Answer
435.92 K
Work Step by Step
Here we use equations 14.7 $U=\frac{3}{2}nRT$ and 15.1 $Q=\Delta U+W$ to find the final temperature of the system. We can write,
$U_{I}=\frac{3}{2}nRT_{I}-(1)$ and $U_{F}=\frac{3}{2}nRT_{F}-(2)$
(2)-(1),
$$\Delta U=(U_{F}-U_{I})=\frac{3}{2}nR(T_{F}-T_{I})$$
$T_{F}=(\frac{2}{3nR})\Delta U+T_{I}-(3)$
We can write,
$Q=\Delta U+W=>\Delta U=Q-W-(4)$
(4)=>(3),
$$T_{F}=(\frac{2}{3nR})(Q-W)+T_{I}$$
Let's plug known values into this equation.
$T_{F}=\frac{2}{3(3\space mol)(8.31\space J/mol\space K)}\{+2438\space J-(-962\space J)\}+345\space K=435.92\space K$
Final temperature = 435.92 K
