Final answer:
To determine the pressure in a 12.0-L cylinder with 40.6 g of oxygen gas at 327 K, the ideal gas law is used. After calculating the moles of O₂ and using the constant R with appropriate units, the pressure is found to be approximately 2.843 atm.
Step-by-step explanation:
The student is asking to calculate the pressure inside a 12.0-liter cylinder filled with 40.6 grams of oxygen gas at a temperature of 327 K. To solve this problem, we can use the ideal gas law equation PV = nRT, where P represents pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvins.
First, we need to convert the mass of oxygen gas into moles. The molar mass of O₂ is approximately 32.00 g/mol. So, the number of moles of oxygen (n) is:
n = Mass / Molar mass = 40.6 g / 32.00 g/mol = 1.269 mol (rounded to three significant figures)
The ideal gas constant (R) in units of L·atm/K·mol is approximately 0.0821. Now we can rearrange the ideal gas law to solve for pressure (P):
P = nRT / V
Thus, the pressure of the gas is:
P = (1.269 mol × 0.0821 L·atm/K·mol × 327 K) / 12.0 L
P = 2.843 atm (rounded to three significant figures)
Therefore, the pressure in the 12.0-L cylinder filled with 40.6 g of oxygen gas at 327 K is approximately 2.843 atm.