1. A polyatomic gas with n degrees of freedom has a mean energy per molecule given by

1. 1. nkT

      N

   
   
   

   2. nkT

      2N

   
   
   

   3. nkT

      2

   
   
   

   4. 3kT

      2

      
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   According to law of equipartition of energy, the energy per degree of freedom is
   

   	1	kT
   	2

   
   For a polyatomic gas with n degrees of freedom, the mean energy per molecule
   

   =	1	nkT
   	2

   
   

   Correct Option: C

   According to law of equipartition of energy, the energy per degree of freedom is
   

   	1	kT
   	2

   
   For a polyatomic gas with n degrees of freedom, the mean energy per molecule
   

   =	1	nkT
   	2

   
   

---

2. One mole of an ideal monoatomic gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is
   [Given the gas constant R = 8.3 J/ mol. K]

1. 1. 198.7 J

   
   
   

   2. 29 J

   
   
   

   3. 215.3 J

   
   
   

   4. 124 J

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Cp =	5	R and Cv =	3	R
   	2		2

   
   We know that Q_v = nC_v∆T and Q_p = nC_p∆T
   

   ⇒	Qv	=	3	.
   	Qp		5

   
   ⇒ . Given Q_p = 207 J ⇒ Q_v ≅ 124 J
   

   Correct Option: D

   Cp =	5	R and Cv =	3	R
   	2		2

   
   We know that Q_v = nC_v∆T and Q_p = nC_p∆T
   

   ⇒	Qv	=	3	.
   	Qp		5

   
   ⇒ . Given Q_p = 207 J ⇒ Q_v ≅ 124 J
   

---

---

3. For a certain gas the ratio of specific heats is given to be γ = 1.5. For this gas

1. 1. C_V = 3R/J

   
   
   

   2. C_P = 3R/J

   
   
   

   3. C_P = 5R/J

   
   
   

   4. C_V = 5R/J

   
   
   

---

1. View Hint View Answer Discuss in Forum

   γ =	CP	=	15	=	3	⇒ CV =	2	CP
   	CV		10		2		3

   
   

   CP - CV =	R	⇒ CP -	2	CP =	R
   	J		3		J

   
   

   ⇒	CP	=	R	⇒ CP =	3R
   	3		J		J

   
   

   Correct Option: B

   γ =	CP	=	15	=	3	⇒ CV =	2	CP
   	CV		10		2		3

   
   

   CP - CV =	R	⇒ CP -	2	CP =	R
   	J		3		J

   
   

   ⇒	CP	=	R	⇒ CP =	3R
   	3		J		J

   
   

---

4. For hydrogen gas, C_p – C_v = a and for oxygen gas, C_p – C_v = b, so the relation between a and b is given by

1. 1. a = 16 b

   
   
   

   2. 16 b = a

   
   
   

   3. a = 4 b

   
   
   

   4. a = b

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Both are diatomic gases and C_p – C_v = R for all gases.
   

   Correct Option: D

   Both are diatomic gases and C_p – C_v = R for all gases.
   

---

---

5. If for a gas, R/C_v = 0.67 , the gas is made up of molecules which are

1. 1. diatomic

   
   
   

   2. mixture of diatomic and polyatomic molecules

   
   
   

   3. monoatomic

   
   
   

   4. polyatomic

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Since	R	= 0.67 ⇒ CV =	3	R, hence gas is monoatomic.
   	CV		2

   
   

   Correct Option: C

   Since	R	= 0.67 ⇒ CV =	3	R, hence gas is monoatomic.
   	CV		2

   
   

---