Question

If \(12.0 \mathrm{g}\) of \(\mathrm{O}_{2}\) is required to inflate a balloon to a certain size at \(27^{\circ} \mathrm{C},\) what mass of \(\mathrm{O}_{2}\) is required to inflate it to the same size (and pressure) at \(5.0^{\circ} \mathrm{C} ?\)

Short answer

11.13 grams of \( \mathrm{O}_2 \) are required at 5.0°C.

Step-by-step solution

Understand the Given Information

We are given the initial conditions where 12.0 grams of \( \mathrm{O}_2 \) (oxygen) are needed to inflate a balloon at 27°C. We need to find out how much \( \mathrm{O}_2 \) (in grams) is needed to inflate the same balloon under the same pressure but at 5°C.

Apply the Ideal Gas Law Concept

We will use the concept of the Ideal Gas Law, which is \( PV = nRT \). Here \( P \), \( V \), and \( R \) remain constant for the problem, allowing us to relate temperature and moles. Since \( P \), \( V \), and \( R \) are constant, the relationship \( \frac{n_1}{T_1} = \frac{n_2}{T_2} \) applies, where \( n \) is the number of moles and \( T \) is temperature in Kelvin.

Convert Temperature to Kelvin

Convert the temperatures from Celsius to Kelvin: \[ T_{1} = 27 + 273.15 = 300.15 \, K \] and \[ T_{2} = 5 + 273.15 = 278.15 \, K \].

Calculate the Moles of Oxygen

Determine the moles of \( \mathrm{O}_2 \) initially: \[ n_1 = \frac{12.0 \, \text{g}}{32.00 \, \text{g/mol}} = 0.375 \, \text{mol} \] (32.00 g/mol is the molar mass of \( \mathrm{O}_2 \)).

Use the Proportional Relationship

Use the proportional relationship \( \frac{n_1}{T_1} = \frac{n_2}{T_2} \) to find the number of moles at the new temperature: \[ n_2 = \left( \frac{n_1 \times T_2}{T_1} \right) = 0.375 \, \text{mol} \times \frac{278.15}{300.15} \approx 0.3477 \, \text{mol} \].

Convert Moles Back to Grams

Calculate the required mass of \( \mathrm{O}_2 \) at the new temperature: \[ \text{mass} = n_2 \times 32.00 \, \text{g/mol} = 0.3477 \, \text{mol} \times 32.00 \, \text{g/mol} = 11.13 \, \text{g} \].

Conclusion

Thus, 11.13 grams of \( \mathrm{O}_2 \) are needed to inflate the balloon at 5.0°C if 12.0 grams are needed at 27.0°C.

Key concepts

Temperature conversion

When working with gas calculations, it's crucial to convert temperatures from Celsius to Kelvin, because the Kelvin scale is used in the Ideal Gas Law. The Kelvin scale starts at absolute zero, meaning it doesn’t dip into negative values, which makes it a natural fit for calculations involving gases. To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. This is shown in the solution process, where:

• 27°C was converted to 300.15 K,
• 5°C was converted to 278.15 K.

These conversions ensure that temperature scales align with the physical properties of gases, offering an absolute measure that is directly proportional to the internal energy of the gas.

Proportional relationships in gases

Gases exhibit some unique and predictable behaviors under changing conditions, which is where the Ideal Gas Law, stated as \(PV = nRT\), comes into play. In various scenarios, certain components of this equation may remain constant. For this exercise, the pressure \(P\), volume \(V\), and gas constant \(R\) remain unchanged. This simplifies the equation down to a focus on moles \(n\) and temperature \(T\).

• When pressure, volume, and \(R\) are constant, the number of moles \(n\) is inversely proportional to the temperature \(T\).
• Expressed mathematically, this is \(\frac{n_1}{T_1} = \frac{n_2}{T_2}\), allowing us to solve for unknowns when one condition changes, such as a temperature shift.

This principle helps us understand how gas quantities will adjust when only the temperature is varied, making it easier to calculate how much more or less gas is needed at different temperatures.

Molar mass calculation

Calculating molar mass is essential when you need to understand how much of a substance is present. Molar mass links the macroscopic amounts (like grams) to the number of moles (a microscopic concept). For oxygen gas (\(\text{O}_2\)), which consists of two oxygen atoms, each with an atomic mass of approximately 16 amu (atomic mass unit), you get:

• Molar mass of \(\text{O}_2 = 32.00 \text{ g/mol}\).

Using this molar mass, you can convert grams to moles and vice versa, which is a critical step in gas law calculations. For example:

• Given 12.0 g of \(\text{O}_2\), you calculate the moles by dividing by the molar mass: \(n_1 = \frac{12.0 \, \text{g}}{32.00 \, \text{g/mol}} = 0.375 \, \text{mol} \).
• To adjust for new conditions or calculate new mass, adjust the moles accordingly using the molar mass.

Understanding and applying molar mass conversions is required to ensure accurate calculations in chemistry, especially while dealing with gases.