## College Physics (4th Edition)

We can rank the gases in order of total translational kinetic energy, from highest to lowest: $b \gt a = c = d \gt e = f$
The total translational kinetic energy is the number of molecules $N$ multiplied by the average translational kinetic energy of the molecules $\overline{KE}$. We can write an expression for the total translational kinetic energy: $N\times~\overline{KE} = \frac{3}{2}~NkT$ $KE = \frac{3}{2}~PV$ Let $P_0 = 50~kPa$ Let $V_0 = 2~L$ We can find an expression for the total translational kinetic energy in each case. (a) $KE= \frac{3}{2}~(2P_0)~(2V_0) = 4\times \frac{3}{2}~P_0~V_0$ (b) $KE= \frac{3}{2}~(4P_0)~(2V_0) = 8\times \frac{3}{2}~P_0~V_0$ (c) $KE= \frac{3}{2}~(P_0)~(4V_0) = 4\times \frac{3}{2}~P_0~V_0$ (d) $KE= \frac{3}{2}~(2P_0)~(2V_0) = 4\times \frac{3}{2}~P_0~V_0$ (e) $KE= \frac{3}{2}~(2P_0)~(V_0) = 2\times \frac{3}{2}~P_0~V_0$ (f) $KE= \frac{3}{2}~(P_0)~(2V_0) = 2\times \frac{3}{2}~P_0~V_0$ We can rank the gases in order of total translational kinetic energy, from highest to lowest: $b \gt a = c = d \gt e = f$