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Chapter 5: Problem 58

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

Expert verified
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

01

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
02

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
03

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.

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Most popular questions from this chapter

A sealed balloon is filled with 1.00 \(\mathrm{L}\) helium at $23^{\circ} \mathrm{C}$ and 1.00 atm . The balloon rises to a point in the atmosphere where the pressure is 220 . torr and the temperature is $-31^{\circ} \mathrm{C}$ . What is the change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220 . torr?

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