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 [tex] 2.78\times10^{3}\,J[/tex] , 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 [tex]\Delta U [/tex] is the sum of the heat [tex] Q[/tex] and the work [tex]W [/tex]:
[tex\Delta U=W+Q [/tex] (1)
The heat lost in BD is [tex] -2.78\times10^{3}\,J[/tex] and because it is lost has negative sign, the work is the area under the curve BD that is a rectangle with length [tex]\Delta V [/tex] and height [tex]P [/tex], using those on (1)
[tex] \DeltaU=P\DeltaV-2.78\times10^{3}\,J[/tex]
[tex] \DeltaU=(202650\,Pa)(0.008\,m^{3})-2.78\times10^{3}\,J[/tex]
(8L=0.008 [tex]m^{3} [/tex] and 2.0 atm =202650 Pa )
[tex] \DeltaU=-1158.8\,K[/tex]
That means internal energy decreses.
