A tank containing 50 mol% ethanol in water at 298 K has been opened to the atmos

Question

A tank containing 50 mol% ethanol in water at 298 K has been opened to the atmosphere. A subsequent explosion has occurred, but the operator -responsible for venting the tank is claiming that the ethanol vapour in the tank would have been outside its flammability limits. Given the data below, check whether or not the vapour in the vented tank is within the flammability limits. Data: Flammable limits if ethanol in air: 3.3 to 18.8 % by volume. Vapour pressure ethanol at 298 K = 7.63 kPa. Activity coefficient of ethanol in water at a mole fraction of 0.5 = 1 17 a- 298 K). {4.46 % (v/v) ethanol vapour (neglecting water vapour), therefore flammable}

Explanation / Answer

The data given in the question is as below,

liquid mole fraction of ethanol is 50% i.e. Xethanol = 0.5

Vapor pressure of ethanol at 298K is 7.63 KPa = 0.0753 atm.

So, as per raoults law, Pethanol = Xethanol * V.P.ethanol

Pethanol = 0.5 * 0.0753 = 0.03765 atm

Also vapor phase mole fraction of ethanol will be Yethanol = Pethanol / Ptotal

Yethanol = 0.03765 / 1 = 0.03765 = 3.765 % mole fraction.

Converting this to %v/v will be 4.46%.

This concentration is above LEL & below UEL.

LEL (lower exposion limit) means the lowest concentration of the gas or vapor required in air for producing the flash & UEL (Upper explosion limit) means the highest concentration of the gas or vapor in air which can produce the flash. Above UEL, its too rich to burn. Hence if the concentration is between LEL & UEL, then it can produce flash at that given conditions. For the given case, as the concentration lies between LEL & UEL, it will be flammable mixture & can produce flame and thus explosion.

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