Answer:
10.2 mL.
Explanation:
From the question given above, the following data were obtained:
Original Volume (V₁) at temperature θ₁ = 500 mL
Initial temperature (θ₁) = 5 °C
Final temperature (θ₂) = 22 °C
coefficient of volume expansion (γ) = 1200×10¯⁶ / °C
Rise in volume (ΔV) =?
Next, we shall determine the volume (V₂) of the methanol at 22 °C. This can be obtained as follow:
Original Volume (V₁) at temperature θ₁ = 500 mL
Initial temperature (θ₁) = 5 °C
Final temperature (θ₂) = 22 °C
coefficient of volume expansion (γ) = 1200×10¯⁶ / °C
Volume 2 (V₂) at temperature θ₂ =?
γ = V₂ – V₁ / V₁ (θ₂ – θ₁)
1200×10¯⁶ = V₂ – 500 / 500 (22 – 5)
1200×10¯⁶ = V₂ – 500 / 500 × 17
1200×10¯⁶ = V₂ – 500 / 8500
Cross multiply
V₂ – 500 = 1200×10¯⁶ × 8500
V₂ – 500 = 10.2
Collect like terms
V₂ = 10.2 + 500
V₂ = 510.2 mL
Thus, the volume of the methanol at 22 °C is 510.2 mL
Finally, we shall determine the rise in volume of methanol as illustrated below:
Original Volume (V₁) at temperature θ₁ = 500 mL
Volume (V₂) at temperature θ₂ = 510.2 mL
Rise in volume (ΔV) =?
ΔV = V₂ – V₁
ΔV = 510.2 – 500
ΔV = 10.2 mL
Thus, 10.2 mL of the methanol will overflow when the temperature reaches 22 °C.
