Chemistry, 15.01.2021 19:40 MattPong

A chemist is studying the properties of a gas under various conditions. He observes that when the gas is at room temperature and low pressure, it behaves as an ideal gas. When the gas is cooled to 10 kelvin and is placed under high pressure, however, it deviates significantly from an ideal gas. Explain these observations

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Answer from: aaqa3735

An ideal gas is a theoretical concept and a real gas behaves in an ideal manner under conditions which includes a high temperature and a low pressure such that the gas has high kinetic energy, and the intermolecular forces between the molecules are weak

The Ideal Gas Law is P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles of the gas

R = The Universal Gas Constant

T = The temperature of the gas

For a gas cooled to 10 Kelvin and placed under high pressure, the interaction between individual gas molecules increases, and the kinetic energy of the gases is much lower and more comparable to the inter molecular forces between the gas molecules, which in turn produces observable changes from the initial ideal behavior of the gas such that the gas behavior deviates from the Ideal Gas Laws appreciably and are better modelled by the Van der Waals Equations which takes into account, the volume the gas particles occupy and the intermolecular forces between the molecules

The Van der Waals equation is presented as follows;

$P = \dfrac{R \cdot T}{V - b} - \dfrac{a}{V^2}$

Where;

V = The molar volume

a = The gas constant a represents the attractive forces between the gas particles

b = Represent the volume occupied by the particles of the gas

Explanation:

Answer from: Quest

yes

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Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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A chemist is studying the properties of a gas under various conditions. He observes that when the ga...