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den301095 [7]
4 years ago
9

SELECT ALL THAT APPLY. Which of the following statements about Charles's law are true?

Chemistry
2 answers:
Serjik [45]4 years ago
7 0
A. Real gases may be expected to deviate from Charles's law at high pressures.

C. Real gases may be expected to deviate from Charles's law near the liquefaction temperature.
Ludmilka [50]4 years ago
3 0

Answer is:

A. Real gases may be expected to deviate from Charles's law at high pressures.

C. Real gases may be expected to deviate from Charles's law near the liquefaction temperature.

At high pressure gas molecules are close to one another and near the liqueffaction temperature energy is very low.

Charles' Law (The Temperature-Volume Law) - the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature:  

V₁/T₁ = V₂/T₂.  

When temperature goes down, the volume also goes down.


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Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

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⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

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