Answer
1200 K, 4800 K, 2400 K, 9600 K
Work Step by Step
Here we use equation 14.6 $\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT$ to find the temperatures of the gas in each tank.
$$\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT=>T=\frac{mv_{rms}^{2}}{3k}$$
*For tank A :-
$T_{A}=\frac{mv_{rms}^{2}}{3k}$ ; Let's plug known values into this equation.
$T_{A}=\frac{(3.32\times10^{-26}kg)(1223\space m/s)^{2}}{3(1.38\times10^{-23}J/K)}=1200\space K$
*For tank B :-
$T_{B}=\frac{m(2v_{rms})^{2}}{3k}$ ; Let's plug known values into this equation.
$T_{B}=\frac{(3.32\times10^{-26}kg)(2\times1223\space m/s)^{2}}{3(1.38\times10^{-23}J/K)}=4800\space K$
*For tank C :-
$T_{C}=\frac{2mv_{rms}^{2}}{3k}$ ; Let's plug known values into this equation.
$T_{C}=\frac{2(3.32\times10^{-26}kg)(1223\space m/s)^{2}}{3(1.38\times10^{-23}J/K)}=2400\space K$
*For tank D :-
$T_{D}=\frac{2m(2v_{rms})^{2}}{3k}$ ; Let's plug known values into this equation.
$T_{D}=\frac{2(3.32\times10^{-26}kg)(2\times1223\space m/s)^{2}}{3(1.38\times10^{-23}J/K)}=9600\space K$
