The melting point of cadmium at 1 atm (101,325 N/m2) is 321°C (594 K) and its heat of fusion is 13.66 cal/g (57.15 xio J/kg). The volume change on the melting of cadmium is +0.0064 c.c./g (+0.0064 xl0~3 m3/kg). Calculate 2 the melting point of cadmium, (a) at 2 atm (202,650 N/m ) pressure, and (b) in vacuo.
Question
The melting point of cadmium at 1 atm (101,325 N/m2) is 321°C (594 K) and its heat of fusion is 13.66 cal/g (57.15 xio J/kg). The volume change on the melting of cadmium is +0.0064 c.c./g (+0.0064 xl0~3 m3/kg). Calculate 2 the melting point of cadmium, (a) at 2 atm (202,650 N/m ) pressure, and (b) in vacuo.
Solution
To solve this problem, we need to use the Clausius-Clapeyron equation, which describes the phase transition between two states of matter (in this case, solid and liquid). The equation is as follows:
ΔP/ΔT = ΔH/TΔV
where ΔP is the change in pressure, ΔT is the change in temperature, ΔH is the heat of fusion, T is the absolute temperature, and ΔV is the change in volume.
(a) To find the melting point of cadmium at 2 atm pressure, we first need to find the change in pressure (ΔP). Since we're going from 1 atm to 2 atm, ΔP = 2 atm - 1 atm = 1 atm = 101,325 N/m^2.
Next, we plug the given values into the Clausius-Clapeyron equation:
101,325 N/m^2 / ΔT = (57.15 x 10^6 J/kg) / (594 K * 0.0064 x 10^-3 m^3/kg)
Solving for ΔT gives us ΔT = 0.94 K. Since the melting point increases with pressure, the new melting point at 2 atm is 594 K + 0.94 K = 594.94 K, or 321.94°C.
(b) In vacuo means at zero pressure. To find the melting point of cadmium at 0 atm, we again use the Clausius-Clapeyron equation. This time, ΔP = 1 atm - 0 atm = 1 atm = 101,325 N/m^2.
Plugging the values into the equation gives us:
101,325 N/m^2 / ΔT = (57.15 x 10^6 J/kg) / (594 K * 0.0064 x 10^-3 m^3/kg)
Solving for ΔT gives us ΔT = 0.94 K. Since the melting point decreases with pressure, the new melting point at 0 atm is 594 K - 0.94 K = 593.06 K, or 319.91°C.
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