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Gas is trapped inside of a cell with volume 140 m3. The gas exerts 7600 Pa of pressure against the walls of the cell. A machine then expands the cell and gives off 490 kJ of heat to the gas during the process. If the internal energy changed by 140 kJ, what is the final volume of the cell? Assume the pressure stays constant.
a. 57 m3
b. 94 m3
c. 660 m3
d. 220 m3
e. 190 m3

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  

                     = -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].

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