Answer:
- the final equilibrium temperature in the cylinder is 85.67 °C
- Also the answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats.
Explanation:
Given That;
Compartment Nitrogen:
Volume V1 = 1 m³, Pressure P1 = 500 kPa , Temperature T = 120°C
Compartment of helium:
Volume V1H = 1 m³, Pressure P1H = 500 kPa, Temperature T1H = 40°C
From the ideal specific heat of gases
-Nitrogen
The gas k (constant) and the k (constant) volume specific heats are;
R = 0.2968 kJ
/kg.K
Cv = 0.743 kJ/kg.K
also From the ideal specific heat of gases
-helium
R = 2.0769 kJ /kg.K
Cv = 3.1156 kJ/kg.K
we know that
PV = mRT
Mass of the nitrogen
mN2 = (P1V1 /RT1)_N2
mN2 = (500)(1) / (0.2968)(393)
= 4.29kg
mass of helium
mHe = (P1HVIH /RT1H)_He
mHe = (P1V1 /RT1)_N2
mHe = (500)(1) / (2.0769)(313)
= 0.769kg
Taking the whole contents of the cylinder,
the 1st law relation can be expressed as;
Ein - Eout = ΔEsystem
0 = ΔU = (ΔU)_N2 + (ΔU)_He
0 = [mcV(T2 -T1)]_N2 + [mcV(T2 - Ti)]He
Since T2 =T_F
0 = [mcV(T_F -T1)]_N2 + [mcV(T_F - Ti)]He
we Substitute
(4.29)(0.743)(T_F - 120) + (0.769)(3.1156)(T_F - 40) = 0
T_F = 85.67 °C
therefore the final equilibrium temperature in the cylinder is 85.67 °C
Also the answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats.