Question

How do you do this problem.

Answer 1

Complete question:

An ideal gas is slowly compressed at a constant pressure of2.0 atm from 10.0 L to 2.0 L, path B to D in Fig 19-12. If the heat lost from the gas in the process BD is $2.78\times10^{3}\,J$ , what is the change in internal energy of the gas?

(The missing figure is attached below)

Answer:

The change in internal energy of the gas is -1158.8 joules

Explanation:

To solve this problem we're going to use the first law of thermodynamics, that states the change in the internal energy of a system $\Delta U$ is the sum of the heat $Q$ and the work $W$:

[tex\Delta U=W+Q [/tex] (1)

The heat lost in BD is $-2.78\times10^{3}\,J$ and because it is lost has negative sign, the work is the area under the curve BD that is a rectangle with length $\Delta V$ and height $P$, using those on (1)

$\DeltaU=P\DeltaV-2.78\times10^{3}\,J$

$\DeltaU=(202650\,Pa)(0.008\,m^{3})-2.78\times10^{3}\,J$

(8L=0.008 $m^{3}$ and 2.0 atm =202650 Pa )

$\DeltaU=-1158.8\,K$

That means internal energy decreses.