A piece of aluminum has a volume of 1.55 x 10-3 m3. The coefficient of volume expansion for aluminum is = 69 x 10-6(C°)-1. The temperature of this object is raised from 25.9 to 406 oC. How much work is done by the expanding aluminum if the air pressure is 1.01 x 105 Pa?

To find the work done by the expanding aluminum, we need to calculate the change in volume and use it to determine the work done.

1. Calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 406 °C - 25.9 °C
ΔT = 380.1 °C

2. Calculate the change in volume using the coefficient of volume expansion:
ΔV = coefficient of volume expansion × initial volume × ΔT
= (69 x 10^-6 C^(-1)) × (1.55 x 10^-3 m^3) × (380.1 °C)
= (69 x 10^-6) × (1.55 x 10^-3) × (380.1)
= 0.0398298 m^3

3. Calculate the work done using the equation:
Work = force × distance
In this case, the applied force is the pressure multiplied by the change in volume, and the distance is assumed to be constant.

Work = pressure × change in volume
= (1.01 x 10^5 Pa) × (0.0398298 m^3)
≈ 4,028.998 J

Therefore, approximately 4,028.998 Joules of work is done by the expanding aluminum at an air pressure of 1.01 x 10^5 Pa.