Question

Азот объемом 8,31 м3 при давлении 3∙105 Па и температуре 30˚С находится в сосуде. Определите его массу. Молярная масса азота 0,014 кг/моль.

Answer 1

Ответ:

По закону Менделеева-Клапейрона

Объяснение:

PV=(m/M)*R*T

p-давление-3*10^5

V- объём=8.31м^3

M-молярная масса= 0.014 кг/моль

R-универсальная газовая постоянная=8.31  Дж⁄(моль∙К)

T-абсолютная температура (в кельвинах) T=t+273=303K

Решение:   3*10^5*8.31=(m/0.014)*8.31*303

m~13.86 кг

Ответ 13.86 кг

Answer 2

Calculation of Nitrogen Mass

To determine the mass of nitrogen in the given volume, pressure, and temperature, we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure of the gas (in Pascal) - V is the volume of the gas (in cubic meters) - 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 Kelvin)

First, let's convert the given values to the appropriate units: - Volume (V): 8.31 m^3 - Pressure (P): 3 × 10^5 Pa - Temperature (T): 30°C = 303 K - Molar mass of nitrogen: 0.014 kg/mol

Now, we can rearrange the ideal gas law equation to solve for the number of moles (n):

n = PV / RT

Substituting the given values:

n = (3 × 10^5 Pa) × (8.31 m^3) / ((8.314 J/(mol·K)) × (303 K))

Simplifying the equation:

n = 9.99 mol

Finally, we can calculate the mass of nitrogen using the number of moles and the molar mass:

Mass = n × Molar mass

Mass = 9.99 mol × 0.014 kg/mol

Mass = 0.13986 kg

Therefore, the mass of nitrogen in the given volume, pressure, and temperature is approximately 0.13986 kg.

Please note that the calculation assumes ideal gas behavior and neglects any other gases or impurities that may be present in the given volume.