1. A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62°C, the efficiency of the engine is doubled. The temperatures of the source and sink are​​
   

1. 1. 99°C, 37°C ​

   
   
   

   2. 80°C, 37°C ​

   
   
   

   3. 95°C, 37°C ​

   
   
   

   4. 90°C , 37°C

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​Initially the efficiency of the engine was	1		1	which increases to
   	6		3

   
   when the sink temperature reduces by 62° C.
   

   η =	1	= 1 -	T2	when T2 = sink temperature ​T1 = source temperature
   	6		T1

   
   

   ⇒ T2 =	5	T1
   	6

   
   Secondly,
   

   	1	= 1 -	T2 - 62	= 1 -	T2	+	62	= 1 -	5	+	62
   	3		T1		T1		T1		6		T1

   
   or, T₁ = 62 × 6 = 372K = 372 – 273 = 99°C
   

   & T2 =	5	× 372 = 310 K = 310 - 273 = 37° C
   	6

   Correct Option: A

   ​Initially the efficiency of the engine was	1		1	which increases to
   	6		3

   
   when the sink temperature reduces by 62° C.
   

   η =	1	= 1 -	T2	when T2 = sink temperature ​T1 = source temperature
   	6		T1

   
   

   ⇒ T2 =	5	T1
   	6

   
   Secondly,
   

   	1	= 1 -	T2 - 62	= 1 -	T2	+	62	= 1 -	5	+	62
   	3		T1		T1		T1		6		T1

   
   or, T₁ = 62 × 6 = 372K = 372 – 273 = 99°C
   

   & T2 =	5	× 372 = 310 K = 310 - 273 = 37° C
   	6

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2. A mass of diatomic gas (γ = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C. The pressure of the gas in final state is
   

1. 1. 28 atm

   
   
   

   2. 68.7 atm

   
   
   

   3. 256 atm

   
   
   

   4. 8 atm

   
   
   

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1. View Hint View Answer Discuss in Forum

   T₁ = 273 + 27 = 300K ​
   T₂ = 273 + 927 = 1200K​ ​
   For adiabatic process, ​P^1–γ T^γ = constant ​
   ⇒ P₁​^1–γ T₁^γ = P₂​^1–γ T₂^γ
   

   ⇒		P2		1-γ	=		T1		γ
   		P1					T2

   
   

   ⇒		P1		1-γ	=		T2		γ
   		T2					T1

   
   

   ⇒		P1		1-1.4	=		1200		1.4
   		P2					300

   
   

   ⇒		P1		-0.4	= (4)1.4
   		P2

   
   

   		P2		0.4	= (4)1.4
   		P1

   
   
   = P₁ (2⁷) = 2 × 128 = 256 atm

   Correct Option: C

   T₁ = 273 + 27 = 300K ​
   T₂ = 273 + 927 = 1200K​ ​
   For adiabatic process, ​P^1–γ T^γ = constant ​
   ⇒ P₁​^1–γ T₁^γ = P₂​^1–γ T₂^γ
   

   ⇒		P2		1-γ	=		T1		γ
   		P1					T2

   
   

   ⇒		P1		1-γ	=		T2		γ
   		T2					T1

   
   

   ⇒		P1		1-1.4	=		1200		1.4
   		P2					300

   
   

   ⇒		P1		-0.4	= (4)1.4
   		P2

   
   

   		P2		0.4	= (4)1.4
   		P1

   
   
   = P₁ (2⁷) = 2 × 128 = 256 atm

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