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Thermodynamics Solved examples




Question 1.
What is true of Isothermal Process
(a) ΔT >0
(b) ΔU=0
(c) ΔQ=ΔW
(d) PV=constants
Solution
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
Total heat supplied =Work done + Change in internal energy
So work done=2140-1580=560 J
Let s be the distance moved then
the work done 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 monatomic gas occupy a volume V.The gas is adiabatically expanded to a volume 2V.
(a)Calculate the ratio of final pressure to the initial pressure
(b)Calculate the final temperature
(c)Change in internal energy
(d)Calculate the molar specific heat capacity of the process
Solution
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 monatomic 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 efficiency =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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