Question

Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains 1m3 of N2 gas at 500 kPa and 160 degrees C while the other side contains 1m3 of He gas at 500 kPa and 60 degrees C. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move?

Answer 1

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.