Spherical tanks A and B shown here are connected by a pipe and valve. Tank A has

Question

Spherical tanks A and B shown here are connected by a pipe and valve. Tank A has a diameter of 2.29 m and tank B a diameter of 4.46 m. Initially, tank A contains 285 kg of helium and tank B is completely evacuated. The valve is now opened and remains open. After sufficient time, the pressure and temperature in the two tanks become equal.

What is the initial specific volume of the helium?
What is the final specific volume of the helium in tank A?
What is the final mass contained in tank A?
What is the final mass contained in tank B?

Explanation / Answer

initial specific volume of the helium = Volume/ mass = 4/3 (2.29/2)3 / 285 = 0.022m3/kg

total volume = 4/3 (2.29/2)3 + 4/3 (4.46/2)3 = 52.74

mass contained in tank A = [4/3 (2.29/2)3 / 52.74 ] x 285 = 33.99kg

mass contained in tank B = 285 - 33.97 = 251kg

now, final specific volume of the helium in tank A =  4/3 (2.29/2)3 / 33.99 = 0.185m3/kg

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