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yuradex [85]
3 years ago
5

A plastic bottle is closed outdoors on a cold day when the temperature is −15.0°C and is later brought inside where the temperat

ure is 21.5°C. What is the pressure of the air in the bottle after it reaches room temperature, assuming the air in the bottle was at a pressure of 1.00 atm upon reaching thermal equilibrium with the outdoor temperature?
Chemistry
1 answer:
Lana71 [14]3 years ago
3 0

Answer:

Pressure = 1.14 atm

Explanation:

Hello,

This question requires us to calculate the final pressure of the bottle after thermal equilibrium.

This is a direct application of pressure law which states that in a fixed mass of gas, the pressure of a given gas is directly proportional to its temperature, provided that volume remains constant.

Mathematically, what this implies is

P = kT k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =........= Pn / Tn

P1 / T1 = P2 / T2

P1 = 1.0atm

T1 = -15°C = (-15 + 273.15)K = 258.15K

P2 = ?

T2 = 21.5°C = (21.5 + 273.15)K = 294.65K

P1 / T1 = P2 / T2

P2 = (P1 × T2) / T1

P2 = (1.0 × 294.65) / 258.15

P2 = 1.14atm

The pressure of the gas after attaining equilibrium is 1.14atm

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<span>Since,
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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
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<u>Chemistry</u>

<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.51 × 10²⁶ atoms Xe

[Solve] moles Xe

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 1.51 \cdot 10^{26} \ atoms \ Xe(\frac{1 \ mol \ Xe}{6.022 \cdot 10^{23} \ atoms \ Xe})
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<u>Step 4: Check</u>

<em>Follow sig fig rule and round. We are given 3 sig figs.</em>

250.747 mol Xe ≈ 251 mol Xe

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