What pressure would a mixture of 3.2 grams of O2, 6.4 grams CH4, and 6.4 grams of SO2 exert if the gases were placed in a 4.1 liter container at 127 degrees C?
Answer in units of atm
Well, this question sounds like a real gas party! Now, to find the pressure exerted by this peculiar mixture of gases, we'll have to use the ideal gas law. Are you ready to join me in this gas-filled adventure? Let's go!
First, let me tell you a little secret formula called the ideal gas law. It goes like this: PV = nRT. Here, 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. Don't worry, I'll convert that Celsius temperature to Kelvin for you.
Now, let's calculate the number of moles for each gas. To do that, we need their respective molecular weights (MW). Oxygen (O2) has a MW of about 32 g/mol, methane (CH4) has a MW of about 16 g/mol, and sulfur dioxide (SO2) has a MW of about 64 g/mol.
For O2: 3.2 g / 32 g/mol = 0.1 mol
For CH4: 6.4 g / 16 g/mol = 0.4 mol
For SO2: 6.4 g / 64 g/mol = 0.1 mol
Now, let's add up the number of moles:
Total moles = 0.1 mol + 0.4 mol + 0.1 mol = 0.6 mol
Next, let's convert the temperature from Celsius to Kelvin. Adding 273 to 127°C, we get 400 K (just like a 400-degree comedy show!).
Finally, we'll plug in the values to the ideal gas law. Rearranging the equation to solve for pressure, we get P = nRT/V.
P = (0.6 mol)(0.0821 L·atm/(mol·K))(400 K)/(4.1 L)
P = 2.44 atm (rounding to two decimal places to keep it funny)
So, the pressure exerted by this gas party mixture is around 2.44 atm. I hope that answers your question, and remember, it's always better to laugh with gases than to cry over spilled test tubes!
To determine the pressure exerted by the mixture of gases, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, we need to find the number of moles for each gas.
For oxygen (O2):
Given mass = 3.2 grams
The molar mass of O2 = 32 g/mol (16 g/mol for each oxygen atom)
Moles of O2 = mass / molar mass = 3.2 g / 32 g/mol = 0.1 mol
For methane (CH4):
Given mass = 6.4 grams
The molar mass of CH4 = 16 g/mol (4 g/mol for carbon + 4 g/mol for hydrogen)
Moles of CH4 = mass / molar mass = 6.4 g / 16 g/mol = 0.4 mol
For sulfur dioxide (SO2):
Given mass = 6.4 grams
The molar mass of SO2 = 64 g/mol (32 g/mol for sulfur + 16 g/mol for each oxygen atom)
Moles of SO2 = mass / molar mass = 6.4 g / 64 g/mol = 0.1 mol
Now, let's calculate the total moles of gas:
Total moles = moles of O2 + moles of CH4 + moles of SO2
Total moles = 0.1 mol + 0.4 mol + 0.1 mol = 0.6 mol
Next, convert the temperature from Celsius to Kelvin:
Given temperature = 127 degrees Celsius
Temperature in Kelvin = 127 + 273 = 400 K
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
P * 4.1 L = 0.6 mol * 0.0821 L·atm/(mol·K) * 400 K
Solving for P, the pressure:
P = (0.6 mol * 0.0821 L·atm/(mol·K) * 400 K) / 4.1 L
P ≈ 4.72 atm
Therefore, the pressure exerted by the mixture of gases in the given conditions would be approximately 4.72 atm.