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 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
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 ,V
_{1}=V,V
_{2}=2V
Now PV
^{y}=constant
P
_{1}V
_{1}^{y}=P
_{2}V
_{2}^{y}
P
_{2}/P
_{2}=(V
_{1}/V
_{2})
^{y}
=.5
^{5/3}
ALso
T
_{1}V
_{1}^{y-1}=T
_{2}V
_{2}^{y-1}
or T
_{2}=300/2
^{5/3}=189K
Change in internal energy=nC
_{v}ΔT
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*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 efficiency =Heat supplied from source/Heat absorbed at high temperature
so Heat supplied from source=6*10
^{2}*(1/5)=1.2*10
^{2}cal
link to this page by copying the following text
Also Read
- Notes
- Assignments
- Revision Notes
Go back to Class 11 Main Page using below links
Class 11 Maths
Class 11 Physics
Class 11 Chemistry
Class 11 Biology