Thermodynamics A 1 m 3 rigid tank has propane at 100 kPa and 300 K. The tank is

Question

Thermodynamics

A 1 m3 rigid tank has propane at 100 kPa and 300 K. The tank is connected to another 0.5 m3 rigid tank which has propane at 250 kPa and 400 K by a ball valve. The valve is opened and both tanks come to a uniform state at 325 K.

1-      State all assumptions you need to solve the problem.

2-      Calculate the final pressure of propane in both tanks?

3-      Calculate the mass of propane before and after opening the valve in each tank.

4-      Sketch the process on a P-V diagram.

5-      Find the compressibility factor of propane in each tank before opening the valve.

6-      Calculate the change in internal energy and enthalpy of propane in each tank.

Explanation / Answer

1) I will assume the propane to behave as ideal gas

2) the moles of propane in the first tank will be calculated by ideal gas law

PV = nRT

R = Gas constant = 8.314 J /K mol = 8.314 L KPa / K mol

P = 100 kPA

T = 300 K

V = 1m^3 = 1000 L

n1 = P1V1 / RT1 = 100 X 1000 / 8.314 X 300 = 40.09 moles

1 mole = 44 g / mole

40.09 moles = 44 X 40.09 = 1.763 Kg

Similarly moles in tank 2 will be

n2 = P2V2/ RT2 = 250 X 0. 5X 1000 / 8.314 x 400 = 37.58 moles

Mass in tank 2 = 37.59 X 44 = 1.653 Kg

total moles = 37.58 + 40.09 = 77.67

Volume = V1+V2 = 1.5 m^3 = 1500 L

So final pressure = nRT / Volume = 77.67 X 8.314 X 325 / 1500 = 139.91 KPa

3) As calculated

Mass in tank2 = 1.653 Kg

Mass in tank 1 = 1.763

Initial mass = Final mass = 1.653 + 1.763 = 3.416 Kg

4)

5) From tables the comprssibility factor fro propane

a) in tank 1 = 0.99

b) in tank 2 = 0.98

almost 1 (as for ideal gas, we have assumed)

6) change in internal energy

The total work done by the surrounding is zero (as volume is constant)

There is not heat transfer so the internal energy before the mixing = internal energy after the mixing

Internal energy of tank 1 + internal energy of tank 2 = final internal energy

Internal energy = nCv Temperature

Cv =1.6794

internal energy of tank 1 = moles X Cv X 1.6794 = 40.09 X 1.6794 X 300 = 20.198 KJ

Internal energy of tank 2 = 37.58 X 1.6794 X 400 = 25.24 KJ

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