What is the density of propane (CH3CH2CH3) gas at 25°C and 65.0 cm Hg? (1atm=76 cm Hg)
To find the density of propane gas at a given temperature and pressure, you 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, convert the given pressure from cm Hg to atm:
65.0 cm Hg * (1 atm / 76 cm Hg) = 0.8553 atm
Next, convert the given temperature from °C to Kelvin:
25°C + 273.15 = 298.15 K
Now, let's rearrange the ideal gas law equation to solve for density:
PV = nRT
Divide both sides by V:
P = (nRT) / V
Since density (ρ) is defined as mass (m) divided by volume (V):
ρ = m / V
We can substitute the value of V from the ideal gas law equation:
ρ = m / (nRT / P)
Finally, we need to know the molar mass of propane to find m/n and substitute it into the equation. The molar mass of propane (C3H8) is approximately 44.1 g/mol.
Now we have all the values we need to calculate the density of propane gas:
ρ = m / (nRT / P)
Let's calculate it step by step:
1. Calculate the value of n:
n = m / M
Where:
m = mass of propane
M = molar mass of propane
2. Calculate the value of m/n:
m/n = m / (molar mass) = (mass of propane) / (molar mass)
3. Substitute the known values into the equation:
ρ = [(mass of propane) / (molar mass)] / (RT / P)
4. Calculate the density:
ρ = (mass of propane * P) / (molar mass * RT)
Plug in the values:
P = 0.8553 atm
molar mass (M) = 44.1 g/mol
R = 0.0821 L·atm/(mol·K)
T = 298.15 K
ρ = (mass of propane * 0.8553 atm) / (44.1 g/mol * 0.0821 L·atm/(mol·K) * 298.15 K)
Solving this equation will give you the density of propane gas.
To find the density of propane gas at a given temperature and pressure, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
First, let's convert the given pressure from cm Hg to atm:
65.0 cm Hg ÷ 76 cm Hg/atm = 0.855 atm (rounded to three decimal places)
Next, convert the temperature from Celsius to Kelvin:
25°C + 273.15 = 298.15 K (rounded to five decimal places)
Now we need to rearrange the ideal gas law equation to solve for density (ρ). The equation becomes:
ρ = (PM) / (RT)
Where:
ρ = Density (in g/L)
P = Pressure (in atm)
M = Molar mass of the gas (in g/mol)
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
The molar mass of propane (CH3CH2CH3) can be calculated by adding up the atomic masses of its constituent elements:
1 carbon (C) atom: 12.01 g/mol
8 hydrogen (H) atoms: 1.008 g/mol each
So the total molar mass (M) of propane is:
M = (1 × 12.01) + (8 × 1.008) = 44.11 g/mol (rounded to two decimal places)
Now we can plug in the values into the equation:
ρ = (0.855 atm × 44.11 g/mol) / (0.0821 L·atm/(mol·K) × 298.15 K)
Calculating this expression will give you the density of propane gas at 25°C and 65.0 cm Hg.
Use the modified universal gas equation
P*molar mass = density*RT
P must be in atm and T in kelvin.