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

Answer:

ΔV= -5.833×10⁻³

Negative sign indicates that volume decreases

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)

$=(\frac{-770J}{1.32*10^{5}Pa } )\\= -5.833*10^{-3}$

ΔV= -5.833×10⁻³

Negative sign indicates that volume decreases

Answer 2

Answer:

$\Delta V=-5.83*10^{-3}m^3$

The volume of the system decrease

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 $\Delta V$:

$\Delta U=Q-P\Delta V\\\Delta V=\frac{Q-\Delta U}{P}\\\Delta V=\frac{3220J-3990J}{1.32*10^{5}Pa}\\\Delta V=-5.83*10^{-3}m^3$

The negative sign indicates that volume decrease