Question

As the model in the picture shows, I am confused that why the work by gas during the expansion is -P times deltaV (P represents the external pressure). In my view, this should be the work done on the gas by environment. And the work done by gas should be the pressure of the gas (i.e. the internal pressure) which is changed by the volume of gas. (PS. The picture is from a chapter about thermochemistry in my General Chemistry textbook)

Answer 1

actually the sign convention in physics and chem varies:

Step-by-step explanation:

In chem , work done by the system (gas) is taken as negative.

Work done on the system (gas) is taken as positive.

this is because when the gas does work , its internal energy decreases and thus it looses energy upon doing work , so it is taken as negative.

delta V is the change in volume.

P is the ext. pressure.

Now , to make it simple , when external pressure is applied to the gas , it gets compressed and thus it (gas) does work by exerting more pressure on the walls of the cylinder, which is why it is -P.

Answer 2

Here's what I get

Step-by-step explanation:

Assume that the piston is weightless and frictionless and has an area of 0.01 m², and that the cylinder contains 0.001 m³ of air at 1 bar.

Then the gas is slowly heated, making the gas expand and force the piston up until the volume is 0.002 m³.

The gas is doing work, because it is pushing back against the pressure of the atmosphere in order to expand. The sign is negative, because of the thermodynamic convention that anything going out of the system is negative.

How much work has the gas done?

w = F × d, and

p = F/A, so

F = p × A. Then

w =p × A × d

A × d is the change in volume (ΔV) swept out by the piston as the gas expands. Thus

w = -pΔV (negative sign because of the convention)

= -1 × 10⁵ Pa × 0.001 m³ =-100 Pa·m³ = -100 J

The gas has done 100 J of work by expanding against an external pressure.