36. A mixture of ideal gases has the following composition by mass :
    

    N2	O2	CO2
    60%	30%	10%

    
    If the universal gas constant is 8314 J/kmolK, the characteristic gas constant of the mixture (in J/kgK) is _______.

1. 1. 274.996

   
   
   

   2. 247.996

   
   
   

   3. 374.996

   
   
   

   4. 174.996

   
   
   

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1. View Hint View Answer Discuss in Forum

   Gas constant of mixture,
   

   Rm =	Universal gas constant
   	Average molar mass

   
   

   Rm =	Universal gas constant
   	Average molar mass

   
   

   Average molar mass =	100						= 30.233 kg/kmol
   	60	+	30	+	10
   	28		32		44

   
   

   Rm =	8314	= 274.996 J/kg -K
   	30.233

   
   

   Correct Option: A

   Gas constant of mixture,
   

   Rm =	Universal gas constant
   	Average molar mass

   
   

   Rm =	Universal gas constant
   	Average molar mass

   
   

   Average molar mass =	100						= 30.233 kg/kmol
   	60	+	30	+	10
   	28		32		44

   
   

   Rm =	8314	= 274.996 J/kg -K
   	30.233

   
   

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37. The internal energy of an ideal gas is a function of
    

1. 1. temperature and pressure
      

   
   
   

   2. volume and pressure
      

   
   
   

   3. entropy and pressure
      

   
   
   

   4. temperature only

   
   
   

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1. View Hint View Answer Discuss in Forum

   temperature only

   Correct Option: D

   temperature only

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38. A piston-cylinder device initially contains 0.4 m³ of air (to be treated as an ideal gas) at 100 kPa and 80°C. The air is now isothermally compressed to 0.1 m³. The work done during this process is ______kJ.
    (Take the sign convention such that work done on the system is negative)
    

1. 1. 55.45 kJ

   
   
   

   2. -21.45 kJ

   
   
   

   3. -55.45 kJ

   
   
   

   4. -15.45 kJ

   
   
   

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1. View Hint View Answer Discuss in Forum

   For isothermal process
   

   W = P1V1ln		V2
   		V1

   
   

   W = 100 × 0.4 ln		0.1		= -55.45 kJ
   		0.4

   Correct Option: C

   For isothermal process
   

   W = P1V1ln		V2
   		V1

   
   

   W = 100 × 0.4 ln		0.1		= -55.45 kJ
   		0.4

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39. The first law of thermodynamics takes the form W = –∆H when applied to
    

1. 1. A closed system undergoing a reversible adiabatic process
      

   
   
   

   2. An open system undergoing an adiabatic process with negiigible changes in kinetic and potential energies
      

   
   
   

   3. A closed system undergoing a reversible constant volume process
      

   
   
   

   4. A closed system undergoing a reversible constant pressure process.

   
   
   

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1. View Hint View Answer Discuss in Forum

   Steady flow equation is given be,
   H₁ + q = H₂ + W
   Adiabatic q = 0
   H₁ – H₂ = W
   W = -∆H

   Correct Option: B

   Steady flow equation is given be,
   H₁ + q = H₂ + W
   Adiabatic q = 0
   H₁ – H₂ = W
   W = -∆H

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40. In the case of a refrigeration system undergoing an irreversible cycle, ∮	δQ	is ....(< 0/ = 0/ > 0)
    	T

1. 1. ∮	δQ	< 0
      	T

   
   
   

   2. ∮	δQ	> 0 clausius inequality
      	T

   
   
   

   3. δQ > 0

   
   
   

   4. δQ = 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∮	δQ	< 0 clausius inequality
   	T

   Correct Option: A

   ∮	δQ	< 0 clausius inequality
   	T

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