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Naddika [18.5K]
3 years ago
15

Goodyear tires are listed with 29.72 psi pressure limits in the summer, with 10.22 L of air. In the winter, however, the volume

decreases to 7.19 L because of the colder weather. What would the be the new pressure?
Chemistry
1 answer:
serg [7]3 years ago
6 0

Answer:

42.24 psi.

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 29.72 psi

Initial volume (V₁) = 10.22 L

Final volume (V₂) = 7.19 L

Final pressure (P₂) =?

The final pressure can be obtained by applying the Boyle's law equation as follow:

P₁V₁ = P₂V₂

29.72 × 10.22 = P₂ × 7.19

303.7384 = P₂ × 7.19

Divide both side by 7.19

P₂ = 303.7384 / 7.19

P₂ = 42.24 psi

Therefore, the final pressure is 42.24 psi

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The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond
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a)- The chemical equation for the corresponden equilibrium of Ka1 is:

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

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

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Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

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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

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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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