Question

При температуре 546 С и давлении 6 атм воздух занимает объём 20 м3. Какова масса этого воздуха? Плотность воздуха при нормальных условиях 1,3 кг/м3

Answer 1

Ответ:

m ≈ 51,8 кг

Объяснение:

Дано:

$T =$ 819 K

$p =$ 607 950 Па

$V =$ 20 м³

$\rho_{n} =$ 1,3 кг/м³

$R =$ 8,31  Дж / (моль · К)

$M =$  0,029 кг/моль

Найти:

$m \ - \ ?$

----------------------------

Решение:

$pV = \dfrac{mRT}{M}$ - закон Менделеева - Клапейрона

$\boxed{m = \dfrac{MpV}{RT}}$

m = (0,029 кг/моль * 607 950 Па * 20 м³) / (8,31  Дж / (моль · К) * 819 K) ≈ 51,8 кг

Ответ: m ≈ 51,8 кг.

Answer 2

Calculating the Mass of Air at Given Conditions

To calculate the mass of the air at the given conditions, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15: - Temperature (T) = 546°C + 273.15 = 819.15 K - Pressure (P) = 6 atm - Volume (V) = 20 m^3

Using the ideal gas law, we can solve for the number of moles of air (n): ``` n = PV / RT ``` Where: - P = Pressure = 6 atm - V = Volume = 20 m^3 - R = Ideal gas constant (approximately 0.0821 atm·m^3/mol·K) - T = Temperature = 819.15 K

Now, let's calculate the number of moles of air using the given values.

Calculation

``` n = (6 atm) * (20 m^3) / (0.0821 atm·m^3/mol·K * 819.15 K) ``` ``` n ≈ 0.146 moles ```

Next, we can calculate the mass of the air using the molar mass of air and the number of moles.

The molar mass of air is approximately 29 g/mol.

Mass Calculation

``` Mass = Number of moles * Molar mass Mass = 0.146 moles * 29 g/mol ``` ``` Mass ≈ 4.234 g ```

Therefore, the mass of the air at the given conditions is approximately 4.234 grams.