# A person accidentally swallows a drop of liquid oxygen, O2 (l), which has a density of 1.149 g/mL. Assuming the drop has a volume of 0.050 mL, what volume (mL) of gas will be produced in the person’s stomach at body temperature (37 °C) and a pressure of 1.0 atm?

## The answer should be 0.0457 L

## To determine the volume of gas produced in the person's stomach, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure (1.0 atm)

V = volume of the gas (unknown)

n = number of moles of gas (unknown)

R = ideal gas constant (0.0821 L·atm/(mol·K))

T = temperature in Kelvin (37°C + 273.15 = 310.15 K)

First, we need to find the number of moles of gas produced. We can use the molar mass of oxygen to convert the mass of the liquid oxygen to moles.

The molar mass of oxygen (O2) is:

2 * atomic mass of oxygen

2 * 16.00 g/mol = 32.00 g/mol

Now we can calculate the number of moles (n):

mass (g) = density (g/mL) * volume (mL)

mass (g) = 1.149 g/mL * 0.050 mL

mass (g) = 0.05745 g

n = mass (g) / molar mass (g/mol)

n = 0.05745 g / 32.00 g/mol

n = 0.0018 mol

Substituting the known values into the ideal gas law equation, we have:

(1.0 atm) * V = (0.0018 mol) * (0.0821 L·atm/(mol·K)) * (310.15 K)

Simplifying the equation, we find:

V = (0.0018 mol * 0.0821 L·atm/(mol·K) * 310.15 K) / 1.0 atm

V ≈ 0.044 L or 44 mL

Therefore, approximately 44 mL of gas will be produced in the person's stomach.

## To solve this problem, we can use the Ideal Gas Law formula, which is expressed as: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

To find the volume of gas produced, we need to determine the number of moles of O2(g) produced.

First, let's convert the given density of liquid oxygen to grams:

Density of liquid oxygen = 1.149 g/mL

Volume of the drop of liquid oxygen = 0.050 mL

Mass of the drop of liquid oxygen = Volume x Density = 0.050 mL x 1.149 g/mL

Now, let's calculate the number of moles of O2:

Molar mass of O2 = 32.00 g/mol

Number of moles of O2 = Mass / Molar mass = (0.050 mL x 1.149 g/mL) / 32.00 g/mol

Next, we need to convert the number of moles of O2 to volume at body temperature and 1.0 atm pressure.

We can use the Ideal Gas Law to calculate the volume:

PV = nRT

Rearranging the equation to solve for V:

V = (nRT) / P

Now, let's substitute the values into the equation:

V = (Number of moles of O2 x R x Temperature) / Pressure

Temperature needs to be in Kelvin, so let's convert 37 °C to Kelvin:

Temperature in Kelvin = 37 + 273.15

Finally, substitute the values into the equation to find the volume of gas produced:

V = [(Number of moles of O2) x (Ideal Gas Constant) x (Temperature in Kelvin)] / (Pressure)

Calculate the value to find the volume of gas produced in the person's stomach at body temperature and a pressure of 1.0 atm.

## mass = volume x density = 0.05 mL x 1.149 g/mL = ? g O2.

n = mols O2 = grams/molar mass = ?

Then PV = nRT. Remember T must be in kelvin. V will be in L so convert to mL. Post your work if you get stuck.