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Darina [25.2K]
3 years ago
6

A rubber balloon was filled with helium at 25.0˚C and placed in a beaker of liquid nitrogen at -196.0˚C. The volume of the cold

helium was 15.6 cm3. Assuming ideal gas behavior and isobaric conditions, what was the volume of the helium at 25.0˚C?
Chemistry
1 answer:
Ksenya-84 [330]3 years ago
3 0

Answer:

The volume of helium at 25.0 °C is 60.3 cm³.

Explanation:

In order to work with ideal gases we need to consider absolute temperatures (Kelvin). To convert Celsius to Kelvin we use the following expression:

K = °C + 273.15

The initial and final temperatures are:

T₁ = 25.0 + 273.15 = 298.2 K

T₂ = -196.0 + 273.15 = 77.2 K

The volume at 77.2 K is V₂ = 15.6 cm³. To calculate V₁ in isobaric conditions we can use Charle's Law.

\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}} \\V_{1}=\frac{V_{2}}{T_{2}} \times T_{1}=\frac{15.6cm^{3} }{77.2K} \times 298.2K=60.3cm^{3}

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Explanation:

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In this case, we can discard the <u>propanone</u> because this molecule don't have the ability to form <u>hydrogen bonds</u>. (Let's remember that to have hydrogen bonds we need to have a hydrogen bond to a <u>heteroatom</u>, O, N, P or S).

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Answer:

P2≈393.609Kpa so I think the answer is 394 kPa

Explanation:

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