Question

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?

Answer 1

Step-by-step explanation:

The given data is as follows.

Moles of propylene = 100 moles,
`C_(p)` = 100 J/mol K

`T_(i)` = 300 K,
`T_(f)` = 800 K

`V_(i)` = 2
`m^(3)`,
`V_(f)` = 0.02
`m^(3)`

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.

`m * C_(p) \Delta T` = W

Putting the given values into the above formula as follows.

`m * C_(p) \Delta T` = W

W =
`100 moles * 100 J/mol K * (800 K - 300 K)`

=
`5 * 10^(6)` 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
`m^(3)`.