Question

Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the heat transferred during this process, in kJ/kg. [use the room temperature specific heat for nitrogen] [for nitrogen Tcr=126.2 K, Pcr=3.39 MPa]

Answer 1

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83
`(KJ)/(kg)`

Step-by-step explanation:

Given Data

Initial pressure
`P_(1)` = 100 k pa

Initial temperature
`T_(1)` = 25 degree Celsius = 298 Kelvin

Final pressure
`P_(2)` = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒
`V_(1)` =
`V_(2)` ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒
`(P_(2) )/(P_(1))` =
`(T_(2) )/(T_(1))`

⇒ Put all the values in the above formula we get the final temperature

⇒
`T_(2)` =
`(300)/(100)` × 298

⇒
`T_(2)` = 894 Kelvin

(A). Work done during the process is given by W = P × (
`V_(2) -V _(1)`)

From equation (1),
`V_(1)` =
`V_(2)` so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m
`C_(v)` (
`T_(2)` -
`T_(1)` )

Where m = mass of the gas = 1 kg

`C_(v)` = specific heat at constant volume of nitrogen = 0.743
`(KJ)/(kg k)`

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83
`(KJ)/(kg)`

Therefore the value of heat transfer during the process Q = 442.83
`(KJ)/(kg)`