Please answer both question An insulated piston-cylinder device contains 7 liter
ID: 1601861 • Letter: P
Question
Please answer both question An insulated piston-cylinder device contains 7 liters of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 400 kJ, a. Determine the work produced from the resistor. An ideal gas undergoes a sequence of mechanically reversible processes in a closed system beginning with an initial state of 70 degree C and 1 bar, which is then compressed adiabatically to 150 degree C. The second step of the sequence cools the gas from 150 to 70 degree C at constant pressure, and finally, it is expanded isothermally to its original state. You may assume 1 mol of gas. Calculate w, Q, Delta U, and Delta H for each of the three processes and for the entire cycle. Take c_v = (3/2)R and c_p = (5/2)RExplanation / Answer
Answer of question no -1
From first law of thermodynamics we know that
U = Q – W
For insulated system Q=0
Work done are as follows,
Electrical W E,
Paddle work W PW
Work done by the system = P 2 V 2 - P 1 V 1
U = 0 – (W E+ W PW + P 2 V 2 - P 1 V 1)
U2-U1=0 – ( W E+ W PW + P 2 V 2 - P 1 V 1 )
-(W E+ W PW )=U2-U1+ P 2 V 2 - P 1 V 1= m( h 2 - h 1 )
Since pedal and electrical work are done on the system so both will be negative sign.
(W E+ W PW )= m( h 2 - h 1 )--------------------(i)
Given pressure P 1=175 k Pa and X 1 =0 for saturation state
From enthalpy chart
h 1 = h f = 487.01 KJ/ kg
V1 = Vf = 0.00157 m3/Kg
Given pressure P 2=175 k Pa and X 2 =0.5 (Since half of the liquid evaporated)
h 2 = hf + X2 hfg =487.01 + 0.5 (2213.1)
hfg=2213.1 KJ/kg (From enthalpy chart)
h2=1593.6 KJ/Kg
Mass of the liquid = Volume of the water (Given) / V1
= 0.007 m3/0.001057 m3/Kg = 6.623 Kg
Putting the value of m,h2, h1 & W PW in equation (i)
W PW = 400 KJ (Given)
W E+ W PW = m( h 2 - h 1 )
WE=6.623 (1593.6 -487.01) – 400=6929 KJ
So work done by resistor =6929 KJ
Answer of question no -2
Given:
1. System as 1 mol of gas.
2. CV = 12.471, CP = 20.785 J/mol
For R= 8.314 J/mol.K
P1=1.0 Bar
a) Process 1, Adiabatic Compression (Heat Transfer Q=0)
Given, T= 80 K
U = CV T= 12.471 x 80 = 998 J
H = CP T = 20.785 x 80= 1663 J
W= U= 998 J (For adiabatic Compression)
Q=0 (For adiabatic Compression)
U= 998 J
H=1663 J
b) Process 2, Constant Pressure Process
T= 80 K (Since third process is isothermal T3=T1)
Q = H = CP T = 20.785 x 80= -1663 J
U = CV T = 12.471x 80 = -998 J
W = U – Q = 665 J
W= U-Q= 665 J (For Constant Pressure process)
Q=-1663 J
U= -998 J
H=-1663 J
c) Process 3, Isothermal Process
Q=-W= RT loge (P3/P1)
P3=P2 As process 2 is constant pressure process.
We know that for adiabatic process
P2=P1 (T2/T1) (/ -1)
Given,
= 1.66 (As gas is monoatomic)
T2=273+150=423 K
T1=273+70=343 K
P1= 1 Bar
So, P2=1.689 Bar= P3, T=70 Deg C, P1=1 Bar
Q=-W= RT loge (P3/P1) =1495.681 J
W= -1495.681 J
Q=+ 1495.681 J
U= 0 (For Isothermal Process)
H= 0 (For Isothermal Process)