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Consider the transition between two forms of solid \(\operatorname{tin}, \operatorname{Sn}(s, \text { gray }) \rightarrow \operatorname{Sn}(s, \text { white }) .\) The two phases are in equilibrium at 1 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 \(350 .\) bar.

Short Answer

Expert verified
The equilibrium temperature between gray and white tin phases at a pressure of 350 bar is approximately -468.34°C.

Step by step solution

01

Understand the Clapeyron Equation

The Clapeyron Equation is given by: \[\frac{dP}{dT} = \frac{\Delta S}{\Delta V}\] Here, \(\Delta S\) represents the change in molar entropy, and \(\Delta V\) represents the change in molar volume. Also, \(dP\) represents the change in pressure, and \(dT\) represents the change in temperature. We will use this equation to find the equilibrium temperature at the new pressure (350 bar).
02

Calculate the change in molar entropy and molar volume

First, calculate the change in molar entropy. This is given by the difference between the molar entropies of gray and white tin: \[\Delta S = S_{white} - S_{gray} = 51.18 \ J/molK - 44.14 \ J/molK = 7.04 \ J/molK\] Next, we need to find the change in molar volume. Recall that the molar volume is given by the ratio of the molar mass to the density. The molar mass of tin is the same in both phases, so we can calculate the change in molar volume as: \[\Delta V = V_{m,white} - V_{m,gray} = \frac{M_{Sn}}{\rho_{white}} - \frac{M_{Sn}}{\rho_{gray}}\] Here, \(\rho_{white}\) and \(\rho_{gray}\) represent the densities of white and gray tin, respectively, and \(M_{Sn}\) is the molar mass of tin, which is approximately 118.71 g/mol.
03

Calculate the molar volumes and change in molar volume

Calculate the molar volume for each phase by dividing the molar mass by the respective densities, and then find the change in molar volume: \[V_{m,white} = \frac{118.71 \ g/mol}{7280 \ kg/m^3 \times 1000 \ g/kg} = 1.63 \times 10^{-5} \ m^3/mol\] \[V_{m,gray} = \frac{118.71 \ g/mol}{5750 \ kg/m^3 \times 1000 \ g/kg} = 2.06 \times 10^{-5} \ m^3/mol\] \[\Delta V = V_{m,white} - V_{m,gray} = -4.30 \times 10^{-6} \ m^3/mol\]
04

Use the Clapeyron Equation to find the equilibrium temperature

Now we can use the Clapeyron Equation to find the equilibrium temperature at the new pressure. Rearrange the equation as follows: \[dT = \frac{\Delta V}{\Delta S} dP\] Integrate the equation with respect to P from the initial pressure (1 bar) to the final pressure (350 bar), and with respect to T from the initial temperature (18°C) to the final temperature (Tf): \[\int_{T_0}^{T_f} dT = \int_{P_0}^{P_f} \frac{\Delta V}{\Delta S} dP\] Since \(\Delta V\) and \(\Delta S\) are constant: \[T_f - T_0 = \frac{\Delta V}{\Delta S} (P_f - P_0)\] Now we plug in the known values and solve for \(T_f\): \[T_f - 18 = \frac{-4.30 \times 10^{-6} \ m^3/mol}{7.04 \ J/molK} (350 \times 10^5 \ Pa - 1 \times 10^5 \ Pa)\] \[T_f = 18 + \frac{(-4.30 \times 10^{-6}) (349 \times 10^5)}{7.04} = 18 - 213.19 = -195.19\]
05

Convert the answer to the correct units and conclude

The answer we found is in Kelvin, and we should convert it to Celsius. The conversion formula is: \[T(°C) = T(K) - 273.15\] So the equilibrium temperature at 350 bar is: \[-195.19 \ K - 273.15 = -468.34 °C\] Therefore, the equilibrium temperature between gray and white tin phases at 350 bar is approximately -468.34°C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Transition
A phase transition involves a change from one state of matter to another, such as solid to liquid, liquid to gas, or even one solid form to another. In the context of tin, we observe a transition between gray tin and white tin.
In such a transition, it is crucial that the two forms coexist at equilibrium, meaning their chemical potentials are equal. This is vital because it defines the specific conditions under which one phase becomes more stable than the other. This is governed by thermodynamics, often involving temperature and pressure changes.
This transition of tin is reversible and depends heavily on thermodynamic variables. By studying it, one can understand how alterations in pressure and temperature favor the transformation from one solid phase to another. Understanding these phase changes provides crucial insights into material properties and their applications.
Molar Entropy
Molar entropy is a measure of the disorder or randomness at a molecular level within a given amount of substance, typically expressed in units of J/(K·mol).
Entropy, as a concept, provides a gauge of the dispersal of energy within a system. For tin, the molar entropies are different for gray and white phases, indicating variation in their molecular organizations.
  • Gray tin has a molar entropy of 44.14 J/(K·mol).
  • White tin has a molar entropy of 51.18 J/(K·mol).
  • The difference in molar entropies (\(\Delta S \)) results from the phase transition, signifying the change in disorder levels between these forms.
Entropy changes are instrumental in predicting the feasibility and direction of phase transitions. Higher entropy in white tin suggests that it has a greater degree of molecular freedom at equilibrium compared to gray tin.
Molar Volume
Molar volume indicates how much space a mole of a substance occupies and is calculated by dividing the molar mass by its density.Tin, transitioning from gray to white form, experiences a noticeable shift in molar volume due to the different densities of these phases.
For tin:
  • Gray tin has a lower molar volume because of its higher density.
  • White tin's higher molar volume stems from its slightly less dense structure, yielding 1.63 × 10-5 m3/mol compared to gray tin's 2.06 × 10-5 m3/mol.
The change in molar volume, \(\Delta V\), has direct implications in the Clapeyron Equation. It provides deeper insights into the compressibility and structural packing differences among crystal forms, reflecting how pressure might affect phase stability.
Density of Tin
Density is mass per unit volume and aids in determining how closely-packed atoms or molecules are within a material.
Tin exhibits different densities in gray and white forms, playing a pivotal role in understanding its physical properties. Here are the specifics for tin:
  • Gray tin, denser at 5750 kg/m3, indicates tightly packed atoms, presenting a more compact structural arrangement.
  • White tin, less dense at 7280 kg/m3, displays a looser structure, leading to a different set of material qualities.
The disparity in densities between the two phases affects their molar volumes and, in turn, the Clapeyron Equation's parameters used to predict phase transition behaviors. Recognizing how density shifts influence material attributes can drive innovations in material science and applications.

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Most popular questions from this chapter

Carbon tetrachloride melts at \(250 .\) K. The vapor pressure of the liquid is 10,539 Pa at \(290 .\) K and 74,518 Pa at 340. K. The vapor pressure of the solid is \(270 .\) Pa at \(232 \mathrm{K}\) and 1092 Pa at \(250 .\) K. a. Calculate \(\Delta H_{\text {vaporization}}\) and \(\Delta H_{\text {sublimation}}\). b. Calculate \(\Delta H_{\text {fusion}}\). c. Calculate the normal boiling point and \(\Delta S_{\text {vaporization}}\) at the boiling point. d. Calculate the triple point pressure and temperature.

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