Theheat ofthe reaction Sn (grey) = Sn (white) J/kg) is 500 cal/mole (2,092 J/mol) at the transition temperature of 13°C (286 K). The atomic weight of tin is 118.7 and the densities of grey and white tin are 5.75 g/c.c. (5.75 x IO3 kg/m3) and 7.28 g/c.c.(7.28 xlO3 kg/m3 ) respectively.Calculate the change in the transition temperature if the pressure on the grey tin phase is increased by 100 atm (101,325 x 10 N/m ) .

To solve this problem, we need to use the Clausius-Clapeyron equation, which describes the relationship between the pressure and temperature in phase transitions. The equation is:

ΔP/ΔT = ΔH / TΔV

where ΔP is the change in pressure, ΔT is the change in temperature, ΔH is the enthalpy change of the transition (which is equal to the heat of the reaction at constant pressure), T is the absolute temperature, and ΔV is the change in volume.

First, we need to calculate ΔV, the change in volume per mole. This can be found using the densities and the molar mass:

ΔV = V_white - V_grey = (M / ρ_white) - (M / ρ_grey) = (118.7 g/mol / 7.28 g/cm³) - (118.7 g/mol / 5.75 g/cm³) = 0.00263 cm³/mol

We need to convert this to m³/mol for consistency with the other units:

ΔV = 0.00263 cm³/mol * (1 m / 10² cm)³ = 2.63 x 10^-6 m³/mol

Now we can substitute the known values into the Clausius-Clapeyron equation:

ΔP/ΔT = ΔH / TΔV = (2,092 J/mol) / (286 K * 2.63 x 10^-6 m³/mol) = 2.79 x 10^8 Pa/K

Finally, we can find ΔT, the change in temperature, by rearranging the equation and substituting the given value for ΔP:

ΔT = ΔP / (ΔH / TΔV) = (100 atm * 101,325 Pa/atm) / (2.79 x 10^8 Pa/K) = 36.3 K

So, the transition temperature would increase by about 36.3°C if the pressure on the grey tin phase is increased by 100 atm.