Calculate the mass of nitrogen present in a volume of 3000 cm^3 if the temperature of the gas is 22 degrees Celsius and the absolute pressure is 2.00 * 10^-13 atm, a partial vacuum easily obtained in laboratories. The molar mass of nitrogen (N2) is 28 g/mol. What is the density (in kg/m^3) of the N2

1 mole ocupies 22.4 liters at 273K and 1 atm. At your condtions, 1 mole will occupy (295/273)*1/(2*10^-13)= 5.40*10^12 times as much volume, or 1.21*10^14 liters.

You say you have 3 liters, so you have
3/1.21*10^14 = 2.48*10^-14 mole
Multiply that by the molar mass (28 g/mole) for the answer.

The answer provided by DrWLS is the grams. You are also asked for the density.

density = mass/volume = mass/3000 cc .
You will need to convert this to kg/m^3

To calculate the mass of nitrogen present in a volume of gas, we need to use the ideal gas law equation:

PV = nRT

Where:
P = absolute pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature of the gas in Kelvin

First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 22 + 273.15
T(K) = 295.15 K

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

Given:
P = 2.00 * 10^-13 atm
V = 3000 cm^3 = 0.003 m^3
R = 8.314 J/(mol·K)
T = 295.15 K

Substituting the values:
n = [(2.00 * 10^-13) * (0.003)] / (8.314 * 295.15)
n = 2.42 * 10^-15 mol

Next, we can calculate the mass of nitrogen (N2) using the molar mass:
Mass = molar mass × number of moles

Given:
Molar mass of nitrogen (N2) = 28 g/mol
Number of moles (n) = 2.42 * 10^-15 mol

Mass = 28 × (2.42 * 10^-15)
Mass = 6.776 * 10^-14 g

Finally, we can convert the mass from grams to kilograms and the volume from cm^3 to m^3 to calculate the density:

Density = mass / volume

Given:
Volume of nitrogen (N2) = 3000 cm^3 = 0.003 m^3
Mass of nitrogen (N2) = 6.776 * 10^-14 g

Density = (6.776 * 10^-14) / 0.003
Density = 2.2593 * 10^-11 kg/m^3

Therefore, the density of nitrogen (N2) in a volume of 3000 cm^3 is approximately 2.2593 * 10^-11 kg/m^3.

To calculate the mass of nitrogen present in a given volume, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant
T = temperature (in Kelvin)

First, let's convert the temperature from Celsius to Kelvin by adding 273.15:

Temperature (K) = 22 + 273.15 = 295.15 K

Next, we need to convert the given volume from cm^3 to liters:

Volume (L) = 3000 cm^3 ÷ 1000 = 3 L

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

n = PV / RT

Given that the pressure is 2.00 * 10^-13 atm and the molar gas constant, R, is 0.0821 L·atm/(K·mol), we can calculate:

n = (2.00 * 10^-13 atm) * (3 L) / (0.0821 L·atm/(K·mol) * 295.15 K) ≈ 2.22 * 10^-17 mol

Now we can calculate the mass of nitrogen by multiplying the number of moles by the molar mass:

Mass of nitrogen = (2.22 * 10^-17 mol) * (28 g/mol) ≈ 6.22 * 10^-16 g

To convert the mass to kg, divide by 1000:

Mass of nitrogen = 6.22 * 10^-16 g ÷ 1000 ≈ 6.22 * 10^-19 kg

Finally, to find the density of N2, divide the mass by the volume:

Density = (6.22 * 10^-19 kg) / (3 L) = 2.07 * 10^-19 kg/L

To convert the density to kg/m^3, multiply by 1000:

Density = (2.07 * 10^-19 kg/L) * 1000 = 2.07 * 10^-16 kg/m^3

Therefore, the density of nitrogen (N2) in the given conditions is approximately 2.07 * 10^-16 kg/m^3.