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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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An element has 3 electrons, 3 protons and 4 neutrons. What will be its atomic number and mass number? (​
stiv31 [10]

Answer:

its atomic number will be 3 and it's mass number will be 7. it is lithium

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2 years ago
Calculate the energy needed to raise the temperature of 42 grams of water from 25.0 degrees Celsius to 48 degrees Celsius. The s
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Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Lelechka [254]

Answer:

ΔG°rxn = -69.0 kJ

Explanation:

Let's consider the following thermochemical equation.

N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ

Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.

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8 0
3 years ago
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

Regards.

7 0
2 years ago
4.5x10'25 atoms of nickel equal how many moles
olga2289 [7]
1 mole of any particles = 6.02* 10²³ particles

4.5*10²⁵ atoms Ni* 1 mol Ni/6.02*10²³ Ni ≈ 74.75≈ 75 mol Ni
 
8 0
2 years ago
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