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valentina_108 [34]
3 years ago
7

If the pressure on a gas at -73°C is doubled but its volume is held constant, what will its final temperature be in degrees Cels

ius?
Chemistry
1 answer:
Gre4nikov [31]3 years ago
4 0

Answer:

127°C

Explanation:

This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.

We convert our value to K → -73°C + 273 = 200 K

The moles are the same, and the volume is also the same:

P₁ / T₁ = P₂ / T₂

But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂

P₁ / 200K = 2P₁ / T₂

1 /2OOK = (2P₁ / T₂) / P₁

See how's P₁ term is cancelled.

200K⁻¹ = 2/ T₂

T₂ = 2 / 200K⁻¹  → 400K

We convert the T° to C → 400 K - 273 = 127°C

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Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

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K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

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x = 0.09 mol

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