8.5\ m^3` `P = 2.1\ atm = 2.1 times 1.013 times 10^5\ Pa` `n = 1300\ mol` `N = 1300 times 6.02 times 10^23` molecules Known: `k = 1.38 times 10^(-23)\ J/K` `m_N_2 = 28\ u = 28 times 1.66 times 10^(-27)\ kg` Solution `PV = NkT => T = (PV)/(Nk)` `v_(rms) = sqrt((3kT)/m) = sqrt((3k((PV)/(Nk)))/m) = sqrt((3PV)/(Nm))` `v_(rms)=sqrt((3(2.1 times 1.013 times 10^5\ Pa)(8.5\ m^3))/((1300 times 6.02 times 10^23)(28 times 1.66 times 10^(-27)\ kg)))`
`v_(rms)= 3.9 times 10^2\ m//s`