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A 22 liter container holds 9.00 grams of oxygen gas at 23 °C. What is the pressure in the container

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User Akay
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Answer:

2.15 atmospheres

Step-by-step explanation:

In order to determine the pressure of the oxygen gas within the container, one may utilize the ideal gas law equation:

PV=nRT

Where:

P represents the pressure (in atmospheres)

V represents the volume (in liters)

n represents the number of moles

R represents the ideal gas constant (0.0821 mol·K·L·atm)

T represents the temperature in Kelvin

Initially, it is necessary to calculate the number of moles of oxygen gas using the ideal gas law equation. The volume (V) is given as 22 liters, the temperature (T) is given as 23°C (or 23 + 273.15 K), and the number of moles (n) can be determined as follows:

n = (PV) / RT

Substituting the given values:

n = (22 L) (0.0821 mol·K·L·atm) / [(23 + 273.15) K]

Next, it is provided that the container contains 9.00 grams of oxygen gas. To determine the number of moles from the mass, one may use the molar mass of oxygen (O2), which is approximately 32 grams per mole. Thus:

n = 9.00 g / 32.00 g/mol

Having obtained the value of n, one may now solve for P:

P = (nRT) / V

Substituting the values:

P = [(9.00 g / 32.00 g/mol) (0.0821 mol·K·L·atm) (23 + 273.15 K)] / 22 L

Evaluating this expression yields the pressure (in atmospheres):

P ≈ 2.15 atm (rounded to two decimal places)

Therefore, the pressure of the oxygen gas within the container is approximately 2.15 atmospheres.

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User Samshers
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