Thermodynamics Solved examples

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 =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.
Calculate the ratio of final pressure to the initial 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 PV^y=constant
P₁V₁^y=P₂V₂^y
P₂/P₂=(V₁/V₂)^y
=.5^5/3

ALso
T₁V₁^y-1=T₂V₂^y-1
or T₂=300/2^5/3=189K

Change in internal energy=nC_vΔT
For monatomic 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*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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