Question

a weather balloon slowly expands as energy is transferred as heat from the outside air. if the average net pressure is 1.5×10^5m^3, how much work is done by the expanding gas?

Answer 1

There is only one pressure this situation would be a "constant pressure" process

Answer 2

W =
`-8.1 * 10^(-2) J`

Step-by-step explanation:

The rest of the question actually states that:

**If the average net pressure is 1.5x10^3Pa and the balloons volume increases by 5.4x10^-5 m^3, how much work is done by the expanding gas.**

Work done, W against an external pressure,
`P_(ext)`, which causes a change in volume, ΔV of an object is given as:

`W = -P_(ext)` ΔV

`P_(ext)` =
`1.5 * 10^5 Pa`

ΔV =
`5.4 * 10^(-5) m^3`

=>
`W = -1.5 * 10^3 * 5.4 * 10^(-5)\\\\W = -8.1 * 10^(-2) J`

Note: The negative sign denotes that the work is done against external pressure.