 on August 9th, 2012

Dear Colleagues

An interesting little (simple) brain teaser for you – no matter whether what discipline or type of engineering professional you are.

The Challenge
A tank at atmospheric pressure contains 1 kg of air. The tank is then pressurized with an additional 3 kgs of air. What is the resultant gauge pressure (in bars) in the tank after this 3kgs of air has been added? Absolute pressure = gauge pressure + atmospheric pressure

A. 1 bar
B. 3 bar
C. 4 bar

You could use the Universal Gas Law in your deliberations:

PV = nRT

Where V is volume; P is absolute pressure; n is the number of moles (‘molecules’); R is a constant and T is the temperature.

Assume atmospheric pressure is 1 bar (the initial pressure) – naturally this is a number which could change depending on what altitude you are at.

Solution (suggested – you could do it in a myriad of different ways)
There is initially 1kg of air occupying V0 initial volume.

With 3kgs of air added; there is now a total of 4kg of air in the tank (thus the final number of moles or particles is effectively 4 times that of the initial number).

The volume (V0) and temperature (TO) of the tank still stays the same.

Thus:

P0 V0  = n0 R T0 (initial state)
P1  V0  =  n1 R T0  (final state where n1 = 4 x n0)
Thus – dividing each side of equation 1 and 2 we get:

P1  V0 / P0 V0  =  4 x n0 R T0  /  n0 R T0

P1  V0 / P0 V0  =  4 x n0 R T0  /  n0 R T0

Thus P1 = 4 P0

Hence, the final pressure P1 = 4 bar absolute or 3 bar gauge pressure (thus answer B above)

Do you agree?

My humble appreciation to Dr Rodney Jacobs and David Spitzer (the flow guru) for their contributions and critique.

Albert Einstein remarked: I never teach my pupils. I only attempt to provide the conditions in which they can learn.

Yours in engineering learning

Steve

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