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 atm, and that in the He balloon is 2 atm. The \(\mathrm{H}_{2}\) 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?

Short answer

(a) The \(\mathrm{H}_2\) balloon has more molecules. (b) The He balloon has greater mass.

Step-by-step solution

Understanding the Problem

We are given two balloons, one with hydrogen (\(\mathrm{H}_2\)) and one with helium (He). We need to find out which one has more molecules and which has more mass. The size, pressure, and temperature of each balloon are given.

Use Ideal Gas Law for Molecules

To find the number of molecules, we use the ideal gas law: \[ PV = nRT \]where \( n \) is the number of moles. We're given pressures \( P \), and temperatures \( T \), but volume \( V \) is implied by balloon size. The hydrogen balloon is twice the size of the helium balloon, so let the helium balloon volume be \( V \) and the hydrogen balloon volume be \( 2V \). Remember to convert the temperatures to Kelvin: - \(-5^{\circ}C = 268K\) (for \( \mathrm{H}_2 \)) - \(23^{\circ}C = 296K\) (for He).

Calculate Molecules in Helium Balloon

For helium: \[ P_1V = n_1RT_1 \]Substitute values: \[ 2 \times V = n_1 \times R \times 296 \]Thus, \[ n_1 = \frac{2V}{R \times 296} \]

Calculate Molecules in Hydrogen Balloon

For hydrogen: \[ P_2(2V) = n_2R(T_2) \]Substitute values: \[ 1 \times 2V = n_2 \times R \times 268 \]Simplify: \[ n_2 = \frac{2V}{R \times 268} \]

Compare Number of Molecules

To compare \( n_1 \) and \( n_2 \): Compare \[ \frac{2V}{R \times 296} \] with \[ \frac{2V}{R \times 268} \]. Since \( 268 < 296 \), \( n_2 > n_1 \). Thus, the hydrogen balloon contains more molecules.

Calculate Mass from Number of Moles

The mass of each gas is calculated using the number of moles and the respective molar masses:\[ m = n \times M \]- Molar mass of He = 4 g/mol, - Molar mass of \( \mathrm{H}_2 \) = 2 g/mol. Since we determined \( n_2 > n_1 \) in the \( \mathrm{H}_2 \) balloon, calculate their respective masses:- Mass of helium balloon: \(4 \times n_1\),- Mass of hydrogen balloon: \(2 \times n_2\).If \( n_2 > n_1 \), then \( 2 \times n_2 \) is likely smaller than \( 4 \times n_1 \), implying more mass in helium.

Final Step: Conclusion

(a) The \( \mathrm{H}_2 \) balloon contains the greater number of molecules.(b) The He balloon contains the greater mass of gas.

Key concepts

Ideal Gas Law

The ideal gas law is a fascinating tool that allows us to understand the behavior of gases under varying conditions. This law is represented as \( PV = nRT \), where:

• \( P \) is the pressure the gas exerts,
• \( V \) is the volume it occupies,
• \( n \) is the number of moles,
• \( R \) is the universal gas constant, and
• \( T \) is the temperature in Kelvin.

To apply this law correctly, we need to make sure each unit matches the standard units used in the equation. For instance, temperatures must be converted to Kelvin, and pressures typically in atmospheres or pascals.
In our balloon exercise, we have hydrogen and helium balloons with known pressures, volumes related to their size, and temperatures. By converting temperatures to Kelvin, we cater to the requirement of the ideal gas law, ensuring accurate comparisons are made in the number of molecules each balloon contains.

Molecular Count Comparison

Comparing the number of molecules in different gas samples relies directly on their mole count, derivable from the ideal gas law. Since the variable \( n \) denotes the number of moles and correlates directly with the number of molecules (as one mole contains approximately \( 6.022 \times 10^{23} \) molecules), this calculation becomes crucial.In our scenario, the hydrogen balloon exhibits twice the volume of the helium balloon. Applying the ideal gas law, as outlined above, we observe that for a fixed volume and a defined temperature in Kelvin, the pressure difference indeed affects the mole count.
Since the hydrogen balloon operates at a lower temperature, i.e., \(-5^{\circ}C\) or \(268K\), it ends up having more moles because the denominator, temperature \( T \), is smaller, thus increasing \( n \) when expressed as \( n = \frac{PV}{RT} \). This means the hydrogen (\( \mathrm{H}_2 \) balloon contains more molecules than the helium balloon.

Molar Mass Calculation

The molar mass calculation is essential when determining the mass from the number of moles. Molar mass (\( M \)) is the mass of one mole of a substance, often expressed in grams per mole (g/mol).For helium (He), the molar mass is \( 4 \) g/mol, whereas, for hydrogen (\( \mathrm{H}_2 \)), it is \( 2 \) g/mol.
In our exercise, even though hydrogen had more molecules (or moles), its molar mass is half that of helium. This relationship is crucial because the mass of the gas in each balloon is the product of its mole count and its molar mass (\( m = n \times M \)).Thus, because the helium balloon has fewer moles yet a higher molar mass than hydrogen, it ultimately possesses the greater mass of gas, emphasizing the significant difference between molecular count and actual mass.