(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
The answer is b) the highest occupied orbital is a “d”orbital.
Transition metals are metals where the highest energy electrons partially fill the d subshells. There are some elements with complete d subshells but on forming cations they have incomplete d subshells.
These transition metals have some properties that are different from the other metals .
The answer is (4) solid to gas. The entropy is the measurement of disorder. The entropy of CO2 under different status will increase from solid to liquid to gas. So the answer is (4).
It's something to do with the weathering process.
Eg. At night it's really cold, and in the day it's really hot.
Because the temperature keeps changing dramatically day in and day out,
the sand dunes start to be sort of hacked or re-shaped, or even formed.
