Express the approximate density of carbon dioxide gas at 0 degrees C and 2.00 atm pressure in grams per liter.
I use a modified form of the gas equation.
P*molar mass = density(in g/L)*R*T
Substitute and solve for density.
To calculate the approximate density of carbon dioxide gas at 0°C and 2.00 atm pressure in grams per liter, we need to use the ideal gas law and the molar mass of carbon dioxide. Here's how you can do it:
1. Start by writing down the given values:
- Temperature (T) = 0°C = 273.15 K (convert Celsius to Kelvin)
- Pressure (P) = 2.00 atm (given)
2. Convert the pressure from atm to Pascals (Pa):
- 1 atm = 101,325 Pa, so multiply the given pressure by 101,325:
P = 2.00 atm × 101,325 Pa/atm = 202,650 Pa
3. The ideal gas law states: PV = nRT, where:
- P = pressure in Pa
- V = volume in m^3 (we'll convert it to liters later)
- n = number of moles of gas
- R = ideal gas constant = 8.314 J/(mol·K)
- T = temperature in Kelvin
4. Rearrange the ideal gas law to solve for the volume:
V = (nRT) / P
5. Find the molar mass of carbon dioxide (CO2):
- Carbon (C) has a molar mass of 12.01 g/mol
- Oxygen (O) has a molar mass of 16.00 g/mol
- CO2 consists of 1 carbon and 2 oxygen atoms, so the molar mass of CO2 is:
Molar mass (CO2) = 1(12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol
6. Convert the molar mass from grams/mol to kilograms/mol:
- 44.01 g/mol = 0.04401 kg/mol
7. Plug in the values into the rearranged ideal gas law, making sure to use consistent units:
V = (nRT) / P
V = (1 mol × 0.04401 kg/mol × 8.314 J/(mol·K) × 273.15 K) / 202,650 Pa
V = (0.9856 J) / 202,650 Pa
V = 4.863 × 10^-6 m^3 (cubic meters)
8. Convert the volume from m^3 to liters (1 m^3 = 1000 L):
V = 4.863 × 10^-6 m^3 × 1000 L/m^3 = 0.004863 L
9. Finally, calculate the density by dividing the molar mass by the volume:
Density = molar mass / volume
Density = 0.04401 kg/mol / 0.004863 L
Density ≈ 9.06 g/L
Therefore, the approximate density of carbon dioxide gas at 0 degrees C and 2.00 atm pressure is approximately 9.06 grams per liter.