Problem 1: In an acetylene production plant, there are three different types of
ID: 717851 • Letter: P
Question
Problem 1: In an acetylene production plant, there are three different types of cylinders produced. The behavior of the acetylene is described by the following equation of state.
Where a, b, and R are constants, V is the molar volume of the acetylene. The three types of cylinders have volume of 50 L and a temperature of 30 oC, but they contain different weights of acetylene. The first type of cylinders contains 2 kg, and a pressure of 200atm. The second type contains 1.3 kg, and a pressure of 120 atm. What will be the pressure of the third type if it contains 1 kg of acetylene?
Problem 2: Given propane at 200 K and 4 bar, determine the change in molar volume with an adjustment to 350 K at the same pressure, comparing results from the ideal gas equation and the generalized compressibility-factor correlation.
Problem 3: Ethylene gas and steam at 230 C and atmosphere pressure are fed into a reaction process as an equimolar mixture. The process produces ethanol by the reaction: C2H4 (g) + H2O (g) C2H6OH (l) If the product stream exits at 40 C and contains a mol fraction of 0.1 water vapor, what is the heat transferred with the process per mol of product produced?
Explanation / Answer
Ans 1
The given equation
P = RT/aV2 + b/V
Gas constant R = 0.0821 L-atm/mol·K
Temperature T = 30 + 273 = 303 K
V = molar volume = volume of gas/moles of gas
For first cylinder
Moles of gas = mass/molecular weight
= (2 kg x 1000g/kg) / (26.04 g/mol)
= 76.81 mol
V = 50 L / 76.81 mol = 0.651 L/mol
Pressure P = 200 atm
200 = 0.0821*303/a*0.6512 + b/0.651
200 = 58.698/a + 1.536b
Let x = 1/a
200 = 58.698x + 1.536b
For second cylinder
Moles of gas = mass/molecular weight
= (1.3 kg x 1000g/kg) / (26.04 g/mol)
= 49.92 mol
V = 50 L / 49.92 mol = 1.002 L/mol
Pressure P = 120 atm
120 = 0.0821*303/a*1.0022 + b/1.002
120 = 24.777/a + 0.998b
Let x = 1/a
120 = 24.777x + 0.998b
200 = 58.698x + 1.536b
Solve the above two equations simultaneously
x = 0.744499 = 1/a
a = 1.3432 L2-atm/mol2
b = 101.757 L/mol
For third type of cylinder
Moles of gas = mass/molecular weight
= (1kg x 1000g/kg) / (26.04 g/mol)
= 38.402 mol
V = 50 L / 38.402 mol = 1.302 L/mol
P = 0.0821*303/1.3432*1.3022 + 101.757/1.302
= 10.925 + 78.154
= 89.1 atm