Question

Please consider 1 kg of NH3 that was compressed from 2.5 bar and 30°C to 5.0 bar in a well-insulated compressor. Please determine the (a) final temperature in C and (b) the work required in kJ.

Answer 1

Step-by-step explanation:

(a) The given data is as follows.

mass = 1 kg = 1000 g (as 1 kg = 1000 g)

Molar mass of
`NH_(3)` = 17 g/mol

`P_(1)` = 2.5 bar =
`2.5 * 10^(5) Pa` (as 1 bar =
`10^(5) Pa`)

`P_(2)` = 5 bar =
`5 * 10^(5) Pa`

`T_(1) = 30^(o)C` = 30 + 273 = 303 K

For adiabatic process,
`PV^(\gamma)` = constant = k

`\gamma` = 1.33 =
`(C_(p))/(C_(v))`

`P_(1)V_(1)^(\gamma) = P_(2)V_(2)^(\gamma)`

`((V_(2))/(V_(1)))^(\gamma) = (P_(1))/(P_(2))`

`(V_(2))/(V_(1)) = ((P_(1))/(P_(2)))^{(1)/(\gamma)}`

`V_(2) = ((2.5 * 10^(5))/(5 * 10^(5)))^{(1)/(1.33)} * (nRT_(1))/(P_(1))` (as PV = nRT)

=
`((2.5)/(5))^{(1)/(1.33)} * (58.82 * 8.314 J/mol K * 303 K)/(2.5 * 10^(5))`

= 0.352
`m^(3)`

Also, w =
`(P_(1)V_(1) - P_(2)V_(2))/(\gamma - 1)`

=
`(2.5 * 10^(5) * 0.5927 - 5 * 10^(5) * 0.352)/(1.33 - 1)`

= -84318.2 J

As 1 kJ = 1000 J. So, -84318.2 J = -84.318 kJ

Hence, the work required in kJ is -84318.2 J.

(b) It is known that for adiabatic system Q = 0,

`\Delta U = Q - w`

`nC_(v)dT` = -w

dT =
`(-w)/(nC_(v))`

=
`(84318.2)/(58.82 * 1.6)`

= 895.93 K

We known that dT =
`T_(1) - T_(2)`

so, 895.93 = 303 K -
`T_(2)`

`T_(2)` = (895.93 - 303)K

= 592.93 K

= (592.93 - 273.15)^{o}C

=
`319.78^(o)C`

Hence, the final temperature is
`319.78^(o)C`.