Question

Определите объём азота массой 64 г, находящегося в сосуде при атмосферном давлении и температуре 300 К. Найти плотность азота. Объём в ответе перевести в литры.

Answer 1

Ответ:

Объяснение:

Дано:

m = 64 г = 64·10⁻³ кг

M = 28·10⁻³ кг/моль

p = 1·10⁵ Па

T = 300 К

_______________

V - ?

ρ -?

По уравнению Клапейрона-Менделеева:

p·V = m·R·T / M

Объем:

V = m·R·T / (M·p)

V = 64·10⁻³·8,31·300 / (28·10⁻³·1·10⁵) ≈ 57·10⁻³ м³   или 57 л

Учтем, что плотность:

ρ = m / V

Тогда:

ρ =  0,064 / (57·10⁻³) ≈  1,12 кг/м³

Answer 2

Calculating the Volume of Nitrogen

To calculate the volume of nitrogen, 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

Given: - Mass of nitrogen (m) = 64 g - Atmospheric pressure (P) = 1 atm - Temperature (T) = 300 K - Molar mass of nitrogen (M) = 28.02 g/mol (from external source)

First, we need to find the number of moles of nitrogen using the formula:

n = m/M

Where: - m = mass of nitrogen - M = molar mass of nitrogen

Let's calculate the number of moles of nitrogen: n = 64 g / 28.02 g/mol

n ≈ 2.284 mol

Now, we can use the ideal gas law to find the volume of nitrogen: V = nRT / P

Substitute the values: V = (2.284 mol)(0.0821 L·atm/mol·K)(300 K) / 1 atm

V ≈ 55.3 L

So, the volume of nitrogen at atmospheric pressure and 300 K is approximately 55.3 liters.

Finding the Density of Nitrogen

The density of a gas can be calculated using the formula:

Density (ρ) = mass / volume

Given: - Mass of nitrogen = 64 g - Volume of nitrogen = 55.3 L

Let's calculate the density of nitrogen: ρ = 64 g / 55.3 L

ρ ≈ 1.16 g/L

So, the density of nitrogen is approximately 1.16 g/L.

I hope this helps! If you have further questions or need additional assistance, feel free to ask!