a steel cylinder holds 1.50 g of ethanol what is the pressure of an ethanol vapor if the cylinder has a volume of 251 cm3 and the teperature is 250 degrees celcius?
To determine the pressure of the ethanol vapor in the cylinder, we can use the ideal gas law equation: PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R is the ideal gas constant, and T represents temperature.
However, before we can use the ideal gas law equation, we need to calculate the number of moles of ethanol. We can use the molar mass of ethanol to convert the given mass (1.50 g) to moles.
The molar mass of ethanol (C2H5OH) is calculated as follows:
(2 * atomic mass of carbon) + (6 * atomic mass of hydrogen) + atomic mass of oxygen.
C: 12.01 g/mol
H: 1.01 g/mol
O: 16.00 g/mol
Molar mass of ethanol = (2*12.01) + (6*1.01) + 16.00 = 46.07 g/mol
Now, we can convert the mass of ethanol to moles using the molar mass:
moles = mass / molar mass
moles = 1.50 g / 46.07 g/mol
Next, we need to convert the given volume from cm3 to liters since the ideal gas law requires volume in liters:
volume = 251 cm3 / 1000 cm3/L
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
P * volume = moles * R * temperature
To solve for pressure (P), rearrange the equation:
P = (moles * R * temperature) / volume
The ideal gas constant, R, is 0.0821 L∙atm / (mol∙K).
Now we have all the necessary values to calculate the pressure:
P = (moles * 0.0821 * 523 K) / 0.251 L
Substitute the values into the equation and calculate:
P = (0.0326 mol * 0.0821 * 523 K) / 0.251 L
P ≈ 88.83 atm
Therefore, the pressure of the ethanol vapor in the cylinder is approximately 88.83 atm.