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

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

HC Verma Solutions of Ch-27  Vol-2 Concept of Physics for Class-12
(Page-76)

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 CAand CB be the molar heat capacities for the two processes,

(a) CA = CB

(b) CA < CB

(c) CA > CB

(d)  CA and CB cannot be defined.

Answer-1 :-

The option (c) CA > CB is 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. ΔUA = ΔUB.

But as , ΔWA =ΔWB this gives ΔQA = 2ΔQB.

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 ΔQA = 2ΔQB, CA > CB.

Question-2 :-

For a solid with a small expansion coefficient,

(a) Cp − Cv = R

(b) Cp = Cv

(c)  Cp is slightly greater than Cv

(d)  Cp is slightly less than Cv

Answer-2 :-

The option(c)  Cp is slightly greater than Cv is correct
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, Cp is slightly greater than Cv.

Answer-5 :-

The option (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,

hc verma specific heat capacity of gasas img 4

For an ideal gas ,

CP – CV = R = 8.314 J – mol -1 K-1 ≃ 2 calories mol-1K-1

⇒ CV =CP -R

⇒ CV = (7-2) calories – mol-1 K-1

⇒ CV = 5 calories – mol-1 K-1

hc verma specific heat capacity of gasas img 5

⇒ 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 oC 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.

The figure 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 = Cv

(c) C > Cv

(d) C < Cv

Answer-6 :-

The option (c)  C > Cv is correct
Explanation:

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

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 Cp, as pressure is constant in this process and the molar heat capacity of process CB is Cv, as volume is constant in it.
Internal energy, U, is a state function, i.e. it doesn’t depend on the path followed. Therefore,

UAB = UACB

WAB > WACB

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

⇒ WAB + UAB > WACB + UACB

⇒ C > CV + CP

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 > CV

Question-7 :-

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

(a) C = CpThe molar heat capacity for the process shown in the figure is

(b)  C = Cv

(c)  C > Cv

(d)  C = 0.

Answer-7 :-

The option (d)  C = 0. is correct
Explanation:

C = 0.he defined process is

p=k/Vg

⇒ pVg =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)

The option (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,

pV = p‘V’

hc verma specific heat capacity of gasas img 8

= -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 :-

The option (a) 0.36% is correct
Explanation:
Let p and p, be the initial and final pressures of the system and V and Vbe the initial and final volumes of the system. p, is 0.5% more than p and the process is adiabatic. So,
hc verma specific heat capacity of gasas img 9

⇒ 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 pA and pBrespectively.

(a) pA > pB

(b) pA = pB

(c) pA < pB

(d) The relation between pA and pB cannot be deduced.

Answer-10 :-

The option (a) pA < pB is correct
Explanation:

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 pA and pBrespectively.

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 Vo. 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, pA < pB.

Let ∆Wa and ∆Wb be the work done by the systems A and B, respectively, in the previous question.

(a) ∆Wa > ∆Wb

(b) ∆Wa = ∆Wb

(c) ∆Wa < ∆Wb

(d) The relation between ∆Wa and ∆Wb cannot be deduced.

Answer-12 :-

The option (c) ∆Wa < ∆Wb is correct
Explanation:

Let ∆Wa and ∆Wb be the work done by the systems A and B, respectively, in the previous question.

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,  ∆Wa < ∆Wb.

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