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A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a Soave- Redlich-Kwong gas and with Cp* = 100 J/(mol·K), from 300 K and 2 m^3 to 800 K and 0.02 m^3 by using less than 5 MJ of work. Is this possible?

Relax

Respuesta :

Explanation:

The given data is as follows.

          Moles of propylene = 100 moles,    [tex]C_{p}[/tex] = 100 J/mol K

          [tex]T_{i}[/tex] = 300 K,          [tex]T_{f}[/tex] = 800 K

          [tex]V_{i}[/tex] = 2 [tex]m^{3}[/tex],   [tex]V_{f}[/tex] = 0.02 [tex]m^{3}[/tex]

Therefore, the assumptions will be as follows.

  • The given system is very well insulated.
  • The work is done on the system because the given process is a compression process.
  • Assume that there is no friction so, work done on the system is equal to the heat energy liberated.

                    [tex]m \times C_{p} \Delta T[/tex] = W

Putting the given values into the above formula as follows.

                  [tex]m \times C_{p} \Delta T[/tex] = W

         W = [tex]100 moles \times 100 J/mol K \times (800 K - 300 K)[/tex]

              = [tex]5 \times 10^{6}[/tex] J

              = 5 MJ

Hence, this shows that a minimum of 5 MJ work needs to be done.

Since, work is very less. Hence, it will not compress the given system to 800 K and 0.02 [tex]m^{3}[/tex].      

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