Class 12-science H C VERMA Solutions Physics Chapter 4 - Laws of Thermodynamics
Laws of Thermodynamics Exercise 62
Solution 1
(a) Vessel is thermally insulated thus no transfer of heat can take place
∴ Heat transferred is:
∆Q=0
(b)Work done on system is:
(c) We know
From above parts
And
Solution 2
(a) Heat transfer does not take place but work is done which is then converted into heat.
Thus, heat given to liquid Q=0
(b) Work done on liquid,
W=P.E = mgh
= 84J
(c) We know
Solution 3
We know
dQ=dU+dW
dU=-dW [As dQ=0]
dU=200J
Solution 4
We know,
=100J
Solution 5
(a) Work done on gas is given as:
dW=-4.5J
(b) We know,
dQ=dU+dW
dU=-dW [As dQ=0]
=4.5J
Solution 6
According to question:
Thus, change in volume is 0 i.e.
And work done by gas is given as:
W=0
Solution 7
(a) For AB path:
=0.30J
b) For ACB path:
=0.45J
c) For ADB path:
= 0.15J
Solution 8
We know,
Or
=50J
And for ADC path
=60J
Laws of Thermodynamics Exercise 63
Solution 9
For ACB path
And
Also We know,
dQ=dU+dW
=210-10
dU=200J ____1
Now, for ADB path
dU = 200J [from 1]
and
=31J
Also we know,
dQ=dU+dW = 200+31
=231J or 55cal
Solution 10
Heat absorbed is in form of work done which in this case is area covered in graph.
Heat = Area of circle
=31.4J
Solution 11
Figure shows cyclic process thus dU=0.
dW=10J
Also we know
dQ =dU+dW
2.4J=10
J=4.17J/cal
Solution 12
According to graph,
=6000J
Also we know,
2625J=6000+5000
Solution 13
According to figure,
=52.5J
And
dQ =-70cal
=-294J
Also we know
dQ = dU+dW
dU=dQ-dW
=-294-52.5
dU=-346.5J
Solution 14
We know
dU=UPdV
dU=15J
and dW =PdV =
=10J
Also,
dQ=dU+dW
=15+10
dQ=25J
Solution 15
We know
dW=PdV
dW=4J
AlsodQ = dV+dW
dU=dQ-dW
=10-4
=6J
Solution 16
(a) According to question
And
According to figure,
Area AC > Area AB
∴ Heat is extracted from the system.
b) Heat = Area ABC
=25000J
Solution 17
It is a cyclic process thus,
dU=0
and
Also we know
dQ=dU+dW
Solution 18
It is a cyclic process
Also
Also dQ=dU+dW
dQ=2305.3J
Solution 19
We know
dQ=dU+dW
dU=dQ-dW
dU=(33600-0.02)J
Solution 20
dW=1699J
and
=29200
Also we know,
dQ=dU+dW
dU=dQ-dW
=29200-1699=27501
Laws of Thermodynamics Exercise 64
Solution 21
(a) The wall is fixed and cannot be moved. Thus,
dU=0 and dQ=0
and dW=dQ-dU=0
By ideal gas equation
Also
Now internal energy
And, U=1.5nRT
Also
(b) Final pressure of left is and right is
so,
(c) Temperature is given by equation 3
(d) Loss of heat is given as:
Solution 22
(a) According to figure, wall is fixed
On left part
b) (for left) and (for right)
And
Where,
(c) No. of mole
(d) For right part
dQ=dU+dW
dQ=dU [As dW=0]
(e) dQ =-dU