Multiple choice

A container of capacity 700 ml is filled with two different liquids of volume 200 ml and 500 ml with respective volume expansivities as $1.4\times 10^{-5}/^{0}C$ and $2.1\times 10^{-5}/^{0}C$ . During the heating of the vessel, it is observed that neither liquid overflows nor empty space is created. The volume expansitivity of the container is

  1. $1.9\times 10^{-5}/^{0}C$
  2. $1.9\times 10^{-6}/^{0}C$
  3. $1.6\times 10^{-5}/^{0}C$
  4. $1.6\times 10^{-6}/^{0}C$
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A Correct answer
Explanation

For the volume to remain constant, the expansion of the liquids must equal the expansion of the container. (V1 * alpha1 + V2 * alpha2) * deltaT = V_total * alpha_container * deltaT. (200 * 1.4e-5 + 500 * 2.1e-5) = 700 * alpha_container. (2.8e-3 + 10.5e-3) / 700 = alpha_container. 13.3e-3 / 700 = 0.019e-3 = 1.9e-5.

AI explanation

The total volume expansion of the liquids must equal the volume expansion of the container, so the overall volume expansivity is the weighted average of the two liquids. Calculate this as [(200)(1.4 x 10^-5) + (500)(2.1 x 10^-5)] / 700. This equals (2.8 x 10^-3 + 10.5 x 10^-3) / 700, resulting in 1.9 x 10^-5 per degree C.

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