The vapour pressure of liquid zinc as a function of temperature is given as log P (mm Hg) = - Î é l l - 1.255 log T + 12.34. Calculate the heat of vaporization of zinc at its boiling point 907°C
Question
The vapour pressure of liquid zinc as a function of temperature is given as log P (mm Hg) = - Î é l l - 1.255 log T + 12.34. Calculate the heat of vaporization of zinc at its boiling point 907°C
Solution
To calculate the heat of vaporization of zinc at its boiling point, we can use the Clausius-Clapeyron equation, which is a way of characterizing the phase transition from liquid to gas. The equation is given as:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where:
- P1 and P2 are the pressures at the initial and final states,
- ΔHvap is the heat of vaporization,
- R is the ideal gas constant,
- T1 and T2 are the initial and final temperatures.
However, in this case, we are given a slightly different form of the equation:
log P = -ΔHvap/2.303R * (1/T) + C
Where:
- P is the pressure,
- ΔHvap is the heat of vaporization,
- R is the ideal gas constant,
- T is the temperature,
- C is a constant.
From the given equation, we can see that ΔHvap/2.303R = 1.255. We can solve for ΔHvap using the value of R = 8.314 J/(mol*K):
ΔHvap = 1.255 * 2.303 * R = 1.255 * 2.303 * 8.314 J/(mol*K) = 23.8 kJ/mol
This is the heat of vaporization of zinc at its boiling point of 907°C.
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