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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

  

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