# Obj-1 Specific Heat Capacities of Gases HC Verma Solutions Vol-2 Class-12 Ch-27

**Obj-1 Specific Heat Capacities of Gases** HC Verma Solutions Vol-2 Class-12 Ch-27. Step by Step Solutions of** Objective -1 (MCQ-1) **Questions of Chapter-27 **Specific Heat Capacities of Gases **(Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

**Obj-1 Specific Heat Capacities of Gases **HC Verma Solutions Vol-2 Class-12 Ch-27

Board | ISC and other board |

Publications | Bharti Bhawan Publishers |

Chapter-27 | Specific Heat Capacities of Gases |

Class | 12 |

Vol | 2nd |

writer | HC Verma |

Book Name | Concept of Physics |

Topics | Solution of Objective-1 (MCQ-1) Questions |

Page-Number | 76, 77 |

-: Select Topics :-

Objective-I (Currently Open)

Objective-II (update soon)

Exercise (update soon)

**Specific Heat Capacities of Gases Objective-1 (MCQ-1) Questions**

**Question-1 :-**

Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If C_{A}and C_{B} be the molar heat capacities for the two processes,

(a) C_{A} = C_{B}

(b) C_{A} < C_{B}

(c) C_{A} > C_{B}

(d) C_{A} and C_{B} cannot be defined.

**Answer-1 :-**

**(c) C**

_{A}> C_{B }

**i**s correct

**Explanation:**

According to the first law of thermodynamics, ΔQ = ΔU + ΔW, where Δ Q is the heat supplied to the system when ΔW work is done on the system and ΔU is the change in internal energy produced. Since the temperature rises by the same amount in both processes, change in internal energies are same, i.e. ΔU_{A} = ΔU_{B}.

But as , ΔW_{A} =ΔW_{B} this gives ΔQ_{A} = 2ΔQ_{B}.

Now, molar heat capacity of a gas, C = △Q/n△T, where Δ Q/n is the heat supplied to a mole of gas and ΔT is the change in temperature produced. As ΔQ_{A} = 2ΔQ_{B}, C_{A} > C_{B}.

**Question-2 :-**

For a solid with a small expansion coefficient,

(a) C_{p} − C_{v} = R

(b) C_{p} = C_{v}

(c) C_{p} is slightly greater than C_{v}

(d) C_{p} is slightly less than C_{v}

**Answer-2 :-**

**(c) C**s correct

_{p}is slightly greater than C_{v }i**Explanation:**

For a solid with a small expansion coefficient, work done in a process will also be small. Thus, the specific heat depends slightly on the process. Therefore, C_{p} is slightly greater than C_{v}_{.}

**Question-3 :-**

The value of C_{p} − C_{v}_{ }is 1.00 R for a gas sample in state A and 1.08 R in state B. Let p_{A}and p_{B} denote the pressures and T_{A} and T_{B} denote the temperatures of the states A and B, respectively. It is most likely that

(a) p_{A} < p_{B} and T_{A} > T_{B}

(b) p_{A} > p_{B} and T_{A} < T_{B}

(c) p_{A} = p_{B} and T_{A} < T_{B}

(d) p_{A} > p_{B} and T_{A} = T_{B}

**Answer-3 :-**

**(a) p**

_{A}< p_{B}and T_{A}> T_{B }is correct

**Explanation:**

C_{p} − C_{v }= R for the gas in state A, which means it is acting as an ideal gas in that state, whereas C_{p} − C_{v} = 1.08R in state B, i.e. the behaviour of the gas is that of a real gas in that state. To be an ideal gas, a real gas at STP should be at a very high temperature and low pressure. Therefore, P_{A }< P_{B} and T_{A} >^{ }T_{B }where P_{A }and P_{B} denotes the pressure and T_{A}and T_{B} denotes the temperature of system A and B respectively.

**Question-4 :- (Obj-1 Specific Heat Capacities of Gases HC Verma)**

Let C_{v} and C_{p} denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively. Which of the following is a universal constant?

(a) CP/CP

(b) C_{p}C_{v}

(c) C_{p} − C_{v}

(d) C_{p} + C_{v}

**Answer-4 :-**

**(c) C**is correct

_{p}− C_{v}_{ }**Explanation:**

For an ideal gas, C_{p} − C_{v} = R , where C_{v} and C_{p} denote the molar heat capacities of an ideal gas at constant volume and constant pressure, respectively and R is the gas constant whos value is 8.314 J/K. Therefore, C_{p} − C_{v} is a constant. On the other hand, the ratio of these two varies as the atomicity of the gas changes. Also, their sum and product are not constant.

**Question-5 :-**

70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is

(a) 30 calories

(b) 50 calories

(c) 70 calories

(d) 90 calories

**Answer-5 :-**

**(b) 50 calories**is correct

_{ }**Explanation:**

It is given that 70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. Also, specific heat at constant pressure,

For an ideal gas ,

C_{P} – C_{V} = R = 8.314 J – mol ^{-1} K^{-1} ≃ 2 calories mol^{-1}K^{-1}

⇒ C_{V} =C_{P} -R

⇒ C_{V} = (7-2) calories – mol^{-1} K^{-1}

⇒ C_{V} = 5 calories – mol^{-1} K^{-1}

^{}

⇒ Δ Q = 5 × 2 × (35-30)

⇒ Δ Q = 5 × 2 × 5

⇒ Δ Q = 50 calories

Therefore, 50 calories need to be supplied to raise the temperature of 2 moles of gas from 30-35 ^{o}C at constant volume.

**Question-6 :-**

The figure (figure 27-Q1) shows a process on a gas in which pressure and volume both change. The molar heat capacity for this process is C.

^{}

(a) C = 0

(b) C = C_{v}

(c) C > C_{v}

(d) C < C_{v}

**Answer-6 :-**

**(c) C > C**is correct

_{v }**Explanation:**

^{}

Consider two processes AB and ACB; let W be the work done. C is the molar heat capacity of process AB. Process ACB can be considered as the sum of the two processes, AC and CB. The molar heat capacity of process AC is C_{p}_{,} as pressure is constant in this process and the molar heat capacity of process CB is C_{v}, as volume is constant in it.

Internal energy, U, is a state function, i.e. it doesn’t depend on the path followed. Therefore,

U_{AB} = U_{ACB}

W_{AB} > W_{ACB}

Work done in the *p-V* diagram is the area enclosed under the curve.

⇒ W_{AB} + U_{AB} > W_{ACB} + U_{ACB}

⇒ C > C_{V} + C_{P}

Molar heat capacity is the heat supplied per mole to change the temperature by a degree Kelvin and according to the first law of thermodynamics, dQ = dU + dW, where dQ is the heat supplied to the system in a process.

⇒ C > C_{V}

**Question-7 :-**

The molar heat capacity for the process shown in the figure is

(a) C = C_{p}^{}

(b) C = C_{v}

(c) C > C_{v}

(d) C = 0.

**Answer-7 :-**

**(d) C = 0.**is correct

**Explanation:**

C = 0.he defined process is

p=k/V^{g}

⇒ p^{Vg} =k,

such that the process is adiabatic in which there’s no heat supplied to the system, i.e. Q= 0. Molar heat capacity is the amount of heat supplied to the system per mole to produce a degree change in temperature. Also, in an adiabatic process, no heat exchange is allowed. So, molar heat capacity equals zero, i.e. C = 0.

**Question-8 :-**

In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about

(a) 0.25%

(b) 0.5%

(c) 0.7%

(d) 1%.

**Answer-8 :- (Obj-1 Specific Heat Capacities of Gases HC Verma)**

**(b) 0.5%**is correct

**Explanation:**

Let p and p’ be the initial and final pressures of the system and V and V’ be the initial and final volumes of the system. p’ is 0.5% more than p and the process is isothernal. So, pV = k = p’V’ = constant. Therefore,

p^{V} = p^{‘V’}

^{}

= -0.49 %

So, volume V’ decreases by about 0.5% of V.

**Question-9 :-**

In an adiabatic process on a gas with γ = 1.4, the pressure is increased by 0.5%. The volume decreases by about

(a) 0.36%

(b) 0.5%

(c) 0.7%

(d) 1%

**Answer-9 :-**

**(a) 0.36%**is correct

**Explanation:**

^{,}be the initial and final pressures of the system and V and V

^{, }be the initial and final volumes of the system. p

^{,}is 0.5% more than p and the process is adiabatic. So,

⇒ V’ = 0.99644 V

⇒ V’ =V -0.00356 V

Therefore, V’ is 0.36 % less than V.

**Question-10 :-**

Two samples A and B are initially kept in the same state. Sample A is expanded through an adiabatic process and the sample B through an isothermal process. The final volumes of the samples are the same. The final pressures in A and B are p_{A} and p_{B}respectively.

(a) p_{A} > p_{B}

(b) p_{A} = p_{B}

(c) p_{A} < p_{B}

(d) The relation between p_{A} and p_{B} cannot be deduced.

**Answer-10 :-**

**(a) p**is correct

_{A}< p_{B }**Explanation:**

Let the initial states of samples A and B be i and the final states of samples B and A be fand f’, respectively. Let the final volumes of both be V_{o}. As sample A is expanded through an adiabatic process, its curve in the p-V diagram is steeper than that of sampleB, which is expanded through an isothermal process. Therefore, from the p-V diagram, p_{A} < p_{B}.

**Question-11 :-**

Let T_{a} and T_{b} be the final temperatures of the samples A and B, respectively, in the previous question.

(a) T_{a} < T_{b}

(b) T_{a} = T_{b}

(c) T_{a} > T_{b}

(d) The relation between T_{a}_{ }and T_{b} cannot be deduced.

**Answer-11 :- (Obj-1 Specific Heat Capacities of Gases HC Verma)**

**(a) T**is correct

_{a}< T_{b }**Explanation:**

_{b}. On the other hand, sample A is at the same initial state as B, such that the initial temperature of A is

_{}T

_{b}and it is expanding through an adiabatic process in which no heat is supplied. Therefore, sample A will expand at the cost of its internal energy and its final temperature will be less than its initial temperature.

This implies that T

_{a}< T

_{b}.

**Question-12 :-**

Let ∆W_{a}_{ }and ∆W_{b} be the work done by the systems A and B, respectively, in the previous question.

(a) ∆W_{a}_{ }> ∆W_{b}

(b) ∆W_{a} = ∆W_{b}

(c) ∆W_{a} < ∆W_{b}

(d) The relation between ∆W_{a} and ∆W_{b} cannot be deduced.

**Answer-12 :-**

**(c) ∆W**is correct

_{a}< ∆W_{b }**Explanation:**

In the p-V diagram, the area under the curve w.r.t the V axis is equal to the work done by the system. Since the area under the isotherm is greater than that under the adiabat, the work done by system A is less than that done by system B. Hence, ∆W_{a} < ∆W_{b}.

**Specific Heat Capacities of Gases Objective-1 (MCQ-1) Questions**

### HC Verma Solutions of Ch-27 Vol-2 Concept of Physics for Class-12

**Question-13 :-**

The molar heat capacity of oxygen gas at STP is nearly 2.5 R. As the temperature is increased, it gradually increases and approaches 3.5 R. The most appropriate reason for this behaviour is that at high temperatures

(a) oxygen does not behave as an ideal gas

(b) oxygen molecules dissociate in atoms

(c) the molecules collide more frequently

(d) molecular vibrations gradually become effective

**Answer-13 :-**

**(d) molecular vibrations gradually become effective**is correct

**Explanation:**

Molar specific heat capacity has direct dependence on the degree of freedom of gas molecules. As temperature is increased, the gas molecules start vibrating about their mean position, leading to change (increase) in the degree of freedom and, hence, increasing molar heat capacity.

—: End of Specific Heat Capacities of Gases **Obj-1 (mcq-1) HC Verma** Solutions Vol-2 Chapter-27 :–

Return to — HC Verma Solutions Vol-2 Concept of Physics

Thanks