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sweet [91]
3 years ago
9

6. 7. A hyperbaric chamber has a volume of 200. L. (a) How many moles of oxygen are needed to fill the chamber at room temperatu

re (23°C) and 3.00 atm pressure? b) How many grams of oxygen are needed? (Hint: Since moles have been asked, which equation has the moles listed in the equation. Use that to solve this problem. Also don’t forget to use the equation 1 mole = Formula weight or Molecular weight to calculate the grams of O2).
Chemistry
1 answer:
Otrada [13]3 years ago
6 0

Answer:

a) 24.7 mol

b) 790 g

Explanation:

Step 1: Given data

  • Volume of the chamber (V): 200. L
  • Room temperature (T): 23 °C
  • Pressure of the gas (P): 3.00 atm

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 23°C + 273.15 = 296 K

Step 3: Calculate the moles (n) of oxygen

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 3.00 atm × 200. L/(0.0821 atm.L/mol.K) × 296 K = 24.7 mol

Step 4: Calculate the mass (m) corresponding to 24.7 moles of oxygen

The molar mass (M) of oxygen ga sis 32.00 g/mol. We will calculate the mass of oxygen using the following expression.

m = n × M

m = 24.7 mol × 32.00 g/mol = 790 g

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Answer:

              Final Temperature = 36.54 ⁰C

Explanation:

Lets suppose the gas is acting ideally, then according to Charle's Law, "<em>The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature</em>". Mathematically for initial and final states the relation is as follow,

                                                V₁ / T₁  =  V₂ / T₂

Data Given;

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Solving equation for T₂,

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Putting values,

                         T₂  =  (35 L × 283.15 K) ÷ 32 L

                         T₂  =  309.69 K     ∴ ( 36.54 °C )

Result:

           As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.

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