How many grams of methane gas (CH4) are in a 15.8 liter sample at 1.2 atmospheres and 27°C
Use PV = nRT and solve for n, then
n = grams/molar mass.
To determine the number of grams of methane gas (CH4) in a 15.8 liter sample at 1.2 atmospheres and 27°C, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atmospheres
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
We'll need to convert the temperature from Celsius to Kelvin:
T(Kelvin) = T(°C) + 273.15
T(Kelvin) = 27°C + 273.15 = 300.15 K
Now we can rearrange the ideal gas law equation to solve for the number of moles of methane gas:
n = PV / RT
Substituting the given values:
P = 1.2 atmospheres
V = 15.8 liters
R = 0.0821 L·atm/mol·K
T = 300.15 K
n = (1.2 atm * 15.8 L) / (0.0821 L·atm/mol·K * 300.15 K)
Performing the calculation:
n = 0.760 mol
Finally, we can calculate the mass of methane gas using its molar mass:
Molar mass of CH4 = 12.01 g/mol (carbon) + 4 * 1.01 g/mol (hydrogen)
Molar mass of CH4 = 16.05 g/mol
Mass of CH4 = n * molar mass of CH4
Mass of CH4 = 0.760 mol * 16.05 g/mol
Calculating the result:
Mass of CH4 = 12.204 grams
Therefore, there are approximately 12.204 grams of methane gas in the 15.8 liter sample at 1.2 atmospheres and 27°C.