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3 years ago

A 0.35 L balloon found at 19 C is heated in an oven to 250 C. What is the new volume of the balloon?

Chemistry

2 answers:

Ymorist [56]3 years ago

7 0

Answer:

0.63L

Explanation:

V1 = 0.35L

T1 = 19°C = (19 + 273.15)K = 292.15K

V2 = ?

T2 = 250°C = (250 + 273.15)K = 523.15k

Applying Charle's law, the volume of a given mass of gas is directly proportional to its pressure provided temperature remains constant

V1 / T1 = V2 / T2

V2 = (V1 * T2) / T1

V2 = (0.35 * 523.15) / 292.15

V2 = 0.626L

V2 = 0.63L

Svet_ta [14]3 years ago

5 0

Answer:

V = 0.63 L

Explanation:

To solve this problem, we need to use the Charle's law which is a law that involves temperature and volume, assuming we have a constant pressure. The problem do not state that the pressure is being altered, so we can safely assume that the pressure is constant (Maybe 1 atm).

Now, as the pressure is constant, the Charle's law is the following:

V₁ / T₁ = V₂ / T₂ (1) V is volume in Liter, and T is temperature in Kelvin.

Using this law with the given data, we solve for V₂:

V₂ = V₁T₂ / T₁

Before we use this expression, let's convert the temperatures to Kelvin:

T₁ = 19 + 273 = 292 K

T₂ = 250 + 273 = 523 K

Now, let's calculate the volume of the balloon:

V₂ = 0.35 * 523 / 292

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Answer:

Explanation:

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$W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)$

W = - 7.6 J

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ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

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