11. A gas is taken through the cycle ABCA, as shown in figure. What is the net work done by the gas ?
    
    

1. 1. 1000 J

   
   
   

   2. zero ​

   
   
   

   3. – 2000 J

   
   
   

   4. 2000 J

   
   
   

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1. View Hint View Answer Discuss in Forum

   W_net = Area of triangle ABC
   

   =	1	AC × BC
   	2

   
   

   =	1	× 5 × 10-3 × 4 × 105 = 1000 J
   	2

   Correct Option: A

   W_net = Area of triangle ABC
   

   =	1	AC × BC
   	2

   
   

   =	1	× 5 × 10-3 × 4 × 105 = 1000 J
   	2

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12. A thermodynamic system undergoes cyclic process ABCDA as shown in fig. The work done by the system in the cycle is :​
    
    

1. 1. P₀V₀

   
   
   

   2. 2 ​P₀V₀

   
   
   

   3. 	P0V0
      	2

   
   
   

   4. Zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   Work done by the system in the cycle ​= Area under P-V curve and V-axis
   

   =	1	(2P0 - P0)(2V0 - V0)		-		1		(3P0 - 2P0)(2V0 - V0)
   	2					2

   
   

   =	P0V0	-	P0V0	= 0
   	2		2

   Correct Option: D

   Work done by the system in the cycle ​= Area under P-V curve and V-axis
   

   =	1	(2P0 - P0)(2V0 - V0)		-		1		(3P0 - 2P0)(2V0 - V0)
   	2					2

   
   

   =	P0V0	-	P0V0	= 0
   	2		2

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13. The internal energy change in a system that has absorbed 2 kcals of heat and done 500 J of work is :
    

1. 1. 6400 J

   
   
   

   2. 5400 J

   
   
   

   3. ​7900 J

   
   
   

   4. 8900 J

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to first law of thermodynamics ​Q = ∆U + W
   ∆U = Q – W
   ​= 2 × 4.2 × 1000 – 500
   = 8400 – 500 ​= 7900 J
   

   Correct Option: C

   According to first law of thermodynamics ​Q = ∆U + W
   ∆U = Q – W
   ​= 2 × 4.2 × 1000 – 500
   = 8400 – 500 ​= 7900 J
   

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14. ​Which of the following is not thermodynamical function ?​​
    

1. 1. Enthalpy

   
   
   

   2. Work done

   
   
   

   3. Gibb's energy

   
   
   

   4. Internal energy

   
   
   

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1. View Hint View Answer Discuss in Forum

   Work done is not a thermodynamical function.

   Correct Option: B

   Work done is not a thermodynamical function.

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15. 110 joules of heat is added to a gaseous system whose internal energy is 40 J. Then the amount of external work done is​​​
    

1. 1. 150 J

   
   
   

   2. 70 J

   
   
   

   3. ​110 J

   
   
   

   4. 40 J

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∆Q = ∆U + ∆W
   ​⇒ ∆W = ∆Q  – ∆U = 110 – 40 = 70 J

   Correct Option: B

   ∆Q = ∆U + ∆W
   ​⇒ ∆W = ∆Q  – ∆U = 110 – 40 = 70 J

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