Question 11 of 20intermediate🔧 ApplyNumerical2 marks

The coefficient of volume expansion of glycerine is 49×10−5 K−149 \times 10^{-5} \text{ K}^{-1}. What is the fractional change in its density for a 30 °C rise in temperature?

Correct Answer

The fractional change in density is -1.47×101.47 \times 10⁻² (or -0.0147).

Exercise: EXERCISES | Q: 10.11 | (Page 23)
For More Understanding

Explanation

This question tests understanding of the relationship between volume expansion and density change. From the textbook, the coefficient of volume expansion is defined as αV = (ΔV/V)(1/ΔT). When temperature increases, volume increases but mass remains constant. Since density ρ = m/V, and mass is constant, we get ρ₀V₀ = ρV. This leads to the fractional change in density being equal but opposite to the fractional change in volume: Δρ/ρ₀ = -ΔV/V = -αVΔT. The negative sign indicates density decreases when temperature rises.

Solution Steps

  1. Step 1: Given: Coefficient of volume expansion, αV = 49×1049 \times 10⁻⁵ K⁻¹; Temperature rise, ΔT = 30 °C = 30 K

  2. Step 2: From the definition of coefficient of volume expansion: αV = (ΔV/V)(1/ΔT), so fractional change in volume: ΔV/V = αV × ΔT

  3. Step 3: Since mass remains constant during thermal expansion and density ρ = m/V, we have ρ₀V₀ = ρV. For small changes, fractional change in density equals negative of fractional change in volume: Δρ/ρ₀ = -ΔV/V

  4. Step 4: Substituting: Δρ/ρ₀ = -αV × ΔT = -49×1049 \times 10⁻⁵ × 30 = -147×10147 \times 10⁻⁴ = -1.47×101.47 \times 10⁻²

  5. Step 5: The fractional change in density is -1.47×101.47 \times 10⁻². The negative sign indicates density decreases with temperature rise.

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