Step-by-step explanation:
To calculate the pressure of our oxygen gas sample we will have to use the ideal gas law.
P * V = n * R * T
P = n * R * T/V
If we use the ideal gas constant R in atm*L/(mol*K), P will be the pressure in atm, n the number of moles, T the temperature in K and V the volume in L.
R = 0.082 atm*L/(mol*K)
So we will have to convert the volume in mL to L, the temperature in °C to K and the mass into moles.
T = (273.15 + 12) K
T = 285.15 K
1000 mL = 1 L
V = 140 mL * 1 L/(1000 mL)
V = 0.140 L
To convert the mass into moles we will have to convert from µg to grams and then from grams to moles using the molar mass of O₂.
1 g = 1 *10^6 µg
mass of O₂ = 2.1 µg * 1 g/( 1 *10^6 µg)
mass of O₂ = 2.1 * 10^(-6) g
atomic mass of O = 16.00 amu
molar mass of O₂ = 2 * 16.00 g/mol
molar mass of O₂ = 32.00 g/mol
moles of O₂ = 2.1 *10^(-6) g * 1 mol/(32 g)
moles of O₂ = 6.56 * 10^(-8) moles
n = 6.56 * 10^(-8) moles
Now that we know the different values we can replace them in the ideal gas formula and solve it for P.
P = n * R * T/V
P = 6.56 * 10^(-8) mol * 0.082 atm*L/(mol*K)* 285.15 K/(0.140 L)
P = 1.10 *10^(-5) atm
The pressure is 1.10 *10^(-5) atm but we need it in torr, so to answer our problem we have to convert the pressure from atm to torr.
760 torr = 1 atm
P = 1.10 *10^(-5) atm * 760 torr/(1 atm)
P = 0.00836 torr
Answer: the pressure is 0.00836 torr.