Answer
95.13 K
Work Step by Step
According to equation 14.6 $KE_{Av}=\frac{3}{2}kT$ of each molecule in the gas. From this equation, we can get,
$$T=\frac{2(KE_{Av})}{3k}-(1)$$
Also, $KE_{Av}=\frac{KE_{total}}{N}-(2)$
(2)=>(1),
$T=\frac{2(KE_{total})}{3Nk}-(3)$
We can write,
$$KE_{total}=KE_{bullet}=\frac{1}{2}mv^{2}-(4)$$
(4)=>(3),
$$T=\frac{2(\frac{1}{2}mv^{2})}{3Nk}=\frac{2(\frac{1}{2}mv^{2})}{3(nN_{A})k}=\frac{(mv^{2})}{3(nN_{A})k}=\frac{(mv^{2})}{3(nN_{A})(\frac{R}{N_{A}})}$$
Let's plug known values into this equation.
$T=\frac{(8\times10^{-3}kg)(770\space m/s)^{2}}{3(8.31\space J/mol\space K)(2\space mol)}=95.13\space K$
The temperature of the gas = 95.13 K