Answer
2.098
Work Step by Step
Here we use equation 14.6 $KE=\frac{3}{2}kT$ to find the ratio of average kinetic energies.
$\frac{KE_{krypton}}{KE_{argon}}=\frac{\frac{3}{2}kT_{krypton}}{\frac{3}{2}kT_{argon}}=\frac{T_{krypton}}{T_{argon}}-(1)$
According to the ideal gas law, we can write,
$PV=nRT=>T=\frac{PV}{nR}-(2)$
(2)=>(1),
$\frac{KE_{krypton}}{KE_{argon}}=\frac{\frac{PV}{n_{krypton}R}}{\frac{PV}{n_{argon}R}}=\frac{n_{argon}}{n_{krypton}}=\frac{\frac{m}{M_{argon}}}{\frac{m}{M_{krypton}}}$
$\frac{KE_{krypton}}{KE_{argon}}=\frac{M_{krypton}}{M_{argon}}$
Let's plug known values into this equation.
$\frac{KE_{krypton}}{KE_{argon}}=\frac{83.80\space g/mol}{39.948\space g/mol}=2.098$
