Question

A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 17.0cm wide and 20.4cm high. The maximum safe pressure inside the vessel has been measured to be 2.20MPa. For a certain reaction the vessel may contain up to 0.0985kg of boron trifluoride gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Round your answer to significant digits.

Answer 1

Answer:

• The maximum safe operating temperature the engineer should recommend for this reaction is 590. °C (3 significant figures)

Explanation:

1) Data:

a) Cylinder diameter: d = 17.0 cm

b) Cylinder height: H = 20.4 cm

c) p = 2.20 MPa

d) compound: BF₃ gas

e) mass, m = 0.0985 kg

2) Formulae:

a) Volume of a cylinder, V = π r² H

b) Number of moles, n = mass in grams / molar mass

c) Ideal gas equation: pV = nRT

3) Solution:

a) Volume (V)

i) r = d / 2 = 17.0 cm / 2 = 8.50 cm

ii) V = V = π r² H = π (8.50 cm) ² (20.4 cm) = 4,630 cm³ = 4.630 liter

b) Number of moles (n)

i) molar mass BF₃: 67.82 g/mol

ii) n = mass in grams / molar mass = 98.5 g / 67.82 g/mol = 1.452 mol

c) Temperature

i) pressure conversion: 2.20 MPa = 22.21 atm

ii) pV = nRT ⇒ T = pV / (nR)

• T  = 22.21 atm × 4.630 liter / (1.452 mol × 0.08206 atm-liter /K-mol(

• T = 863 K

• T = 863 - 273 °C = 590. °C ← answer

The result is given with 3 significant figures, since that is the number of significant figures used for the data.

Answer 2

The reaction is to performed in the cylindrical vessel. The volume of the cylindrical vessel for performing the reaction will be:

Volume of cylinder = $\pi$ $\rm r^2$h

• = 22.17 $\times$ 17 $\times$ 17 $\times$ 20.4 $\rm cm^2$
• = 4.630 $\rm cm^2$

For calculating the maximum temperature of the reaction, ideal gas equation is used.

PV = nRT

where, P = pressure

• V = volume
• R = constant
• T = temperature
• n = moles of reactant

• moles = $\rm \dfrac{weight}{molecular\;weight}$

• moles of boron trifluoride = $\rm \dfrac{98.5\;g}{67.82\;g/mol}$
• moles of boron trifluoride = 1.452 mol
• PV = nRT
• T = $\rm \dfrac{PV}{nR}$
• T = $\rm \dfrac{22.21\;atm\;\times\;4.630\;liters}{0.0826\;atm-liter/K-mol\;\times\;1.452\;moles}$
• = 863 K
• = 863 - $\rm 273\;^\circ$C
• = $\rm 590\;^\circ C$.

Therefore, the temperature of the system will be $\rm 590\;^\circ C$.

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