Answer
$\frac{v_2}{v_1}=1.0104 $
Work Step by Step
We take $\nu=v(P, T)$. Its total differential is $$
d v=\left(\frac{\partial v}{\partial T}\right)_P d T+\left(\frac{\partial v}{\partial P}\right)_T d P
$$ which, for a constant pressure process, reduces to $$
d v=\left(\frac{\partial v}{\partial T}\right)_P d T
$$ Dividing by $v$ and using the definition of $\beta$, $$
\frac{d v}{v}=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_P d T=\beta d T
$$ Taking $\beta$ to be a constant, integration from 1 to 2 yields $$
\ln \frac{v_2}{v_1}=\beta\left(T_2-T_1\right)
$$ or $$
\frac{v_2}{v_1}=\exp \left[\beta\left(T_2-T_1\right)\right]
$$ The average value of $\beta$ is $$
\beta_{\mathrm{me}}=\left(\beta_1+\beta_2\right) / 2=\left(49.2 \times 10^{-6}+54.2 \times 10^{-6}\right) / 2=51.7 \times 10^{-6} \mathrm{~K}^{-1}
$$ Substituting the given values, $$
\frac{v_2}{v_1}=\exp \left[\beta\left(T_2-T_1\right)\right]=\exp \left[\left(51.7 \times 10^{-6} \mathrm{~K}^{-1}\right)(500-300) \mathrm{K}\right]=1.0104
$$ Therefore, the volume of copper block will increase by 1.04 percent.
