## Related questions with answers

Consider the transition between two forms of solid tin, $\operatorname{Sn}(s$, gray $) \rightarrow \operatorname{Sn}(s$, white $)$. The two phases are in equilibrium at $1 \mathrm{bar}$ and $18^{\circ} \mathrm{C}$. The densities for gray and white tin are $5750$ and $7280 \mathrm{~kg} \mathrm{~m}^{-3}$, respectively, and the molar entropies for gray and white tin are $44.14$ and $51.18 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, respectively. Calculate the temperature at which the two phases are in equilibrium at $350 . \mathrm{bar}$.

Solution

Verified**We are given the following data:**

$\begin{aligned} &T_i= 18^{\circ} C\\ &=18+273.15=291.15\:\text{K}\\ \\ &P_i= 1\:\text{bar}=1\times10^5\:\text{Pa}\\ \\ &\rho_{Sn,gray}= 5750\:\frac{\text{kg}}{\text{m}^3}\\ \\ &\rho_{Sn,white}= 7280\:\frac{\text{kg}}{\text{m}^3}\\ \\ &S_{Sn,gray}= 44.14\:\frac{\text{J}}{\text{K.mol}}\\ \\ &S_{Sn,white}= 51.18\:\frac{\text{J}}{\text{K.mol}}\\ \\ &P_f=350\:\text{bar}= 350 \times 10^5\:\text{Pa}\\ \\ \end{aligned}$

**Required:**

$\cdot$ The temperature at which the two phases are in equilibrium at $350\:\text{bar}$.

## Create an account to view solutions

## Create an account to view solutions

## More related questions

1/4

1/7