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Sergeeva-Olga [200]
3 years ago
9

Why do real gases deviate from the ideal gas laws at low temperatures?

Chemistry
2 answers:
Debora [2.8K]3 years ago
8 0
As the molecules of gas interact, they are attracted to each other, following ideal laws, but they also collide with one another, and this makes them deviate from them. High pressures and low temperatures can also cause them to deviate, this is because the more pressure there is or the lower temperature there is, the closer together the molecules are, and the more they collide.

Hope this answered your question!
Yakvenalex [24]3 years ago
6 0

The answer is: The volume of a real gas is larger than expected from the ideal gas equation at low temperatures.

An ideal gas is:  

1) made up of molecules which are in constant random motion in straight lines.  

2) all collisions are perfectly elastic, there is no loss of kinetic energy during the collision.  

3) follows ideal gas law: p·V = n·R·T.  

4) the gas particles have negligible volume.  

At low temperatures, volume of the gas is not negligible.

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If the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high.

Spectrometry measures the interaction of light with molecules. The absorbance refers to how much light that interacts with molecules of the substance. The more the concentration of the substance the higher the absorbance of the solution.

Hence, if  the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high. An unusually high absorbance tells us that the solution is too concentrated.

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2 years ago
What exactly is a proton? Is it a member of the periodic table?
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Use the solubility curve above to answer the following: If I add 130 g of potassium iodide to 100 g of water at 10°C, how would
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7 0
3 years ago
Hess’s law
Delvig [45]

From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ

<h3>What is Hess' law?</h3>

Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.

From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:

A---> C = A ---> B + B ---> C

ΔH of A---> C = 30 kJ + 60 kJ

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Therefore, the enthalpy of the reaction A---> C is +90 kJ

The above reaction A---> C can be shown in the enthalpy diagram below:

A -------------------> C (ΔH = +90 kJ)

\ /

\ / (ΔH = +60 kJ)

(ΔH = +30 J) \ /

> B

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