Question

Смесь водорода и азота общей массой m = 290 г при температуре Т = 600 К и давлении p = 2,46 МПа занимает объем V = 30 л. Определить массу m1 водорода и массу m2 азота. Помогите решить плиззз!!!

Answer 1

Дано:
m=290г=0,29кг
T=600К
p=2,46МПа=2,46*106Па
V=30л=3*10-2м3
μ1=2*10-3кг/моль -(молярная масса водорода)
μ2=28*10-3кг/моль -(молярная масса азота)

Решение:Для смеси газов справедлив закон Дальтона:

(1) p=p1+p2

(где p1 и p2- порциальные давления водорода и азота соответственно)

из закона Менделеева-Клайперона можно вычислить давление:

(2) p*V=(m/μ)*R*T

Из (1) и (2), учитывая, что оба газа занимают объем V, получаем:

(3) p*V= (m1/μ1)*R*T + (m2/μ2)*R*T

p*V= ((m1/μ1) + (m2/μ2))*R*T

Так как весь газ весит m, то m1=m-m2, подставим это в (3):

P*V/ (R*T)= (μ2*m - μ2*m2 + μ1*m2)/(μ1*μ2)

m2=((p*V*μ1*μ2)/(R*T)-μ2*m)/(μ1-μ2)

m2=((2,46*106Па*3*10-2м3*2*10-3кг/моль *28*10-3кг/моль)/(8,31Дж/(моль*К)*600К) -28*10-3кг/моль *0,29 кг)/(2*10-3кг/моль - 28*10-3кг/моль)= 0,280 кг=280г

m1=m-m2=0,29кг-0,28кг=0,01кг=10г

Ответ: масса водорода m1=10г, масса азота m2=280г.

Answer 2

Calculation of the Mass of Hydrogen and Nitrogen in a Mixture

To determine the mass of hydrogen (m1) and the mass of nitrogen (m2) in a mixture of hydrogen and nitrogen, we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure of the gas (in this case, 2.46 MPa) - V is the volume of the gas (in this case, 30 L) - n is the number of moles of the gas - R is the ideal gas constant (8.314 J/(mol·K)) - T is the temperature of the gas (in this case, 600 K)

To find the number of moles of each gas, we can rearrange the equation as follows:

n = PV / RT

Let's calculate the number of moles of the mixture first:

n = (2.46 MPa * 30 L) / (8.314 J/(mol·K) * 600 K)

Using the appropriate unit conversions, we can convert the pressure from MPa to Pa and the volume from L to m^3:

n = (2.46 * 10^6 Pa * 0.03 m^3) / (8.314 J/(mol·K) * 600 K)

Now, we can calculate the number of moles of the mixture:

n = 0.0117 mol

Since the mixture is composed of hydrogen and nitrogen, we can assume that the number of moles of hydrogen (n1) plus the number of moles of nitrogen (n2) equals the total number of moles of the mixture:

n1 + n2 = 0.0117 mol

To determine the mass of hydrogen (m1) and the mass of nitrogen (m2), we need to know the molar masses of hydrogen (M1) and nitrogen (M2). The molar mass of hydrogen is approximately 1 g/mol, and the molar mass of nitrogen is approximately 28 g/mol.

Let's calculate the mass of hydrogen (m1):

m1 = n1 * M1 = n1 * 1 g/mol

And the mass of nitrogen (m2):

m2 = n2 * M2 = n2 * 28 g/mol

Substituting the values, we get:

m1 = 0.0117 mol * 1 g/mol

m2 = 0.0117 mol * 28 g/mol

Now, let's calculate the values:

m1 ≈ 0.0117 g

m2 ≈ 0.3276 g

Therefore, the mass of hydrogen (m1) in the mixture is approximately 0.0117 g, and the mass of nitrogen (m2) is approximately 0.3276 g.

Please note that these calculations are based on the ideal gas law and assume ideal gas behavior.