21. The molar specific heats of an ideal gas at constant pressure and volume are denoted by C_p and C_v, respectively. If γ = C_p/C_v and R is the universal gas constant, then C_v is equal to

1. 1. R

      γ - 1

   
   
   

   2. (γ - 1)
      R

   
   
   

   3. γR

   
   
   

   4. 1 + γ
      1 - γ

   
   
   

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1. View Hint View Answer Discuss in Forum

   C_p – C_v = R ⇒ C_p = C_v + R
   

   ∵ γ =	Cp	=	Cv + R	⇒	Cv	+	R
   	Cv		Cv		Cv		Cv

   
   

   ⇒ γ = 1 +	R	⇒	R	= γ - 1
   	Cv		Cv

   
   

   ⇒ Cv =	R
   	&gamm; - 1

   Correct Option: A

   C_p – C_v = R ⇒ C_p = C_v + R
   

   ∵ γ =	Cp	=	Cv + R	⇒	Cv	+	R
   	Cv		Cv		Cv		Cv

   
   

   ⇒ γ = 1 +	R	⇒	R	= γ - 1
   	Cv		Cv

   
   

   ⇒ Cv =	R
   	&gamm; - 1

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22. The molar specific heat at constant pressure of an ideal gas is (7/2) R. The ratio of specific heat at constant pressure to that at constant volume is

1. 1. 8/7

   
   
   

   2. 5/7

   
   
   

   3. 9/7

   
   
   

   4. 7/5

   
   
   

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1. View Hint View Answer Discuss in Forum

   Cp =	7	R; CV = Cp - R =	7	R - R =	5	R =	Cp	=	7/2 R	=	7
   	2		2		2		Cv		5/2 R		5

   Correct Option: D

   Cp =	7	R; CV = Cp - R =	7	R - R =	5	R =	Cp	=	7/2 R	=	7
   	2		2		2		Cv		5/2 R		5

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23. If γ be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of the gas is

1. 1. 25	(γ - 1)
      2

   
   
   

   2. 3γ - 1

      2γ - 1

   
   
   

   3. 2

      γ - 1

   
   
   

   4. 9	(γ - 1)
      2

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that ratio of specific heats,
   

   γ = 1 +	2	or n =	2
   	n		γ - 1

   
   [where n = Degree of freedom]

   Correct Option: C

   We know that ratio of specific heats,
   

   γ = 1 +	2	or n =	2
   	n		γ - 1

   
   [where n = Degree of freedom]

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24. The degree of freedom of a molecule of a triatomic gas is

1. 1. 2

   
   
   

   2. 4

   
   
   

   3. 6

   
   
   

   4. 8

   
   
   

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1. View Hint View Answer Discuss in Forum

   No. of degree of freedom = 3 K – N
   where K is no. of atom and N is the number of relations between atoms. For triatomic gas,
   K = 3, N = ³C₂ = 3
   No. of degree of freedom = 3 (3) – 3 = 6

   Correct Option: C

   No. of degree of freedom = 3 K – N
   where K is no. of atom and N is the number of relations between atoms. For triatomic gas,
   K = 3, N = ³C₂ = 3
   No. of degree of freedom = 3 (3) – 3 = 6

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