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.

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

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml