At a temperature of 23.0 degrees celcius, and a pressure of 2.25 bar what volume would 23.0 g of a compound in the gaseous state with a molecular mass of 207 g/mol occupy?
PV = nRT
n = grams/molar mass = 23.0/207 = ?
Remember T must be in kelvin.
23.0/207?
To find the volume of a gaseous compound, we can use the ideal gas law equation: PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T represents temperature in Kelvin.
First, let's convert the temperature from degrees Celsius to Kelvin. The Kelvin temperature scale is related to Celsius by the equation: K = °C + 273.15.
Temperature in Kelvin = 23.0°C + 273.15 = 296.15 K
Next, we need to calculate the number of moles (n) of the compound using the given mass and molecular mass. The number of moles can be found using the equation: n = mass / molar mass.
n = 23.0 g / 207 g/mol = 0.111 mol (rounded to 3 decimal places)
Now, we can plug the values we have into the ideal gas law equation and solve for the volume (V):
PV = nRT
V = (nRT) / P
V = (0.111 mol * 0.0821 L·atm/mol·K * 296.15 K) / 2.25 bar
V = 2.87 L (rounded to 2 decimal places)
Therefore, 23.0 g of the compound in the gaseous state, with a molecular mass of 207 g/mol, would occupy approximately 2.87 liters of volume at a temperature of 23.0 degrees Celsius and a pressure of 2.25 bar.