Question

A steel cylinder holds 1.50 g of ethanol, C2H5OH. What is the pressure of the ethanol vapor if the cylinder has a volume of 200 cm3 and the temperature is 250°C? (Assume all the ethanol is in the vapor phase at this temperature.)

Answer 1

To calculate the pressure of the ethanol vapor, we'll need to use the ideal gas law, which relates the pressure, volume, temperature, and amount (in moles) of a gas:

PV = nRT

where P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles of the gas, R is the gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin (K).

First, we need to find the number of moles of ethanol in the cylinder. We can do this using the molar mass of ethanol, which is 46.07 g/mol:

n = m/M = 1.50 g / 46.07 g/mol = 0.0326 mol

Next, we need to convert the volume of the cylinder from cm³ to L:

V = 200 cm³ / 1000 cm³/L = 0.2 L

Now we can plug in the values we have into the ideal gas law and solve for P:

P = nRT/V = (0.0326 mol)(0.0821 L·atm/(mol·K))(523 K) / 0.2 L ≈ 1.07 atm

Therefore, the pressure of the ethanol vapor in the cylinder is approximately 1.07 atm (or 1.09 × 10⁵ Pa or 16 psi).