101. An isolated thermodynamic system executes a process. Choose the correct statement (s) from the following
     

1. 1. No heat is transferred
      

   
   
   

   2. No work is done
      

   
   
   

   3. No mass flows across the boundary of the system
      

   
   
   

   4. No Chemical reaction takes place within the system

   
   
   

   5. ( a , b , c )

   
   
   

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1. View Hint View Answer Discuss in Forum

   For an isolated system, Q = 0, W = 0 and mass does not flow through the system.

   Correct Option: E

   For an isolated system, Q = 0, W = 0 and mass does not flow through the system.

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102. Air of mass 1 kg. initially at 300 K and 10 bar, is allowed to expand isothermally till it reaches a pressure of 1 bar. Assuming air as an ideal gas with gas constant of 0.287 kJ/kg K, the change in entropy of air (in kJ/kg K, round off to two decimal places) is _________.

1. 1. 0.66 kJ/kg K

   
   
   

   2. 1.66 kJ/kg K

   
   
   

   3. 0.33 kJ/kg K

   
   
   

   4. - 1.66 kJ/kg K

   
   
   

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1. View Hint View Answer Discuss in Forum

   Air, m = 1 kg
   
   P₁ = 10 bar
   R = 0.287 kJ/kg k Now,
   ∆s = S₂ – S₁
   

   = mcpln			T2		0 - mRln			P2
   			T1					P1

   
   

   = - 1 × 0.287 ln			1
   			10

   
   = 0.66 kJ/kg k
   ∴ ∆s = 0.66 kJ/kg k

   Correct Option: A

   Air, m = 1 kg
   
   P₁ = 10 bar
   R = 0.287 kJ/kg k Now,
   ∆s = S₂ – S₁
   

   = mcpln			T2		0 - mRln			P2
   			T1					P1

   
   

   = - 1 × 0.287 ln			1
   			10

   
   = 0.66 kJ/kg k
   ∴ ∆s = 0.66 kJ/kg k

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103. For an ideal gas with constant properties undergoing a quasi-static process, which one of the following represents the change of entropy (∆s) from state 1 to 2?

1. 1. ds = cpln		T2	- R ln	P2
      		T1		P1

   
   
   

   2. ds = cpln		T2	- R ln	P2
      		T2		P2

   
   
   

   3. ds = cpln		T1	- R ln	P2
      		T2		P1

   
   
   

   4. ds = cpln		T2	+ R ln	P2
      		T1		P1

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∆ S = Cpln		T2	- R ln	P2
   		T1		P1

   
   Tds = dh – vdp
   = C_p dt – vdp
   

   ds =		Cpdt	-	R	dp
   		T		P

   
   

   ds =		cpln	- R ln	P2
   		T		P1

   
   

   Correct Option: A

   ∆ S = Cpln		T2	- R ln	P2
   		T1		P1

   
   Tds = dh – vdp
   = C_p dt – vdp
   

   ds =		Cpdt	-	R	dp
   		T		P

   
   

   ds =		cpln	- R ln	P2
   		T		P1

   
   

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104. An ideal gas undergoes a process from state 1 (T₁ = 300 K, p₁ = 100 kPa) to state 2 (T₂ = 600 K, p₂ = 500 kPa). The specific heats of the ideal gas are: c_p = 1 kJ/kg-K and c_v = 0.7 kJ/kg-K. The change in specific entropy of the ideal gas from state 1 to state 2 (in kJ/kg-K) is _________ (correct to two decimal places)

1. 1. 0.3 kJ/kg-K

   
   
   

   2. 0.2 kJ/kg-K

   
   
   

   3. 0.4 kJ/kg-K

   
   
   

   4. - 0.3 kJ/kg-K

   
   
   

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1. View Hint View Answer Discuss in Forum

   Ideal gas
   State - 1: T₁ = 300 K, P₁ = 100 kPa
   State - 2: T₂ = 600 K, P₂ = 500 kPa,
   c_p = 1 kJ/kg-K, c_p – c_v = R, c_v = 0.7 kJ/kg-K
   ⇒ c_p – c_v = 1 – 0.7 = R
   R = 0.3 kJ/kg– K
   Change in specific entropy
   

   = 1 ×ln		600	- R ln	P2
   		300		P1

   
   

   Correct Option: A

   Ideal gas
   State - 1: T₁ = 300 K, P₁ = 100 kPa
   State - 2: T₂ = 600 K, P₂ = 500 kPa,
   c_p = 1 kJ/kg-K, c_p – c_v = R, c_v = 0.7 kJ/kg-K
   ⇒ c_p – c_v = 1 – 0.7 = R
   R = 0.3 kJ/kg– K
   Change in specific entropy
   

   = 1 ×ln		600	- R ln	P2
   		300		P1

   
   

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105. One kg of an ideal gas (gas constant R = 287 J/ kgK) undergoes an irreversible process from state-1 (1 bar, 300 K) to state-2 (2 bar, 300 K). The change in specific entropy (s₂ – s₁) of the gas (in J/kgK) in the process is _______.

1. 1. - 198.93 J/kg K

   
   
   

   2. 198.93 J/kg K

   
   
   

   3. 199.93 J/kg K

   
   
   

   4. - 199.93 J/kg K

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∆S = mCpln		T2	- R ln	P2
   		T1		P1

   
   ∆ S = – 287ln2
   ∆S = – 198.93 J/kg K

   Correct Option: A

   ∆S = mCpln		T2	- R ln	P2
   		T1		P1

   
   ∆ S = – 287ln2
   ∆S = – 198.93 J/kg K

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