1 of 1 Problem 1: In an acetylene production plant, there are three different ty

Question

1 of 1 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. RT b 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 °C, 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?

Explanation / Answer

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

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