Question

2 moles of an ideal gas is compressed from 25 dm^3 to 13 dm^3 at constant external pressure of 4.052 bar calculate work done in dm^3 bar​

Answer 1

Answer:

Explanation:

Work done = P∆V

                   = 4.052(13-25)

                   = 4.052(-12)

                   = -48.624dm³

Answer 2

Given:

Number of moles $=2 moles$

Initial volume $=25dm^{3}$

Final volume $=13dm^{3}$

Pressure $=4.052bar$

To find:

Work done

Explanation:

The external force applied on any system that produces certain displacement is known as the work done on the system.

$W=F$Δ$s$

We know, Pressure is the force applied per unit area.

$P=\frac{F}{A}$  

$F=P$ × $A$

Hence,

$W=P$ × $A$ Δ$s$

The product of area and side of a side is the volume.

$W=P$ Δ$V$

Suppose we take a certain initial volume of liquid in a vessel and apply energy to it, then, the work done by the heat energy to expansion or compression of the volume of the liquid in the vessel at a given pressure is given by this equation that is $W=P$ Δ$V$

Solution:

According to the question, we have,

$P=4.052 bar$       $;$ $V_{1} =25dm^{3}$        $;$ $V_{2} =13dm^{3}$

We get the work done as

$W=P(V_{2} -V_{1} )$

$W=4.052(13-25)$

$W=-48.624dm^{3} bar$

The negative sign indicates work done is the work of compression since the volume decreases as the pressure increases.

Final answer:

The work done on the system is -48.624 dm³bar.