Question

A system gains 3220 J of heat at a constant pressure of 1.32 × 105 Pa, and its internal energy increases by 3990 J. What is the change in the volume of the system, and is it an increase or a decrease?

Answer 1

`\Delta V=-5.83*10^(-3)m^3`

The volume of the system decrease

Step-by-step explanation:

According to the first law of thermodynamics:

`\Delta U=Q-W`

Here
`\Delta U` is the change in internal energy, Q is the heat gained and W is the work done by the system.

In other hand, the work done by a system at constant pressure is given by:

`W=P\Delta V`

Here P is the pressure and
`\Delta V` the change in the volume of the system. So, replacing this in the first equation and solving for
`\Delta V`:

`\Delta U=Q-P\Delta V\\\Delta V=(Q-\Delta U)/(P)\\\Delta V=(3220J-3990J)/(1.32*10^(5)Pa)\\\Delta V=-5.83*10^(-3)m^3`

The negative sign indicates that volume decrease

Answer 2

ΔV= -5.833×10⁻³

Negative sign indicates that volume decreases

Step-by-step explanation:

Given data

System heat gains Q=3220 J

Pressure P=1.32×10⁵Pa

Internal energy increases ΔU=3990 J

To find

Change in volume ΔV

Solution

First we need to find the work done

So

W=Q-ΔU

W=3220J-3990J

W= -770J

Now for the change in volume at constant pressure

ΔV=(W/P)

`=((-770J)/(1.32*10^(5)Pa ) )\\= -5.833*10^(-3)`

ΔV= -5.833×10⁻³

Negative sign indicates that volume decreases