For krypton Tt= -169 °C, Pt = 133 mm Hg, Tc = -63 °C, Pc = 54 atm, mp = -156.6 °C, bp = -152.3 °C. The density of solid krypton is 2.8 g/cm3, and the density of the liquid is 2.4 g/cm3.
What is the physical phase of krypton under the following conditions:
5.3 atm, -153 °C and 65 atm, 250 K
Getting tested on this tomorrow so I'd appreciate some advice.
What is the physical phase of krypton under the following conditions:
5.3 atm, -153 °C and 65 atm, 250 K
Getting tested on this tomorrow so I'd appreciate some advice.
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You have four data points for Krypton, all of which are needed to solve the problem
- The triple point. At pressures lower than Pt, the liquid doesn't exist. Only the solid and gas phases exist.
- The critical point. At pressures higher than Pc, the liquid and gas phases are indistinguishable and become the supercritical fluid phase. You are missing Tc. You need that to solve part of the problem. Tc=-63.9 deg C for Krypton.
- The density of liquid Krypton is less than the density of solid Krypton. That means that as pressure increases, the melting temperature increases.
- Krypton m.p. = -156.6 deg C, and b.p. =-152.3 deg C. These are standard melting points and boiling points, which means they are valid at 1.00 atm.
You also know that in general
- At pressures between Pt and Pc, all three phases exist.
- At a given temperature, the boiling temperature increases as pressure increases.
At 5.3 atm and -153degC, the pressure is between Pt and Pc, so all three phases exist. -153C is below the boiling point, and since bp increases as pressure increases, you can be sure that Krypton is not a gas at these conditions. But since mp increases with pressure as well, you don't know if Krypton is a solid or a liquid. You can get an idea using an approximation to the Clapeyron equation:
ln (Tm) - ln (Tm, 1atm) = C (P - 1),
where Tm, 1atm is the std melting point, Tm is the melting point at another pressure, C is a constant, P is pressure in atm. You have melting points at two pressures: 133 mmHg (-169C, or 104K), which is 0.175 atm, and mp = -152.3C (or 120.7K) at 1 atm. Plug these in and solve for C. You will need to express temp in Kelvin
ln(104) - ln(120.7) = C (0.175 - 1)
4.644 - 4.793 = C (-0.825)
C = 0.181
Now use the approximation to the Clapeyron equation again, calculating Tm at p = 5.3 atm
ln (Tm) - ln (120.7) = 0.181 (5.3 - 1)
ln (Tm) = 0.778 + 4.793
Tm = 263K, which is -110C
So at 5.3 atm, the melting point is -110C. Your temp is well below that, so Krypton at 5.3 atm and -153C is a solid.
Now let's do the next one. 65 atm is above Pc, and 250K (-23C) is above Pc. So this one is easy. Both temp and pressure are above the critical T and P, so the phase is a supercritical fluid.
- The triple point. At pressures lower than Pt, the liquid doesn't exist. Only the solid and gas phases exist.
- The critical point. At pressures higher than Pc, the liquid and gas phases are indistinguishable and become the supercritical fluid phase. You are missing Tc. You need that to solve part of the problem. Tc=-63.9 deg C for Krypton.
- The density of liquid Krypton is less than the density of solid Krypton. That means that as pressure increases, the melting temperature increases.
- Krypton m.p. = -156.6 deg C, and b.p. =-152.3 deg C. These are standard melting points and boiling points, which means they are valid at 1.00 atm.
You also know that in general
- At pressures between Pt and Pc, all three phases exist.
- At a given temperature, the boiling temperature increases as pressure increases.
At 5.3 atm and -153degC, the pressure is between Pt and Pc, so all three phases exist. -153C is below the boiling point, and since bp increases as pressure increases, you can be sure that Krypton is not a gas at these conditions. But since mp increases with pressure as well, you don't know if Krypton is a solid or a liquid. You can get an idea using an approximation to the Clapeyron equation:
ln (Tm) - ln (Tm, 1atm) = C (P - 1),
where Tm, 1atm is the std melting point, Tm is the melting point at another pressure, C is a constant, P is pressure in atm. You have melting points at two pressures: 133 mmHg (-169C, or 104K), which is 0.175 atm, and mp = -152.3C (or 120.7K) at 1 atm. Plug these in and solve for C. You will need to express temp in Kelvin
ln(104) - ln(120.7) = C (0.175 - 1)
4.644 - 4.793 = C (-0.825)
C = 0.181
Now use the approximation to the Clapeyron equation again, calculating Tm at p = 5.3 atm
ln (Tm) - ln (120.7) = 0.181 (5.3 - 1)
ln (Tm) = 0.778 + 4.793
Tm = 263K, which is -110C
So at 5.3 atm, the melting point is -110C. Your temp is well below that, so Krypton at 5.3 atm and -153C is a solid.
Now let's do the next one. 65 atm is above Pc, and 250K (-23C) is above Pc. So this one is easy. Both temp and pressure are above the critical T and P, so the phase is a supercritical fluid.