Question

A rigid, well-insulated tank is filled initially with 5.0 kg of air at a pressure of 5 bars and a temperature of 500K. A leak develops, and air slowly escapes until the pressure of the air remaining in the tank is 1 bar. Employing the ideal gas model, determine the amount of mass remaining in the tank and its temperature.

Answer 1

Step-by-step explanation:

In an isotropic process, the equation is as follows.

`(T_(2))/(T_(1)) = [(P_(2))/(P_(1))]^{(k - 1)/(k)}`

`T_(2) = T_(1) * [(P_(2))/(P_(1))]^{(k - 1)/(k)}`

As the given data is as follows.

`T_(1)` = 500 K, m = 5 kg

`P_(1)` = 5 bar,
`P_(2)` = 1 bar

Now, putting the given values into the above formula as follows.

`T_(2) = T_(1) * [(P_(2))/(P_(1))]^{(k - 1)/(k)}`

`T_(2) = 500 K * [(5 bar)/(1 bar)]^{(1.4 - 1)/(1.4)}`

= 315.7 K

As according to the ideal gas equation, PV = mRT

So, calculate the volume as follows.

V =
`(mRT)/(P)`

=
`(5 kg * 287 J/kg * 500 K)/(5 * 10^(5))` (as 1 bar = 10^{5} Pa[/tex])

=
`1.435 m^(3)`

Now, we will calculate the value of
`m_(2)` as follows.

`P_(2)V_(2) = m_(2)RT_(2)`

`m_(2) = (P_(2)V_(2))/(RT_(2))`

=
`(1 * 10^(5) * 1.435 m^(3))/(315.7 K * 287 J/kg)`

= 1.58 kg

Thus, we can conclude that the amount of mass remaining in the tank is 1.58 kg and its temperature is 315.7 K.