Problem 105 - Chemistry and Chemical Reactivity

Question

You have two gas-filled balloons, one containing He and the other containing \(\mathrm{H}_{2} .\) The \(\mathrm{H}_{2}\) balloon is twice the size of the He balloon. The pressure of gas in the \(\mathrm{H}_{2}\) balloon is \(1 \mathrm{atm},\) and that in the He balloon is 2 atm. The \(H_{2}\) balloon is outside in the snow \(\left(-5^{\circ} \mathrm{C}\right)\) and the He balloon is inside a warm building \(\left(23^{\circ} \mathrm{C}\right)\) (a) Which balloon contains the greater number of molecules? (b) Which balloon contains the greater mass of gas?

Step-by-Step Solution

Answer

(a) The H₂ balloon contains more molecules. (b) The He balloon contains more mass.

Step 1: Understand the Variables

We have two balloons, one filled with helium (He) and the other with hydrogen (H₂). The balloons have different sizes, pressures, and temperatures. For mathematical clarity, let the volume of the helium balloon be \(V\) and the volume of the hydrogen balloon be \(2V\). The pressure in the helium balloon is \(2 \text{ atm}\), while in the hydrogen balloon it is \(1 \text{ atm}\). The temperatures are different at \(23^{\circ} \text{C} = 296 \text{ K}\) for He and \(-5^{\circ} \text{C} = 268 \text{ K}\) for H₂.

Step 2: Use the Ideal Gas Law

To find the number of molecules (and therefore moles) in each balloon, we apply the ideal gas law \(PV = nRT\). First, solve for \(n\) (moles) in each case: - Helium balloon: \(n_{He} = \frac{P_{He}V}{RT_{He}} = \frac{2V}{R * 296}\) - Hydrogen balloon: \(n_{H2} = \frac{P_{H2} * 2V}{RT_{H2}} = \frac{2V}{R * 268}\).

Step 3: Compare Number of Moles

Simplify the expressions from Step 2: - \(n_{He}\) is proportional to \(\frac{2V}{296R}\) - \(n_{H2}\) is proportional to \(\frac{2V}{268R}\) Since \(\frac{2}{268} \gt \frac{2}{296}\), the hydrogen balloon contains more molecules.

Step 4: Calculate and Compare Mass of Gas

Mass is calculated by multiplying moles by molar mass: - He has a molar mass of 4 g/mol, so \(M_{He} = \frac{2V}{296R} \times 4\) - H₂ has a molar mass of 2 g/mol, so \(M_{H2} = \frac{2V}{268R} \times 2\) Since the factor for \(n_{H2}\) is greater, but each molecule of He is heavier, we need to compute: After straightforward calculation, helium still results in greater mass due to its higher molar mass per molecule.

Key Concepts

Molar Mass

When talking about gases like helium (He) and hydrogen ( H_2 ), it is important to understand molar mass. He has a molar mass of 4 g/mol, meaning each mole of helium weighs 4 grams. On the other hand, H_2 has a molar mass of 2 g/mol because each mole of hydrogen weighs 2 grams.

Molar mass is critical because it allows us to determine the weight of gas in a container if we know the number of moles. The molar mass links the weight of a substance to the number of molecules it contains. This information is useful for calculations related to the ideal gas law, where both the mass and the number of molecules can affect results.

By comparing He and H_2 using their molar masses, you can determine which gas will weigh more overall, even if one contains more molecules. This helps in understanding why helium, despite being in smaller quantities, may still weigh more than the hydrogen in the balloon exercise.

Number of Moles

The concept of moles is at the heart of chemistry, serving as the bridge between the atomic scale and the macroscopic world we interact with. To understand how many moles are present in a gas, we use the ideal gas law. This law gives us the formula PV = nRT, where P is pressure, V is volume, T is temperature in Kelvin, R is the gas constant, and n is the number of moles.
For example, in our helium and hydrogen balloon problem:

• The number of moles in the He balloon is given by \(n_{He} = \frac{P_{He}V}{RT_{He}}\).
• The H_2 balloon moles are \(n_{H2} = \frac{P_{H2} \times 2V}{RT_{H2}}\).

If we suss out the difference, hydrogen ends up with more moles. Why? Because after comparing the calculated values, the number of moles of H_2 surpasses that of He due to the larger volume of the hydrogen balloon and its lower temperature impacting the calculation outcome.

Gas Law Calculations

Gas law calculations allow us to understand the behavior and characteristics of gases under different conditions. The Ideal Gas Law is a key equation: PV = nRT . Here, P is pressure, V is volume, T is the temperature, R is the ideal gas constant, and n is the number of moles.

In applying this to real-world problems, like our two balloons:

• The helium balloon operates at a higher pressure but lower temperature compared to the hydrogen balloon.
• The volume term allows us to understand that the balloons, though differing in conditions, can be quantitatively analyzed for their gas contents.
• By breaking down each equation, we can directly compare moles, demonstrating that despite higher helium pressure, hydrogen has more moles due to its larger size and differing conditions.

Thus, gas law calculations not only reveal how gases will react under various environments but help predict the amount of substance within—essential for understanding mass and molecule count relationship seen in exercises like these.