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]

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

B. The value of heat transfer during the process Q = 442.83

Explanation:

Given Data

Initial pressure = 100 k pa

Initial temperature = 25 degree Celsius = 298 Kelvin

Final pressure = 300 k pa

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

⇒ = ------------- (1)

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

⇒ P ∝ T

⇒ =

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

⇒ = × 298

⇒ = 894 Kelvin

(A). Work done during the process is given by W = P × ()

From equation (1), = 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 ( - )

Where m = mass of the gas = 1 kg

= specific heat at constant volume of nitrogen = 0.743

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

⇒ Q = 442.83

Therefore the value of heat transfer during the process Q = 442.83

false br

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg