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 O₂ gas produced in the person's stomach at body temperature (37°C) and a pressure of 1.0 atm will be 45.7 mL.

Step-by-step solution

Finding the mass of liquid oxygen drop

We are given the density of the liquid oxygen drop, which is 1.149 g/mL, and its volume, which is 0.050 mL. We will use these values to find the mass of this drop. Mass = Density × Volume.

Calculate the mass of the liquid oxygen

We plug the values we have into the formula from Step 1: \( Mass = (1.149 \frac{g}{mL}) \times (0.050 mL) \) By multiplying the values, we get: \( Mass = 0.05745 \ g \)

Finding the number of moles of O₂

Now that we have the mass of the liquid oxygen, we will find the number of moles of O₂. We know the molar mass of O₂ is 32.00 g/mol. We will use the formula: Moles of O₂ = (mass of O₂) / (molar mass of O₂)

Calculate the number of moles of O₂

Plugging in the values from Step 2 and the molar mass of O₂, we get: \( Moles \ of \ O_{2} = \frac{0.05745 \ g}{32.00 \frac{g}{mol}} \) By dividing the mass by molar mass, we get: \( Moles \ of \ O_{2} = 0.0018 \ mol \)

Find the volume of O₂ gas using the ideal gas law

We have the number of moles of O₂ gas and we are given the pressure (1.0 atm) and temperature (37°C or 310 K). The ideal gas law formula is: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm /(K mol)), and T is the temperature in Kelvin. We will solve for V (volume of gas).

Calculate the volume of O₂ gas

Plug in the values we have for P, n, R, and T, and solve for V: \( V = \frac{nRT}{P} \\ V = \frac{(0.0018 \ mol)(0.0821 \frac{L \cdot atm}{K \cdot mol})(310 \ K)}{1.0 \ atm} \) By multiplying and dividing the values, we get: \( V = 0.0457 \ L \)

Express the result in mL

We will convert the volume of O₂ gas from liters to milliliters by multiplying by 1000: Volume of O₂ gas = 0.0457 L × 1000 mL/L Volume of O₂ gas = 45.7 mL Thus, the volume of O₂ gas produced in the person's stomach at body temperature (37°C) and a pressure of 1.0 atm will be 45.7 mL.