Question

A person accidentally swallows a drop of liquid oxygen, \(\mathrm{O}_{2}(l)\) which has a density of 1.149 \(\mathrm{g} / \mathrm{mL}\) . Assuming the drop has a volume of 0.050 \(\mathrm{mL}\) , what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of 1.0 \(\mathrm{atm} ?\)

Short answer

The volume of gas produced in the person's stomach at body temperature (37°C) and a pressure of 1.0 atm is approximately 45.92 mL.

Step-by-step solution

Calculate the mass of liquid oxygen

To find the mass of liquid oxygen, multiply the given volume by its density: mass = (volume) × (density) mass = (0.050 mL) × (1.149 g/mL) mass ≈ 0.05745 g

Convert mass of liquid oxygen to moles

Now, we need to convert the mass of liquid oxygen into the number of moles using the molar mass of oxygen (O2) which is equal to 32 g/mol: moles = (mass) / (molar mass) moles = (0.05745 g) / (32 g/mol) moles ≈ 0.0018 mol

Use the ideal gas law to find the volume of gas

We are given the pressure (P) as 1 atm and the temperature (T) as 37°C which needs to be converted to Kelvin: T(K) = T(°C) + 273.15 T(K) = 37°C + 273.15 T(K) = 310.15 K Now, using the ideal gas law formula, PV = nRT, we can find the volume (V) of the gas: V = (nRT) / P Where R is the ideal gas constant which is equal to 0.0821 L⋅atm/mol⋅K V = (0.0018 mol × 0.0821 L⋅atm/mol⋅K × 310.15 K) / (1.0 atm) V ≈ 0.04592 L or 45.92 mL So, the volume of gas produced in the person's stomach at body temperature (37°C) and a pressure of 1.0 atm is approximately 45.92 mL.

Key concepts

Density

Density is a key concept in understanding how much mass a substance has in a given volume. A substance's density is calculated using the formula \[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]where mass is in grams (g) and volume is in milliliters (mL). In our exercise, the gas has a density of 1.149 g/mL, meaning each milliliter of liquid oxygen weighs 1.149 grams.
This value helps us calculate how much of the substance is present in the tiny volume swallowed. By knowing both the volume and density, we multiply these values to find the mass (0.05745 g). Understanding density is essential since it connects mass and volume, fundamental properties of substances in both solid and liquid states. - **Practical Example**: Think of comparing a rock to a sponge. A small rock may be heavy, while a larger sponge might be light. Their masses are distributed differently in their volumes, reflecting differing densities.

Molar Mass

Molar mass is crucial for converting mass into moles, especially for gases using the ideal gas law. Molar mass gives us the mass of one mole of molecules or atoms, measured in g/mol. For oxygen gas ( O_2 ), the molar mass is 32 g/mol. This value allows us to change grams of a substance into moles, a count of particles for chemical calculations.
By dividing the known mass (0.05745 g) of oxygen by its molar mass (32 g/mol), we convert mass to moles (approximately 0.0018 mol). Moles provide a way to work in the world of chemistry, allowing you to predict how substances behave in reactions and changes of state. - **Why It's Important**: Molar mass serves as a bridge between the macroscopic masses we measure and the microscopic molecules they comprise. This concept is foundational for any chemical calculations, ensuring accurate transformations and reactions.

Gas Volume Calculation

Calculating the volume of a gas involves understanding the ideal gas law, which connects pressure, volume, temperature, and moles of a gas. This law is represented by the equation:\[PV = nRT\]- **P** stands for pressure (in atmospheres, atm)- **V** is volume (in liters, L)- **n** represents the number of moles- **R** is the ideal gas constant (0.0821 L⋅atm/mol⋅K)- **T** is temperature (in Kelvin, K)In this example, we need to find the volume (**V**) of oxygen gas produced when liquid oxygen evaporates at body temperature (37°C, converted to 310.15 K) and standard pressure (1 atm). The equation rearranges to solve for volume: \[V = \frac{nRT}{P}\]Substituting the known values (n = 0.0018 mol, R = 0.0821 L⋅atm/mol⋅K, T = 310.15 K), the volume can be calculated as approximately 45.92 mL.
This calculation demonstrates how gases expand when transitioning from a liquid phase at a tiny volume to a gaseous state. It's a key concept in thermodynamics and chemistry, affecting how gases are used and understood in both laboratory and real-world applications.