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Thermodynamics



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 TA and TB



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 TA on absolute scale A = (273.16XTA)/200
Similarly value of temperature TB on absolute scale B = (273.16XTB)/350
Since TA and TB represent the same temperature
273.16×TA/200 = 273.16×TB/350
Or, TA = 200TB/350 = 4TB/ 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 ,V1=V,V2=2V

Now PVy=constant
P1V1y=P2V2y
P2/P2=(V1/V2)y
=.55/3

ALso
T1V1y-1=T2V2y-1
or T2=300/25/3=189K

Change in internal energy=nCvΔT
For monoatomic gas Cv=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*102 cal of heat at the higher temperature.Calculate the amount of heat supplied to the engine from the source in each cycle

Solutions-5:
T1=227°C =500K
T2=127°C =400K

Efficiency of the carnot cycle is given by
=1-(T2/T1)=1/5

Now also efficency =Heat supplied from source/Heat absorbed at high temperature
so Heat supplied from source=6*102*(1/5)==1.2*102cal



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