Question

A piston cylinder device contains 2 kg of nitrogen gas at 100 kPa and 25 C. We introduce 50 kJ of heat into the system as well as 80 kJ of electrical work. The system expands and produces 20 kJ of work due to such expansion. Find the final temperature of nitrogen in Celsius. Assume constant specific heats at room temperature (300 K).

Answer 1

Answer:

Final Temperature = 350.88 K = 77.88°C

Explanation:

The first law of thermodynamics in this case, can be written as:

Q + E = ΔU + W

where,

Q = heat added to system = 50 KJ

E = electrical work added to system = 80 KJ

W = Work Done by the System = 20 KJ

ΔU = Change in Internal Energy of Gas = ?

Therefore,

50 KJ + 80 KJ = ΔU + 20 KJ

ΔU = 130 KJ - 20 KJ

ΔU = 110 KJ

but, the change in internal energy of a gas is given as:

ΔU = m Cv ΔT = 110 KJ

where,

m = mass of nitrogen = 2 kg

Cv = Specific Heat at Constant Volume of Nitrogen = 1.04 KJ/kg.K

ΔT = Change in Temperature = ?

Therefore,

110 KJ = (2 kg)(1.04 KJ/kg.K)ΔT

ΔT = (110 KJ)/(2.08 KJ/K)

ΔT = 52.88 K

ΔT = Final Temperature - Initial Temperature

52.88 K = Final Temperature - 298 K

Final Temperature = 298 K + 52.88 K

Final Temperature = 350.88 K = 77.88°C