Question

При температуре 17C и давлении 2.3 атм плотность газа равна 3.8кг/м^3. Найти массу киломоля газа.

Answer 1

Не уверен в правильности, давно уже молекулярку не повторял 

Answer 2

Calculation of Molar Mass of Gas

To find the molar mass of a gas, we need to use the ideal gas law equation, which relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law equation is given as:

PV = nRT

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

In this case, we are given the temperature (17°C) and pressure (2.3 atm) of the gas, as well as the density (3.8 kg/m^3). However, we need to convert the temperature to Kelvin before proceeding with the calculation.

Conversion of Temperature to Kelvin

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

T(K) = T(°C) + 273.15

Given that the temperature is 17°C, we can calculate the temperature in Kelvin as follows:

T(K) = 17 + 273.15 = 290.15 K

Now that we have the temperature in Kelvin, we can proceed with the calculation.

Calculation of Molar Mass

To calculate the molar mass, we rearrange the ideal gas law equation to solve for the number of moles (n):

n = PV / RT

Substituting the given values into the equation:

n = (2.3 atm) * (3.8 kg/m^3) / (0.0821 atm·L/mol·K) * (290.15 K)

Simplifying the equation:

n = (2.3 * 3.8) / (0.0821 * 290.15) mol

Calculating the value:

n ≈ 0.036 mol

The molar mass of a gas is defined as the mass of one mole of the gas. To find the mass of one mole of the gas, we divide the mass of the gas by the number of moles:

Mass of one mole of gas = Mass of the gas / Number of moles

Given that the density of the gas is 3.8 kg/m^3, we can calculate the mass of the gas as follows:

Mass of the gas = Density * Volume

Since the volume is not given, we cannot calculate the mass of the gas or the molar mass without additional information.

Please provide the volume of the gas so that we can proceed with the calculation.