What volume of 31.5 grams of molecular element,chlorine gas, occupy at a pressure of 1.29 atm and temperature of 20 degrees cesius, assuming that the behavior of chlorine gas is ideal?
PV=nRT where n=31.5g/molmassCl2
To solve this problem, we need to use the ideal gas law equation, which is given as follows:
PV = nRT
Where:
- P is the pressure (in atm)
- V is the volume (in liters)
- n is the number of moles of gas
- R is the ideal gas constant (0.0821 L * atm / (K * mol))
- T is the temperature (in Kelvin)
First, we need to convert the given temperature from Celsius to Kelvin. This can be done by adding 273.15 to the Celsius temperature:
T = 20°C + 273.15 = 293.15 K
We are given the pressure (P = 1.29 atm) and the mass of the chlorine gas (31.5 grams). To find the number of moles (n), we need to use the molar mass of chlorine.
The molar mass of chlorine gas (Cl2) is the sum of the molar masses of two chlorine atoms (Cl):
Molar mass of Cl2 = (2 × atomic mass of Cl) = 2 × 35.45 g/mol = 70.9 g/mol
Now, we can calculate the number of moles (n) using the given mass:
n = mass / molar mass
n = 31.5 g / 70.9 g/mol ≈ 0.444 mol (rounded to three decimal places)
Now, we can substitute the values into the ideal gas law equation to solve for the volume (V):
PV = nRT
V = (nRT) / P
V = (0.444 mol × 0.0821 L * atm / (K * mol) × 293.15 K) / 1.29 atm
Calculating the result:
V ≈ 9.29 L
Therefore, the volume of 31.5 grams of chlorine gas at a pressure of 1.29 atm and a temperature of 20 degrees Celsius would be approximately 9.29 liters.