How do you solve this problem

A person accidentally swallows a drop of liquid oxygen

O2 (l), which has a density fo 1.149 g/ml. Assumming the drop has a

volumeof 0.035ml. What volume will be in the person's stomach at

bodytemperature (37 C) and a pressure of 1.0atm.

Answer: Volume of gas in the stomach, V = 0.0318L or 31.8mL

Explanation:

The number of moles of oxygen will remain constant even though the liquid oxygen will undergo a change of state to gaseous inside the person's stomach due to an increase in temperature.

Number of moles of oxygen gas = mass/molar mass

molar mass of oxygen gas = 32 g/mol

mass of oxygen gas = density * volume

mass of oxygen gas = 1.149 g/ml * 0.035 ml

mass of oxygen gas = 0.040215 g

Number of moles of oxygen gas = 0.0402 g / (32 g/mol)

Number of moles of oxygen gas = 0.00125 moles

Using the ideal gas equation, PV=nRT

where P = 1.0 atm, V = ?, n = 0.00125 moles, R = 0.082 L*atm/K*mol, T = (37 + 273) K = 310 K

V = nRT/P

V = (0.00125moles) * (0.082 L*atm/K*mol) * (310 K) / 1 atm

V = 0.0318L or 31.8mL