Question

Определите плотность азота ρ при температуре t = 127 °С и давлении p = 0,05 МПа. Молярная масса азота равна Μ = 0,028 кг/моль. Ответ укажите в виде числа, в СИ (без единиц измерения)

Answer 1

По уравнению состояния:
$pV=\frac{m}{M}RT\\pM=\frac{m}{V}RT=\rho RT\\\rho=\frac{pM}{RT}=\frac{50000\times 0.028}{8.31\times(127+273)}=0,42$

Answer 2

Calculation of Nitrogen Density

To calculate the density of nitrogen (ρ) at a given temperature (t) and pressure (p), we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure - V is the volume - n is the number of moles - R is the ideal gas constant - T is the temperature in Kelvin

To convert the given temperature from Celsius to Kelvin, we use the formula:

T(K) = t(°C) + 273.15

Given: - t = 127 °C - p = 0.05 MPa - M = 0.028 kg/mol

Let's calculate the density of nitrogen using the ideal gas law equation.

First, we need to convert the pressure from MPa to Pascal (Pa):

p(Pa) = p(MPa) * 10^6

Substituting the given values:

p(Pa) = 0.05 MPa * 10^6 = 50,000 Pa

Next, we convert the temperature from Celsius to Kelvin:

T(K) = 127 °C + 273.15 = 400.15 K

Now, we can rearrange the ideal gas law equation to solve for density (ρ):

ρ = (p * M) / (R * T)

Where: - R is the ideal gas constant, which is approximately 8.314 J/(mol·K)

Substituting the values:

ρ = (50,000 Pa * 0.028 kg/mol) / (8.314 J/(mol·K) * 400.15 K)

Calculating the result:

ρ ≈ 5.98 kg/m³

Therefore, the density of nitrogen at a temperature of 127 °C and a pressure of 0.05 MPa is approximately 5.98 kg/m³.