Question

A compressor for a jackhammer expands the air in the hammer’s cylinder at a constant pressure of 8. 6 x 10^5 Pa. The increase in the cylinder’s volume is 4. 0^5 x 10^-4 m^3. During the process, 9. 5 J of energy is transferred out of the cylinder as heat.

A. What is the work done by the air


B. What is the change in the air’s internal energy


C. What type of ideal thermodynamic process does this approximate

Answer 1

The work done by the air can be calculated using the formula: work = pressure * change in volume. The change in the air's internal energy can be calculated using the first law of thermodynamics. This process approximates an isobaric process.

Step-by-step explanation:

To calculate the work done by the air in the compressor, we can use the formula:

Work = Pressure * Change in Volume

In this case, the pressure is 8.6 x 105 Pa and the change in volume is 4.0 x 10-4 m3.

So, the work done by the air is:

Work = 8.6 x 105 Pa * 4.0 x 10-4 m3

To calculate the change in the air's internal energy, we can use the first law of thermodynamics:

Change in Internal Energy = Heat - Work

In this case, the heat transferred out of the cylinder is 9.5 J.

So, the change in the air's internal energy is:

Change in Internal Energy = 9.5 J - (8.6 x 105 Pa * 4.0 x 10-4 m3)

Based on the given information, this process approximates an isobaric process.