How do snowflakes form?
2 answers:
Answer:
✍️A snowflake begins to form when an extremely cold water droplet freezes onto a pollen or dust particle in the sky. This creates an ice crystal. As the ice crystal falls to the ground, water vapor freezes onto the primary crystal, building new crystals – the six arms of the snowflake
✍️ All snowflakes contain six sides or points owing to the way in which they form. The molecules in ice crystals join to one another in a hexagonal structure, an arrangement which allows water molecules - each with one oxygen and two hydrogen atoms - to form together in the most efficient way.
✍️ Snowflakes are symmetrical because they reflect the internal order of the water molecules as they arrange themselves in the solid state (the process of crystallization). ... These ordered arrangements result in the basic symmetrical, hexagonal shape of the snowflake.
✍️ At just below freezing temperature (0 C) a snowflake might look like a tiny plate, while a few degrees colder sees snowflakes that are shaped like columns or needles. There are even triangular snowflakes, which are thought to be caused by dust particles mixed in with the crystals.
✍️ All snowflakes are unique, sort of. The short answer is, yes, because each ice crystal has a unique path to the ground. They will float through different clouds of different temperatures and different levels of moisture, which means the ice crystal will grow in a unique way.
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Answer:
(a). Check attachment.
(b). 280.305 J.
(c). 31.81 kpa; 38.26K.
(d). 24.05K.
(e). 24.05k; 40kpa.
(f). -138.6J.
Explanation:
(a). Kindly check the attached picture for the diagram showing the four process.
1 - 2 = adiabatic expansion process.
2 - 3 = Isochoric process.
3 - 4 = isothermal process.
4 - 1 = isochoric process.
(b). Recall that the process from 1 to is an adiabatic expansion process.
NB: b = 5/3 for a monoatomic gas.
Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].
= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.
Thus, the workdone = 280.305 J.
(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.
T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.
(d). The process 2 - 3 is an Isochoric process, then;
T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.
(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.
The pressure can be determine as below;
P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.
(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J
