106. One kg of an ideal gas (gas constant, R= 400 J/ kgK; specific heat at constant volume, c_p = 1000 J/kgK) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100 kJ of work is done on the system by a stirrer. The increase in entropy of the system is _______ J/K.

1. 1. 287.68 J/K

   
   
   

   2. 187.68 J/K

   
   
   

   3. 387.68 J/K

   
   
   

   4. 277.68 J/K

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here,
   m = 1 kg
   R = 400 J/kgK
   C_V = 1000 J/kgK
   P₁ = 1 bar
   T₁ = 300 K
   W = – 100 kJ = – 100 × 10³ J
   Q = 0
   0 = – 100 × 10³ + mCV (T₂ – T₁)
   100 × 10³= 1 × 1000 (T₂ – 300)
   T₂ = 400 K
   V₁ = V₂
   

   ⇒ ln	V2	= 0
   	V1

   
   

   S2 - S1 = mCvln	T2
   	T1

   
   

   S2 - S1 = 1 × 1000 × ln		400
   		300

   
   S₂ – S₁ = 287.68 J/K

   Correct Option: A

   Here,
   m = 1 kg
   R = 400 J/kgK
   C_V = 1000 J/kgK
   P₁ = 1 bar
   T₁ = 300 K
   W = – 100 kJ = – 100 × 10³ J
   Q = 0
   0 = – 100 × 10³ + mCV (T₂ – T₁)
   100 × 10³= 1 × 1000 (T₂ – 300)
   T₂ = 400 K
   V₁ = V₂
   

   ⇒ ln	V2	= 0
   	V1

   
   

   S2 - S1 = mCvln	T2
   	T1

   
   

   S2 - S1 = 1 × 1000 × ln		400
   		300

   
   S₂ – S₁ = 287.68 J/K

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107. One kg of air (R = 287 J/kgK) undergoes an irreversible process between equilibrium state 1 (20°C, 0.9 m³) and equilibrium state 2 (20°C, 0.6 m³). The change in entropy s₂ – s₁ (in J/ kgK) is ______.

1. 1. - 116.36 J /kgK

   
   
   

   2. 116.36 J /kgK

   
   
   

   3. - 117.36 J /kgK

   
   
   

   4. 116.36 J /kgK

   
   
   

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1. View Hint View Answer Discuss in Forum

   – 116.36 J/kg– K
   

   ∆ S = mRln	V2
   	V1

   
   

   = (287) (1) ln		0.6
   		0.9

   
   ⇒ - 116.36 J/ kgK

   Correct Option: A

   – 116.36 J/kg– K
   

   ∆ S = mRln	V2
   	V1

   
   

   = (287) (1) ln		0.6
   		0.9

   
   ⇒ - 116.36 J/ kgK

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108. A closed system contains 10kg of saturated liquid ammonia at 10°C. Heat addition required to convert the entire liquid into saturated vapour at a constant pressure is 16.2 MJ. If the entropy of the saturated liquid is 0.88 kJ/kgK, the entropy (in kJ/kgK) of saturated vapour is____

1. 1. 6.6013 kJ/kg

   
   
   

   2. 5.6013 kJ/kg

   
   
   

   3. 7.6013 kJ/kg

   
   
   

   4. 8.6013 kJ/kg

   
   
   

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1. View Hint View Answer Discuss in Forum

   Tds = du + pdv (closed system)
   283.15 × 10(S₂ – S₁) = 16.2 × 10³
   

   ∴ S2 = 0.88 +	16.2 × 103
   	283.15 × 10

   
   = 6.6013 kJ/kg

   Correct Option: A

   Tds = du + pdv (closed system)
   283.15 × 10(S₂ – S₁) = 16.2 × 10³
   

   ∴ S2 = 0.88 +	16.2 × 103
   	283.15 × 10

   
   = 6.6013 kJ/kg

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109. An amount of 100 kW of heat is transferred through a wall in steady State. One side of the wall is maintained at 127°C and the other side at 27°C. The entropy generated (in W/K) due to the heat transfer through the wall is ______.

1. 1. 83.33 W / K

   
   
   

   2. 82.33 W / K

   
   
   

   3. 80.33 W / K

   
   
   

   4. 84.33 W / K

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∆S1	Q
   	T1

   
   
   

   ∆S2	Q
   	T2

   
   (∆S)_generated = ∆S₁ + ∆S₂
   

   =		Q	-	Q
   		400		300

   
   
   = - 83.33 W / K

   Correct Option: A

   ∆S1	Q
   	T1

   
   
   

   ∆S2	Q
   	T2

   
   (∆S)_generated = ∆S₁ + ∆S₂
   

   =		Q	-	Q
   		400		300

   
   
   = - 83.33 W / K

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110. Which one of the following pairs of equations describes an irreversible heat engine?

1. 1. ∮ δ Q > 0 and ∮	δQ	< 0
      	T

   
   
   

   2. ∮ δ Q < 0 and ∮	δQ	< 0
      	T

   
   
   

   3. ∮ δ Q > 0 and ∮	δQ	> 0
      	T

   
   
   

   4. ∮ δ Q < 0 and ∮	δQ	> 0
      	T

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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