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HACTEHA [7]
2 years ago
9

A balloon is inflated to a volume of 1.25L .If the temperature of the air inside is cooled from 35°C to 15°C , what will be the

final volume of the balloon?
Chemistry
1 answer:
Eddi Din [679]2 years ago
6 0

Answer:

V₂ = 1.17 L

Explanation:

Given data:

Initial volume of balloon = 1.25 L

Initial temperature = 35°C

Final temperature = 15°C

Final volume = ?

Solution:

Initial temperature = 35°C (35+273.15K = 308.15 K)

Final temperature = 15°C (15+273.15 K = 288.15 K)

The given problem will be solve through the Charles Law.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1.25 L × 288.15 K / 308.15 k

V₂ = 360.2  L.K / 308.15 k

V₂ = 1.17 L

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Answer: The volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

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As number of moles is the mass of substance divided by its molar mass.

So, moles of O_{2} (molar mass = 32.0 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

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Substitute the values into above formula as follows.

PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L

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Explanation:

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