Thermodynamics Solved examples

Question 1.
What is true of Isothermal Process
(a) ΔT >0
(b) ΔU=0
(c) ΔQ=ΔW
(d) PV=constants
In an Isothermal Process
Temperature remains constant ΔT =0
Since Internal energy depends on the temperature

From first law of Thermodynamics
Since ΔU=0

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
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
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?
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
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
n=2 T=27°C=300 K ,V1=V,V2=2V

Now PVy=constant

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
T1=227°C =500K
T2=127°C =400K

Efficiency of the carnot cycle is given by

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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