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What is the density of oxygen gas at 5.0 atm and 27°C?

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Final answer:

To find the density of oxygen gas at 5.0 atm and 27°C, we can use the Ideal Gas Law equation. The density is approximately 6.5 g/L.

Step-by-step explanation:

To find the density of oxygen gas at 5.0 atm and 27°C, we can use the Ideal Gas Law equation:

Density = (molar mass of oxygen)/(molar volume)

First, we need to calculate the molar volume of oxygen at the given conditions. The molar volume (V) can be found using the equation:

V = (R*T)/(P)

Where R is the ideal gas constant (0.08206 L·atm/mol·K), T is the temperature in Kelvin (27+273 = 300 K), and P is the pressure in atm (5.0 atm). Substitute the values into the equation: V = (0.08206 L·atm/mol·K * 300 K)/(5.0 atm) = 4.9236 L/mol.

Next, we need to calculate the molar mass of oxygen, which is 32 g/mol.

Finally, substitute the molar mass and molar volume into the density equation:

Density = 32 g/mol / 4.9236 L/mol ≈ 6.5 g/L

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