Study the scenario.
8 grams of C3H4 gas occupies a volume of 8.2 L at 300 K temperature. Assume ideal gas behavior. The ideal gas constant is 0.082 (L·atm)/(K·mol), and the molecular mass of C3H4 is 40 g/mol.
How much pressure is the gas applying to the walls of its container?
0.8 atm
0.4 atm
0.5 atm
0.6 atm
B?
No. Are you using the ideal gas law. PV = nRT
Post your work and I'll find the error.
P = Pressure (atm) V = Volume (L) n = moles R = gas constant = 0.08
so its A?
To find the pressure, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, we need to find the number of moles of C3H4 gas. We can use the given mass and molecular mass of C3H4 to calculate the number of moles:
moles = mass / molecular mass
moles = 8 g / 40 g/mol
moles = 0.2 mol
Next, we can substitute the given values into the Ideal Gas Law equation and solve for pressure (P):
P * 8.2 L = 0.2 mol * 0.082 (L·atm)/(K·mol) * 300 K
P = (0.2 mol * 0.082 (L·atm)/(K·mol) * 300 K) / 8.2 L
P ≈ 0.493 atm
Rounding to one decimal place, we get:
P ≈ 0.5 atm
Therefore, the gas is applying a pressure of approximately 0.5 atm to the walls of its container.
So, the correct answer is:
0.5 atm
To find the pressure the gas is applying to the walls of its container, we can use the ideal gas law equation: PV = nRT.
Given:
- Mass of gas (C3H4) = 8 grams
- Volume (V) = 8.2 L
- Temperature (T) = 300 K
- Ideal gas constant (R) = 0.082 (L·atm)/(K·mol)
- Molecular mass of C3H4 = 40 g/mol
First, we need to calculate the number of moles (n) of the gas using the mass and molecular mass:
n = mass / molecular mass
n = 8 g / 40 g/mol
n = 0.2 mol
Now we can substitute the given values into the ideal gas law equation to solve for pressure (P):
PV = nRT
P * 8.2 L = 0.2 mol * 0.082 (L·atm)/(K·mol) * 300 K
P * 8.2 L = 4.92 (L·atm)
P = 4.92 (L·atm) / 8.2 L
P ≈ 0.6 atm
Therefore, the gas is applying a pressure of approximately 0.6 atm to the walls of its container.
The correct answer is option D: 0.6 atm.