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Consider two 5 L chambers. In one, there are 5.00 g of \( \text{O}_2 \), and in the other, there are 5.00 g of He. Which has the higher pressure at room temperature?

Answer :

Final answer:

Using the Ideal Gas Law and the molar mass of each gas, we can determine that the 5 L chamber containing helium (He) will have a higher pressure at room temperature than the 5 L chamber containing oxygen (O₂), as it has a greater number of moles.

Explanation:

This question is asking which of two gases, O₂ and He, both in 5 L chambers and at room temperature, would exert a greater pressure. To answer this question, we should reference the Ideal Gas Law, which is PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

The number of moles of each gas can be calculated using the molar mass of each element. Oxygen (O₂) has a molar mass of 32 g/mol and helium (He) has a molar mass of 4 g/mol. So there are 5g / 32 g/mol = 0.156 mol of O₂ and 5g / 4 g/mol = 1.25 mol of He.

Because volume (V) and temperature (T) are the same for each gas and R is a constant, the gas with the greater number of moles (n) will exert the greater pressure. Therefore, the chamber containing helium (He) will have a higher pressure.

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Rewritten by : Barada

The gas, having a higher pressure among the two is the gas with greater number of moles which is He(Helium).

The pressure of gas is directly Proportional to the number of moles in the gas... So first of all counting the Number of moles in both the cases:

Number of Moles of He : 5g * [tex]\frac{1 Mole}{4.00 g}[/tex] = 1.25 moles

Number of moles of O2 : 5g * [tex]\frac{1 mole}{32 g}[/tex] = 0.156 moles

Since the number of moles of He is greater than the number of moles of O2, the He will exert more pressure on the container.

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