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26 November, 08:03

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?

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Answers (2)
  1. 26 November, 08:47
    0
    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

    V₂ = 0.63 L
  2. 26 November, 10:26
    0
    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
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