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,‘ thin film 4 00 x 10-5 cm thick of a transpaient material has an index t It is 'Humiliated in air by white light normal to its surface Wh %%taw) the visible spectrum will be intensified in the...

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,‘ thin film 4 00 x 10-5 cm thick of a transpaient material has an index t It is 'Humiliated in air by white light normal to its surface Wh %%taw) the visible spectrum will be intensified in the reflected beam?
6. A hydrogen atom is excited from a state with n = 1 to a state with n (a) Calculate the energy that must be absorbed by the atom (b) Calculate and display on an ener132.1eAtlAtagram the different that may lig emitted if the atom returns to the n = 1 state (c) Calculate the recoil speed of the hydrogen atom, assumed initial a transition from n - 4 to n I in a single quantum jump
r. The common isotope of uranium, . has a half-life of 4.50 x l0' alpha emission. (a) What is the decay constant'' (b) What mass of uranium would be required for an activity of (c) How many alpha-particles are emitted per second by 1.00 g of
s I
A 0.650 MeV x-ray is Compton scattered at 900 by a free electro foil. (a) What is the wavelength of the incident photon') (b) What is the wavelength of the scattered photon? (c) What is the energy of the recoiling electron?
Answered Same Day Dec 29, 2021

Solution

Robert answered on Dec 29 2021
117 Votes
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1) Given data: 
 
No. of moles ‘n’ = 3.45 
P3 = P1 = 1 atm = 100 KPa 
γ = 1.333 
T1 = 300 K 
T2 = 600 K 
T3 = 504.5 K 
Assumptions: 
i) All the processes are reversible. 
ii) Changes in K.E and P.E are negligible. 
iii) Specific heat constants are not varying with temperature. 
 
Now it’s given that  
For Process 1‐2  V = c 
For Process 2‐3   S = c 
For Process 3‐1   P= c 
 
 
 
a) Since the pressure at 1 and 3 are know so we have to find the pressure at 2 only 
 
Now for process 2‐3 
atmP
P
T
T
P
P
22
5.504
600
1
2
3
2
3
2
1333.1
333.1
1
=





⎛=





⎛=







⎟⎟


⎜⎜


−γ
γ
 
 
Now to find out the volume we will apply the ideal gas equation. 
PV=nRT 
 
Here R is universal gas constant and its value is equal to 8.314 KJ/Kmol‐K 
 
So applying ideal gas equation at state 1 we have 
P1*V1 = n*R*T1 
100*V1 = 3.45*8.314*300 
V1 = 86.05 m^3 
 
Now For Process 1‐2 V = c 
So, V2 = V1 = 86.05 m^3 
 
Now applying ideal gas equation at state 3 we have 
P3*V3 = n*R*T3 
100*V3 = 3.45*8.314*504.5 
V3 = 144.707 m^3 
Now making a table 
 
State  1  2  3 
Pressure (atm)  1   2   1  
Volume (m^3)  86.05  86.05  144.707 
 
 
) Net work done by the gas = W12+W23+W31 
W12 = 0 (Since the process is constant volume process) 
( )
( )
KJW
W
KKmolKJCv
RCvNow
TTnCvW
822723
5.504600*97.24*45.323
97.24
1333.1
314.8
1
3223
=
−=
−=

=

=
−=
γ
 
 
Again for process 3‐1 
( )
( )
KJW
W
VVPW
7.586531
05.86707.144*10031
13131
=
−=
−=
 
 
So the net work = 0+8227+5865.7 = 14092.7 KJ 
 
c) Change in entropy for every process 
For process 1‐2 
 
KKJSS
SS
V
VnR
T
TnCvSS
71.5912
05.86
05.86ln*314.8*45.3
300
600ln*97.24*45.312
1
2ln
1
2ln12
=−
+=−
+=−
 
 
 
 
For process 2‐3 
 
S3 – S2 = 0 (Since the process is adiabatic) 
 
For process 3‐1 
 
KKJSS
SS
V
VnR
T
TnCvSS
71.5931
707.144
05.86ln*314.8*45.3
5.504
300ln*97.24*45.331
3
1ln
3
1ln31
−=−
+=−
+=−
 
 
d) To get the efficiency of the cycle we will find out the heat supplied to the gas 
 
Now heat will be supplied during the process 1‐2 because during process 2‐3 heat transfer 
will be zero and during the process 3‐1 heat will be rejected. 
 
So for process 1‐2 applying 1st law of thermodynamics we get 
( )
( )
)(%53.54
5453.0
95.25843
7.14092
12
95.25843300*97.24*45.312
01212
121212
Ans
Q
Wnet
engineheattheofefficiencyTherefore
KJQ
TTnCvQ
WUUQ
=
===
==
+−=
+−=
η
η
 
 
2)...
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