Notes
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Revision Notes
Question-1. What is true of Isothermal Process
a, ΔT >0
b, ΔU=0
c ΔQ=ΔW
d PV=constants
Solution-1:
In an Isothermal Process
Temperature remains constant ΔT =0
Since Internal energy depends on the temperature
ΔU=0
From first law of Thermodynamics
ΔU=ΔQ-ΔW
Since ΔU=0
ΔQ=ΔW
Also PV=nRT
As T is constant
PV= constant
Question-.2 Two absolute scales A and B have triple points of water defined as 200A and 350A. what is the relation between T
A and T
B
Solution-2
Given that on absolute scale
Triple point of water on scale A = 200 A
Triple point of water on scale B = 350 B
Also, triple point of water on Kelvin scale = 273.16 K
Now temperature on scale A and on scale B is equivalent to 273.16 K on absolute temperature scale.
Thus, value of one degree on absolute scale A = (273.16/200) K
Or,
Value of temperature T
A on absolute scale A = (273.16XT
A)/200
Similarly value of temperature T
B on absolute scale B = (273.16XT
B)/350
Since T
A and T
B represent the same temperature
273.16×T
A/200 = 273.16×T
B/350
Or, T
A = 200T
B/350 = 4T
B/ 7
Question 3:A gas is contained in a cylinder with a moveable piston on which a heavy block is placed. Suppose the region outside the chamber is evacuated and the total mass of the block and the movable piston is 102 kg. When 2140 J of heat flows into the gas, the internal energy of the gas increases by 1580 J. What is the distance s through which the piston rises?
Solution:3
Total heat supplied =Workdone + Change in internal energy
So work done=2140-1580=560 J
Let s be the distance moved then
the workdone is given by =Fs
Fs=560
s=560/F
=560/102*10
s=.54 m
Question-4: At 27°C,two moles of an ideal monoatomic gas occupy a volume V.The gas is adiabatically expanded to a volume 2V.
Calculate the ratio of final pressure to the intial pressure
Calculate the final temperature
Change in internal energy
Calculate the molar specific heat capacity of the process
Solution-4
Given
n=2 T=27°C=300 K ,V
1=V,V
2=2V
Now PV
y=constant
P
1V
1y=P
2V
2y
P
2/P
2=(V
1/V
2)
y
=.5
5/3
ALso
T
1V
1y-1=T
2V
2y-1
or T
2=300/2
5/3=189K
Change in internal energy=nC
vΔT
For monoatomic gas C
v=3R/2
Substituting all the values
Change in internal energy==-2764J
As in adiabatic process ΔQ=0,molar specific heat capacity=0
Question-5
An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C.It absorbs 6*10
2 cal of heat at the higher temperature.Calculate the amount of heat supplied to the engine from the source in each cycle
Solutions-5:
T
1=227°C =500K
T
2=127°C =400K
Efficiency of the carnot cycle is given by
=1-(T
2/T
1)=1/5
Now also efficency =Heat supplied from source/Heat absorbed at high temperature
so Heat supplied from source=6*10
2*(1/5)==1.2*10
2cal
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