We mix 119 grams of oxygen gas with 176 grams of argon in a volume of 520 mL at 116 Celsius. What will be the final pressure of the gas mixture? Answer in units of atm.
Well, that's quite a gas-tastic question! Before I provide you with a gas-timate, let's put on our lab coats and do some calculations.
First, we need to find the number of moles of each gas. To do that, we divide the mass of each gas by its molar mass. The molar mass of oxygen is approximately 32 g/mol, and the molar mass of argon is approximately 40 g/mol.
So, for the oxygen gas, we have 119 g / 32 g/mol ≈ 3.72 mol.
For the argon gas, we have 176 g / 40 g/mol ≈ 4.4 mol.
Now that we know the number of moles, we can use the ideal gas law to find the pressure. The ideal gas law is represented by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Remember, we need to convert Celsius to Kelvin by adding 273.15.
Now let's put the pieces together. We have:
P × 520 mL = (3.72 mol + 4.4 mol) × R × (116 Celsius + 273.15)
Solving for P, we get:
P = (8.12 mol × R × 389.15 K) / 520 mL
But hold your lab goggles, we need to make sure that the units match. The ideal gas constant, R, has a value of 0.08206 L·atm/mol·K. So let's substitute the values:
P = (8.12 mol × 0.08206 L·atm/mol·K × 389.15 K) / 520 mL
Now we need to convert mL to L by dividing by 1000:
P = (8.12 mol × 0.08206 L·atm/mol·K × 389.15 K) / 0.52 L
After some mathematical magic, we find:
P ≈ 9.18 atm
So there you have it! The final pressure of the gas mixture should be around 9.18 atmospheres. Remember, though, that this answer is as accurate as a gas station burrito, so please handle it with caution!
To find the final pressure of the gas mixture, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Step 1: Convert the temperature from Celsius to Kelvin.
To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature.
116 Celsius + 273.15 = 389.15 Kelvin
Step 2: Calculate the number of moles of oxygen gas (O2).
To find the number of moles, we use the formula: moles = mass/molar mass.
The molar mass of oxygen gas (O2) is 32 g/mol.
moles of O2 = 119 g / 32 g/mol = 3.72 mol
Step 3: Calculate the number of moles of argon (Ar).
The molar mass of argon (Ar) is 39.95 g/mol.
moles of Ar = 176 g / 39.95 g/mol = 4.41 mol
Step 4: Calculate the total number of moles of gas.
Total moles = moles of O2 + moles of Ar = 3.72 mol + 4.41 mol = 8.13 mol
Step 5: Convert the volume from milliliters (mL) to liters (L).
1 L = 1000 mL
Volume = 520 mL / 1000 mL/L = 0.52 L
Step 6: Convert the pressure to atm.
1 atm = 101.325 kPa (kilopascal)
We can assume that the gas is ideal, so the pressure unit doesn't matter.
Step 7: Substitute the values into the ideal gas law equation and solve for P.
PV = nRT
P * 0.52 L = 8.13 mol * 0.0821 atm L/mol K * 389.15 K
P = (8.13 mol * 0.0821 atm L/mol K * 389.15 K) / 0.52 L
P ≈ 12.35 atm
Therefore, the final pressure of the gas mixture will be approximately 12.35 atm.
To determine the final pressure of the gas mixture, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
To solve this problem, we need to follow a series of steps:
Step 1: Convert the given temperature from Celsius to Kelvin.
To convert Celsius to Kelvin, we use the equation:
Kelvin = Celsius + 273
Therefore, the temperature in Kelvin is:
116 Celsius + 273 = 389 Kelvin
Step 2: Convert the mass of each gas to moles.
To convert mass to moles, we divide the given mass by the molar mass of the substance. The molar mass of oxygen gas (O2) is 32 g/mol, and the molar mass of argon (Ar) is 40 g/mol.
For oxygen gas:
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 119 g / 32 g/mol
moles of oxygen = 3.719 moles
For argon:
moles of argon = mass of argon / molar mass of argon
moles of argon = 176 g / 40 g/mol
moles of argon = 4.4 moles
Step 3: Calculate the total number of moles of gas in the mixture.
Since the moles of oxygen and argon were given separately, we can sum them up to get the total moles of gas in the mixture.
Total moles of gas = moles of oxygen + moles of argon
Total moles of gas = 3.719 moles + 4.4 moles
Total moles of gas = 8.119 moles
Step 4: Convert the volume of the gas to liters.
Since the ideal gas law requires the volume to be in liters, we need to convert the given volume of 520 mL to liters.
1 liter = 1000 milliliters
Volume in liters = 520 mL / 1000 mL/L
Volume in liters = 0.52 L
Step 5: Plug the values into the ideal gas law equation and solve for pressure.
PV = nRT
Plugging in the values:
P * 0.52 L = 8.119 * 0.0821 * 389
Rearranging the equation to solve for pressure:
P = (8.119 * 0.0821 * 389) / 0.52
Evaluating the expression:
P = 242.66 atm
Therefore, the final pressure of the gas mixture is approximately 242.66 atm.
how many moles of gas do you have?
1 mole occupies 22.4L at STP
PV/T remains constant
so find P that works
mols O2 = 119/32 = ?
mols Ar = 176/39.95 = ?
Then PV = nRT
You solve for P.
V = 0.520 L
total n from above
R = 0.08206 L*atm/mol*K
T = 116 C converted to kelvin. K = 273 + C = ?
Post your work if you get stuck.