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?
6 years ago
1 year ago
To calculate the pressure exerted by the gas, 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 of the gas,
R is the ideal gas constant, and
T is the temperature in Kelvin.
First, we need to calculate the number of moles of C3H4 gas:
Number of moles (n) = mass / molecular mass
Given that the mass of C3H4 gas is 8 grams and the molecular mass is 40 g/mol:
n = 8 g / 40 g/mol = 0.2 mol
Next, we need to convert the temperature from degrees Celsius to Kelvin:
Temperature in Kelvin (T) = 300 K
Now, we can substitute the values into the ideal gas law equation:
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 pressure exerted by the gas is approximately 0.6 atm.
11 months ago
To find the pressure applied by the gas, we can use the ideal gas law equation:
PV = nRT
Where:
- P represents the pressure (in atm)
- V represents the volume (in liters)
- n represents the number of moles of gas
- R represents the ideal gas constant (0.082 L·atm/(K·mol))
- T represents the temperature (in Kelvin)
First, we'll need to determine the number of moles of C3H4 gas present. We can use the molecular mass to convert the given mass of the gas to moles.
Number of moles = mass / molar mass
Given:
- Mass of C3H4 gas = 8 grams
- Molar mass of C3H4 = 40 g/mol
Number of moles = 8 g / 40 g/mol
Number of moles = 0.2 mol
Now that we have the number of moles, we can plug the values into the ideal gas law equation:
PV = nRT
P * 8.2 L = 0.2 mol * 0.082 (L·atm)/(K·mol) * 300 K
Simplifying the equation:
8.2P = 0.2 * 0.082 * 300
8.2P = 4.92
P = 4.92 / 8.2
P = 0.6 atm
Therefore, the gas is applying a pressure of 0.6 atm to the walls of its container.