Answer
1.00 L
Work Step by Step
We find:
$\frac{dV}{V} =\beta dT$
$lnV = \int_{T_1}^{T_2}( aT + .5BT^2 + \frac{c}{3}T^3)dT$
$lnV = \int_{0}^{12}( aT + .5BT^2 + \frac{c}{3}T^3)dT$
$V=1.00L$
Thus, we find:
$ ln \frac{V_2}{V} = 12a + 77b + 576c$
Plugging in the values of a, b, c, and V (which was found above), we obtain:
$V_2=1.00L$
