Вопрос
2. A gas mixture under normal conditions has the following mass composition: CO_(2)=25.0% O_(2)=4.0% ;N_(2)=71.0% To what pressure must this mixture be compressed so that its density is rho =1.9kg/m^3
Решения
4.3
(282 Голоса)
Аделаида
мастер · Репетитор 5 лет
Ответ
To solve this problem, we need to use the ideal gas law equation:PV = nRTWhere:P = pressureV = volumen = number of molesR = universal gas constantT = temperatureGiven information:- Mass composition of the gas mixture: - CO2: 25.0% - O2: 4.0% - N2: 71.0%- Density of the gas mixture: ρ = 1.9 kg/m³Step 1: Calculate the molar masses of the individual gases.- Molar mass of CO2 = 44.01 g/mol- Molar mass of O2 = 32.00 g/mol- Molar mass of N2 = 28.02 g/molStep 2: Calculate the average molar mass of the gas mixture.Average molar mass = (0.25 × 44.01) + (0.04 × 32.00) + (0.71 × 28.02) = 36.86 g/molStep 3: Calculate the number of moles of the gas mixture per unit volume.n/V = ρ / (Molar mass of the gas mixture)n/V = 1.9 kg/m³ / 36.86 g/mol = 0.0516 mol/m³Step 4: Calculate the pressure of the gas mixture.P = (n/V)RTP = (0.0516 mol/m³) × (8.314 J/(mol·K)) × (298.15 K)P = 123.6 kPaTherefore, the pressure to which the gas mixture must be compressed so that its density is 1.9 kg/m³ is 123.6 kPa.