Question

Найдите плотность кислорода p при температуре 27 градусов Цельсия и давлении 0,31 МПа ?

Answer 1

Дано      P=0,31*10^6 Па   t=27C       p- ?

PV=m*R*T/M
m=p*V

P=p*R*T/M

p=P*M/R*T=0,31*10^6*32*10^-3/8,31*300=3,98 кг/м3

Answer 2

Calculating Oxygen Density at 27°C and 0.31 MPa

To calculate the density of oxygen (O2) at a temperature of 27 degrees Celsius and a pressure of 0.31 MPa, we can use the ideal gas law equation:

pV = nRT

Where: - p = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature in Kelvin

We need to rearrange the equation to solve for density (ρ), which can be expressed as:

ρ = (p * M) / (R * T)

Where: - ρ = density - p = pressure - M = molar mass of oxygen (32 g/mol) - R = ideal gas constant (8.314 J/(mol*K)) - T = temperature in Kelvin

First, we need to convert the given temperature of 27 degrees Celsius to Kelvin: T(K) = T(°C) + 273.15

Then we can use the ideal gas law to calculate the density of oxygen at the given conditions.

Calculation

Using the ideal gas law equation and the given values: - Temperature (T) = 27°C = 300.15 K - Pressure (p) = 0.31 MPa = 0.31 * 10^6 Pa - Molar mass of oxygen (M) = 32 g/mol - Ideal gas constant (R) = 8.314 J/(mol*K)

We can calculate the density (ρ) of oxygen at the given conditions: ρ = (p * M) / (R * T)

ρ = (0.31 * 10^6 * 32) / (8.314 * 300.15)

Result

The density of oxygen at 27°C and 0.31 MPa is approximately 4.178 kg/m³.

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