Question 4 of 9intermediate🔧 ApplyNumerical3 marks

An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

Correct Answer

The volume of the air bubble at the surface is approximately 5.3 cm³.

Exercise: EXERCISES | Q: 12.5
For More Understanding

Explanation

This problem applies the ideal gas law to a situation where both pressure and temperature change. The key insight is that for a fixed amount of gas, P₁V₁/T₁ = P₂V₂/T₂. Students must recognize that the pressure at the bottom of the lake includes both atmospheric pressure and the pressure due to the water column (P = P₀ + ρgh). The context provides the atmospheric pressure value (1 atm = 1.01×101.01 \times 10⁵ Pa) from the molar volume discussion.

Solution Steps

  1. Step 1: Convert temperatures to Kelvin T₁ = 12 °C = 12 + 273 = 285 K T₂ = 35 °C = 35 + 273 = 308 K

  2. Step 2: Calculate pressure at the bottom of the lake P₁ = P₀ + ρgh = 1.01×101.01 \times 10⁵ + (1000×9.81000 \times 9.8 × 40) P₁ = 1.01×101.01 \times 10⁵ + 3.92×103.92 \times 10⁵ = 4.93×104.93 \times 10⁵ Pa

  3. Step 3: Identify pressure at the surface P₂ = P₀ = 1.01×101.01 \times 10⁵ Pa

  4. Step 4: Apply the ideal gas law for constant amount of gas Using P₁V₁/T₁ = P₂V₂/T₂ V₂ = (P₁V₁T₂)/(P₂T₁)

  5. Step 5: Substitute values and calculate V₂ = (4.93×104.93 \times 10⁵ × 1.0×101.0 \times 10⁻⁶ × 308)/(1.01×101.01 \times 10⁵ × 285) V₂ = 1518.44/287.85×10287.85 \times 10⁻⁶ m³ V₂ ≈\approx 5.28×105.28 \times 10⁻⁶ m³ ≈\approx 5.3 cm³

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