A hollow aluminum cylinder has a volume of 2.500 L at 30.00°C. It is completely

Question

A hollow aluminum cylinder has a volume of 2.500 L at 30.00°C. It is completely filled with acetone at the same temperature. The combination is slowly warmed to a new temperature of 80.00°C. Some of the acetone overflows, and the remaining acetone and the cylinder is slowly cooled back to 30.00°C. How much acetone overflows? What is the final volume of the acetone in the cylinder? The thermal expansion coefficient of aluminum is 25 x 10-6 °C -1 , and the volume expansion coefficient for acetone is 150 x 10-6 °C -1 . [9.00 mL, 2.491 L]

Explanation / Answer

The linear expansion coefficient of aluminum is = 25X10-6 0C-1

So, the volume expansion coefficient of aluminum will be al = 3 = 75X10-6 0C-1

volume expansion coefficient of acetone is ac = 150X10-60C-1

Initial volume of both the aluminium cylinder and acetone is V0 = 2.5 L

the final volume of aluminum cylinder at 800C is,

Val = V0(1 + al)

and the final  volume of acetone at 800C is,

Vac = V0(1 + ac)

So volume of acetone overflown is,

V = Vac - Val =  V0(1 + ac) - V0(1 + al)

or, V = V0(ac - al) = 2.5L(800C - 300C)(150X10-60C-1 - 75X10-60C-1)

or, V = 9.3 mL of acetone overflows,

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Final volume of acetone in the cylinder is the acetone remaining after some of it is overflown

V = initial volume of acetone - volume overflown

or, V = 2.5 L - 9.3 mL

or, V = 2.49 L, of acetone remains in the cylinder.

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