Examkin

Question

Difficulty: MediumThermal Expansion of Liquids and Anomalous Expansion of Water

A glass flask is filled to a mark with a liquid at 0∘C0^\circ\text{C}. If the linear expansivity of the glass is 8.0×10−6 K−18.0 \times 10^{-6} \text{ K}^{-1} and the apparent cubic expansivity of the liquid is 1.80×10−4 K−11.80 \times 10^{-4} \text{ K}^{-1}, what is the real cubic expansivity of the liquid?

  1. A
    1.88×10−4 K−11.88 \times 10^{-4} \text{ K}^{-1}
  2. 2.04×10−4 K−12.04 \times 10^{-4} \text{ K}^{-1}Answer
  3. C
    1.56×10−4 K−11.56 \times 10^{-4} \text{ K}^{-1}
  4. D
    1.72×10−4 K−11.72 \times 10^{-4} \text{ K}^{-1}

Answer

The real cubic expansivity of the liquid is 2.04×10−4 K−12.04 \times 10^{-4} \text{ K}^{-1}.
The real cubic expansivity of a liquid is given by γr=γa+γv\gamma_r = \gamma_a + \gamma_v. Since the container's linear expansivity α=8.0×10−6 K−1\alpha = 8.0 \times 10^{-6} \text{ K}^{-1}, its cubic expansivity is γv=3α=2.40×10−5 K−1=0.24×10−4 K−1\gamma_v = 3\alpha = 2.40 \times 10^{-5} \text{ K}^{-1} = 0.24 \times 10^{-4} \text{ K}^{-1}. Adding this to the apparent cubic expansivity 1.80×10−4 K−11.80 \times 10^{-4} \text{ K}^{-1} gives 2.04×10−4 K−12.04 \times 10^{-4} \text{ K}^{-1}.

Step-by-Step Solution

1
Calculate the cubic expansivity of the glass vessel (γv\gamma_v).
γv=3×α=3×(8.0×10−6 K−1)=2.40×10−5 K−1=0.24×10−4 K−1\gamma_v = 3 \times \alpha = 3 \times (8.0 \times 10^{-6} \text{ K}^{-1}) = 2.40 \times 10^{-5} \text{ K}^{-1} = 0.24 \times 10^{-4} \text{ K}^{-1}
Cubic expansivity of a solid material is three times its linear expansivity.
2
Apply the relationship between real expansivity (γr\gamma_r), apparent expansivity (γa\gamma_a), and vessel expansivity (γv\gamma_v).
γr=γa+γv\gamma_r = \gamma_a + \gamma_v
The real expansion of a liquid is the sum of its observed (apparent) expansion and the expansion of the containing vessel.
3
Substitute the values into the formula to find γr\gamma_r.
γr=1.80×10−4 K−1+0.24×10−4 K−1=2.04×10−4 K−1\gamma_r = 1.80 \times 10^{-4} \text{ K}^{-1} + 0.24 \times 10^{-4} \text{ K}^{-1} = 2.04 \times 10^{-4} \text{ K}^{-1}
Adding the apparent cubic expansivity and the vessel's cubic expansivity yields the true (real) cubic expansivity.

Key Concept

Real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of the vessel (γr=γa+3αv\gamma_r = \gamma_a + 3\alpha_v).
Rate this question

Topics

Thermal Expansion of Liquids and Anomalous Expansion of Water

More questions from this topic