Вопрос
2. A gas mixture under normal conditions has the following mass composition: CO_(2)=32.0% O_(2)=8.0% ;N_(2)=60.0% To what pressure must this mixture be compressed so that its density is rho =1.8kg/m^3
Решения
4.1
(343 Голоса)
Ростислав
мастер · Репетитор 5 лет
Ответ
To solve this problem, we need to use the ideal gas law and the given mass composition of the gas mixture.Given information:- Mass composition of the gas mixture: CO2 = 32.0%, O2 = 8.0%, N2 = 60.0%- Density of the gas mixture: ρ = 1.8 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 = (0.32 × 44.01) + (0.08 × 32.00) + (0.60 × 28.02) = 35.21 g/molStep 3: Use the ideal gas law to find the pressure.PV = nRTP = (nRT) / VWhere:P = PressureV = Volumen = Number of molesR = Universal gas constant (8.314 J/(mol·K))T = Temperature (assumed to be 298 K, standard temperature)Step 4: Calculate the number of moles of the gas mixture.n = m / Mn = (1.8 kg/m³) / (35.21 g/mol) = 0.0511 mol/m³Step 5: Calculate the pressure using the ideal gas law.P = (nRT) / VP = (0.0511 mol/m³ × 8.314 J/(mol·K) × 298 K) / 1 m³P = 123.6 PaTherefore, the pressure to which the gas mixture must be compressed so that its density is 1.8 kg/m³ is 123.6 Pa.