Answer:
a) the initial mass of O₂ is 373.92 gr
b) the mass leaked of O₂ is 104.26 gr
Explanation:
we can assume ideal gas behaviour of oxygen , then we can calculate the mass using the ideal gas equation
P*V = n*R*T ,
where P= absolute pressure , V= volume occupied by the gas , n = number of moles of gas , R= ideal gas constant = 8.314 J/mol K , T= absolute temperature
Initially
P = Pg + Pa ( 1 atm) = 3.20 *10⁵ Pa + 101325 Pa = 4.21*10⁵ Pa
where Pg= gauge pressure , Pa=atmospheric pressure
T = 39 °C= 312 K
V= 7.20 * 10⁻² m³
therefore
P*V = n*R*T → n = P*V/ (R*T)
replacing values
n = P*V/ (R*T) = 4.21*10⁵ Pa*7.20 * 10⁻² m³/(8.314 J/mol K*312 K) = 11.685 mol
since
m= n*M , where m= mass , n= number of moles , M= molecular weight of oxygen
then
m = n*M = 11.685 mol *32.0 g/mol = 373.92 gr of O₂
therefore the initial mass of O₂ is 373.92 gr
for the part B)
P₂= Pg₂ + Pa ( 1 atm) = 1.85*10⁵ Pa + 101325 Pa = 2.86*10⁵ Pa
T₂ = 20.9 °C= 293.9 K
V= 7.20 * 10⁻² m³
therefore
n₂ = P₂*V/ (R*T₂) = 2.86*10⁵ Pa*7.20 * 10⁻² m³/(8.314 J/mol K*293.9 K) = 8.427 mol
m₂ = n*M = 8.427 mol*32.0 g/mol = 269.66 gr of O₂
thus the mass leaked of oxygen is
m leaked = m - m₂ = 373.92 gr - 269.66 gr = 104.26 gr
