---
title: "Percentage air void in bituminous mix"
subject: "Transportation Engineering I"
chapter: 6
key: "semester_6/transportation_engineering_i:n-80c-air-void"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-80c-air-void
site: BCT Engineering
---

# Percentage air void in bituminous mix

Transportation Engineering I · Chapter 6: Highway Materials

- Asked in: 2080 Chaitra, 2 marks
- Video: [MS-2 Chapter 5 Epi No-32 I Percent Air Voids (Pa) Explained | Asphalt Mix Design Volumetrics](https://www.youtube.com/watch?v=WEnNBhlnT8Y) — Lab vala 🔭
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-80c-air-void
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-6

## Question

The weight of a compacted specimen of bituminous mix in air, water and SSD condition is 1231 gm, 715 gm, and 1232 gm, respectively. The maximum theoretical specific gravity of mix is 2.56. Determine the percentage air void in the mix.

## Short answer

Bulk specific gravity G_mb = 1231 / (1232 - 715) = 2.381. Air voids = (2.56 - 2.381) / 2.56 × 100 ≈ 7.0%.

## Answer

**Given:** weight in air $W_a = 1231$ g, in water $W_w = 715$ g, SSD weight $W_{ssd} = 1232$ g, and $G_{mm} = 2.56$.

**Step 1: Bulk specific gravity of the compacted mix**

$$
\begin{aligned}
G_{mb} &= \frac{W_a}{W_{ssd} - W_w} \\
&= \frac{1231}{1232 - 715} = \frac{1231}{517} = 2.381
\end{aligned}
$$

**Step 2: Percentage air voids**

$$
\begin{aligned}
V_v &= \frac{G_{mm} - G_{mb}}{G_{mm}} \times 100 \\
&= \frac{2.56 - 2.381}{2.56}\times 100 = 6.99\%
\end{aligned}
$$

**Answer: Air voids $V_v \approx 7.0\%$** ($G_{mb} = 2.381$). This is more than the usual 3-5% limit for a dense mix (NRS 2070/Asphalt Institute), so the specimen is too porous.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula gives Gmb?
   A. Wa/(Wssd - Ww)
   B. Wa/(Wa - Wssd)
   C. Wssd/Ww
   D. Ww/Wa
   Answer: A. Volume = Wssd - Ww. (Key point: Gmb formula)

2. Gmb for this specimen is
   A. 2.56
   B. 2.381
   C. 1.72
   D. 2.38 x 515
   Answer: B. 1231/517 = 2.381. (Key point: Gmb value)

3. Air voids are about
   A. 70%
   B. 0.7%
   C. 7.0%
   D. 3.5%
   Answer: C. (2.56-2.381)/2.56 x 100 = 6.99%. (Key point: Result)

## Related in Transportation Engineering I

- [Bulk and maximum specific gravity of mix](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Abulk-max-specific-gravity) — Transportation Engineering I, chapter 6
- [Marshall method of bituminous mix design](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Amarshall-mix-design) — Transportation Engineering I, chapter 6, asked 9×

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