Carbon monoxide is a gas at 0 degree Celsius and a pressure of 1.51x10^5 Pa. It is a diatomic gas, each of its molecules consisting of one carbon atom (atomic mass = 12.o u) and one oxygen atom (atomic mass = 16.0 u). Assuming that carbon monoxide is an ideal gas, calculate its density p.
P V = n R T
find n in V= 1 meter^3, then we will find molar mass later
1.51*10^5 * 1 = n *8.31 (273+0)
n = 66.6 mols
now a mol of CO has mass 12+16 = 28 grams = .028 Kg
so we have
.028 * 66 = 1.85 kg /m^3
To calculate the density (p) of carbon monoxide gas, we can use the ideal gas equation:
PV = nRT
Where:
- P is the pressure of the gas (1.51x10^5 Pa)
- V is the volume of the gas (we will calculate this using the molar volume)
- n is the number of moles of gas (we will calculate this using the number of particles and Avogadro's number)
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature of the gas (0 degrees Celsius = 273.15 K)
First, let's calculate the number of moles (n) of carbon monoxide gas. We can use the formula:
n = m/M
Where:
- m is the mass of the gas
- M is the molar mass of the gas
The molar mass (M) of carbon monoxide is calculated by adding the atomic masses of carbon (12.0 u) and oxygen (16.0 u), resulting in a molar mass of 28.0 g/mol.
Using the ideal gas equation, we can rearrange it to solve for the volume (V):
V = (nRT)/P
Now, let's substitute the values into the equation:
n = m/M = m/28.0
V = (nRT)/P = (mRT)/(28.0P)
Finally, let's calculate the density (p), which is defined as mass divided by volume:
p = m/V
Let's substitute the value of V calculated above and solve for p.