+971 52 902 2252 2/17/18, 9:07 PM All Media Solve one problem only and submit: P

Question

+971 52 902 2252 2/17/18, 9:07 PM All Media Solve one problem only and submit: Problem 3-1 Ideal Gases Two rigid tanks (A and B) connected to each other by a valve contain air (as an ideal gas) at specified conditions. Initially the valve is elosed and the state of air in cach tank is as shown in figure. Later the vaive is opened letting the pressure in tanks A and B equalize and the system reach thermal equilibrium with the surroundings of 20°C temperature (i.c. 293 K). a) Calculate the mass of air in tank A (mA)in kg. b) Calculate the volume of air in tank B (Va) in m. c Determine the final equilibrium pressure of air in kPa. d) Calculate the internal energy of air in tank B at state 1 (i.e. Uns) inkJ.(1pt) (Ipt) (Ipt) (2pts) Air m=5kg T 310 K P 200 kE Air T = 298 K P-500 KP

Explanation / Answer

Air is an ideal gas

Part a

Mass of air in tank A

From the ideal gas equation

P1V = mART1

mA = (P1V) /(RT1)

= (500 kPa x 1m3 )/ (0.287 kPa-m3/kg-K) x (298K)

= 5.846 kg

Part b

From the ideal gas equation

P1VB = m1RT1

VB =( m1RT1 /P1)

VB = (5 kg) x (0.287 kPa-m3/kg-K) x (310K)/(200 kPa)

= 2.22425 m3

Part C

Final equilibrium pressure

P2 = mRT2 /V

m = mA + mB

= 5.846 + 5.0

= 10.846 kg

V = VA + VB

= 1 + 2.22425

= 3.22425 m3

P2 = (10.846 kg)(0.287 kPa-m3 /kg-K)(293 K)/3.22425 m3

= 282.872 kPa

Part d

Internal energy of air in tank B at state 1

U = m x Cv x T

= 5 kg x 0.718 kJ/kg-K x 310 K

= 1112.9 kJ

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