Respuesta :
Answer: Option (e) is the correct answer.
Explanation:
The given data is as follows.
     Volume = 140 [tex]m^{3}[/tex],    Pressure = 7600 Pa
     q = 490 kJ,    [tex]\Delta U[/tex] = 140 kJ
As we know that relation between internal energy and work is as follows.
     [tex]\Delta U = q + w[/tex]
        140 kJ = 490 kJ + w
         w = (140 - 490) kJ Â
           = -350 kJ
Now, calculate the final volume using work, pressure and volume change relationship as follows.
     w = [tex]-P \Delta V[/tex]
    -350 kJ = [tex]-7600 Pa \times (V_{2} - 140 m^{3})[/tex]
         [tex]V_{2} - 140 m^{3} = \frac{350000 J}{7600 Pa}[/tex]
         [tex]V_{2} - 140 m^{3} = 46.05 m^{3}[/tex]
(As, 1 J = [tex]1 Pa m^{3}[/tex])
            [tex]V_{2} = 186.05 m^{3}[/tex]
Which is nearest to 190 [tex]m^{3}[/tex].
Thus, we can conclude that the final volume of the cell is 190 [tex]m^{3}[/tex].
