Please can you answer it step by step? An adiabatic piston-cylinder apparatus is
ID: 843745 • Letter: P
Question
Please can you answer it step by step?
An adiabatic piston-cylinder apparatus is set up as shown below. The piston separates the volume within the cylinder into two partitions. Side A is 10 cm long. Side B is 50 cm long. The cros-sectional area is 0.1 m2. The bottome compartment is initially at 20 bar and 250 degree C. The piston is initially held in place by a mass atop the assembly as shown. At time = zero, the mass is removed, allowing the system to continue toward equilibrium. Find the mass of the block. Once the block is removed, find the final temperature, the final pressure and the distance the piston has traveled. If we replace the block after the system has reached equilibrium, how much work is p back into the system?Explanation / Answer
a)
Pressure difference between A & B = 20 - 10 = 10 bar = 10 * (10^5) Pa
so this pressure difference is caused by the mass and the atmospheric pressure,
so, ( mg/ area ) + 101325 = 10* (10^5)
( m * 9.8/0.1) + 101325 = 10* (10^5)
m=9170.153 Kg is the mass of block
b)
Initial volume of system A = 0.1 * 0.1 = 0.01 m^3
Once the block is removed, Final Pressure in the system A = 10* (10^5) + 101325 = 11.01325 * (10^5) Pa
As the process is adiabtic,
we have g = gamma of H2O = adiabatic index H2O = 1.310
so, P1 * V1^ (1.31) = P2 * V2^(1.31)
(20 * (10^5)) * (0.01^1.31) = 11.01325 * (10^5) * (V2^1.31)
v2 = 0.0157688 m^3
distance piston has moved = V2 -V1 ) / cross section area = (0.0157688- 0.01)/ 0.1 = 0.057688 m = 5.7688 cm
P1* V1/T1 = P2* V2/T2
(20 * (10^5)) * 0.01 / ( 250 + 273) = 11.01325 * (10^5) * 0.0157688/ T2
T2 = 454.136 K = 181.136 C
final temperature of system A = T2 = 181.136 C
c)
work put back into system = P1* (V1^g) * ((V2^(1-g)) - (V1^(1-g)))/ ( 1- g)
work = (11.01325 * (10^5) * ( 0.0157688^1.31)) * ((0.1^(1-1.31)) - ( 0.0157688^(1-1.31)))/ ( 1- 1.31)
work = 24422.44619 J is the answer.