Irodov Solutions: The First Law of Thermodynamics: Heat Capacity - 2

Irodov Solutions: The First Law of Thermodynamics: Heat Capacity - 2 | Physics Class 11 - NEET PDF Download

Q. 44. Demonstrate that the process in which the work performed by an ideal gas is proportional to the corresponding increment of its internal energy is described by the equation pVn = const, where n is a constant.

Solution. 44. According to the problem : A α U or dA = aU (where a is proportionality constant)

or,                                 (1)

From ideal gas law, pV= v R T, on differentiating

pdV + Vdp = v RdT                      (2)

Thus from (1) and (2)

or,

Dividing both the sides by pV

On integrating n In V + In p = In C (where C is constant)

or,

Q. 45. Find the molar heat capacity of an ideal gas in a polytropic process pVn = const if the adiabatic exponent of the gas is equal to γ. At what values of the polytropic constant n will the heat capacity of the gas be negative?

Solution. 45. In the polytropic process work done by the gas

(where Ti and Tf are initial and final temperature of the gas like in adiabatic process)

and

By the first law of thermodynamics Q = ΔU + A

According to definition of molar heat capacity when number of moles v = 1 and ΔT = 1 then Q = Molar heat capacity.

Here,

Q. 46. In a certain polytropic process the volume of argon was increased α = 4.0 times. Simultaneously, the pressure decreased β = 8.0 times. Find the molar heat capacity of argon in this process, assuming the gas to be ideal.

Solution. 46. Let the process be polytropic according to the law pVn = constant

Thus,

So,

In the polytropic process molar heat capacity is given by

So,

Q. 47. One mole of argon is expanded polytropically, the polytropic constant being n = 1.50. In the process, the gas temperature changes by ΔT = — 26 K. Find:
(a) the amount of heat obtained by the gas;
(b) the work performed by the gas

Solution. 47. (a) Increment of internal energy for ΔT, becomes

From first law of thermodynamics

(b) Sought work done,

Q. 48. An ideal gas whose adiabatic exponent equals y is expanded according to the law p = αV , where a is a constant. The initial volume of the gas is equal to V0. As a result of expansion the volume increases η times. Find:
(a) the increment of the internal energy of the gas;
(b) the work performed by the gas;
(c) the molar heat capacity of the gas in the process.

Solution. 48. LaW 0f the process is p = α V or pV-1 = α
so the process is polytropic of index n = - 1
As p = αV so, Pi - αVand pf = α η V0

(a) Increment of the internal energy is given by

(b) Work done by the gas is given by

(c) Molar heat capacity is given by

Q. 49. An ideal gas whose adiabatic exponent equals γ is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Find:
(a) the molar heat capacity of the gas in this process;
(b) the equation of the process in the variables T, V;
(c) the work performed by one mole of the gas when its volume increases η times if the initial temperature of the gas is T0

Solution. 49.

where Cn is the molar heat capacity in the process. It is given that

So,

(b) By the first law of thermodynamics, dQ - dU + dA,

or,

So,

or,

(c)

But from part (a), we have

Thus

From part (b); we know

So,  (where T is the final temperature)

Work done by the gas for one mole is given by

Q. 50. One mole of an ideal gas whose adiabatic exponent equals y undergoes a process in which the gas pressure relates to the temperature as p = aTα, where a and α are constants. Find:
(a) the work performed by the gas if its temperature gets an increment ΔT;
(b) the molar heat capacity of the gas in this process; at what value of α will the heat capacity be negative?

Solution. 50. Given p = a Tα (for one mole of gas)

So,

or,

Here polytropic exponent

(a) In the poly tropic process for one mole of gas :

(b) Molar heat capacity is given by

Q. 51. An ideal gas with the adiabatic exponent γ undergoes a process in which its internal energy relates to the volume as U = aVα, where a and α  are constants. Find:
(a) the work performed by the gas and the amount of heat to be transferred to this gas to increase its internal energy by ΔU;
(b) the molar heat capacity of the gas in this process.

Solution. 51.

or,

or,

or

So polytropric index n = 1 - α

(a) Work done by the gas is given by

Hence

By the first law of thermodynamics, Q = ΔU+A

(b) Molar heat capacity is given by

Q. 52. An ideal gas has a molar heat capacity Cv at constant volume. Find the molar heat capacity of this gas as a function of its volume V, if the gas undergoes the following process:

Solution. 52. By the first law .of thermodynamics

Molar specific heat according to definition

We have

After differentiating, we get

So,

Hence

(b)

So,

Q. 53. One mole of an ideal gas whose adiabatic exponent equals γ undergoes a process p = p0 + α /V, w here P0 and α are positive constants. Find:
(a) heat capacity of the gas as a function of its volume;
(b) the internal energy increment of the gas, the work performed by it, and the amount of heat transferred to the gas, if its volume increased from V1 to V2.

Solution. 53.  Using 2.52

(for one mole of gas)

Therefore

Hence

(b) Work done is given by

By the first law of thermodynamics Q = ΔU +A

Q. 54. One mole of an ideal gas with heat capacity at constant pressure Cp undergoes the process T = T0 + αV, where T0  and α are constants. Find:
(a) heat capacity of the gas as a function of its volume;
(b) the amount of heat transferred to the gas, if its volume increased from V1 to V2

Solution. 54. (a) Heat capacity is given by

(see solution of 2.52)

We have

After differentiating, we get,

Hence

(b)

for one mole of gas

Now

By the first law of thermodynamics Q = ΔU + A

Q. 55. For the case of an ideal gas find the equation of the process (in the variables T, V) in which the molar heat capacity varies as:
(a) C = Cv  + αT; (b) C = Cv + βV;  (c) C = Cv + ap, where α, β, and a are constants.

Solution. 55. Heat capacity is given by

Integrating both sides, we get   is a constant.

Or,

and

or,

Integrating both sides, we get

So,

So,

or

or,

or V - a T = constant

Q. 56. An ideal gas has an adiabatic exponent γ. In some process its molar heat capacity varies as C = α/T, where α is a constant. Find:
(a) the work performed by one mole of the gas during its heating from the temperature T0 to the temperature η times higher;
(b) the equation of the process in the variables p, V.

Solution. 56. (a) By the first law of thermodynamics A = Q - ΔU

or,

Given

So,

(b)

Given

or,

or,

or,

Integrating both sides, we get

or,

or,

or,

or,

Q. 57. Find the work performed by one mole of a Van der Waals gas during its isothermal expansion from the volume V1 to V2 at a temperature T.

Solution. 57. The work done is

Q. 58. One mole of oxygen is expanded from a volume V1 = 1.00 1 to V2 = 5.0 l at a constant temperature T = 280 K. Calculate:
(a) the increment of the internal energy of the gas:
(b) the amount of the absorbed heat.
The gas is assumed to be a Van der Waals gas.

Solution. 58. (a) The increment in the internal energy is

But from second law

On the other hand

or,

So,

(b) From the first law

Q. 59. For a Van der Waals gas find:
(a) the equation of the adiabatic curve in the variables T, V;
(b) the difference of the molar heat capacities Cp, — Cv  as a function of T and V.

Solution. 59.  (a) From the first law for an adiabatic

dQ = dU + pd V = 0

From the previous problem

So,

This equation can be integrated if we assume that Cv and b are constant then

or,

(b)  We use

Now,

So along constant p,

Thus

On differentiating,

or,

and

Q. 60. Two thermally insulated vessels are interconnected by a tube equipped with a valve. One vessel of volume V1 =  10 l contains v = 2.5 moles of carbon dioxide. The other vessel of volume V2 = 100 l is evacuated. The valve having been opened, the gas adiabatically expanded. Assuming the gas to obey the Van der Waals equation, find its temperature change accompanying the expansion.

Solution. 60.  From the first law

as the vessels are themally insulated.

As this is free expansion,

But

So,

or,

Substitution gives ΔT = - 3 K

Q. 61. What amount of heat has to be transferred to v = 3.0 moles of carbon dioxide to keep its temperature constant while it expands into vacuum from the volume V1 =  5.0 l to V2 = 10 l ? The gas is assumed to be a Van der Waals gas.

Solution. 61.  (as A = 0 in free expansion).

So at constant temperature.

= 0.33 kJ from the given data.

The document Irodov Solutions: The First Law of Thermodynamics: Heat Capacity - 2 | Physics Class 11 - NEET is a part of the NEET Course Physics Class 11.
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FAQs on Irodov Solutions: The First Law of Thermodynamics: Heat Capacity - 2 - Physics Class 11 - NEET

 1. What is the first law of thermodynamics?
Ans. The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed in an isolated system. It can only be transferred or converted from one form to another.
 2. How is heat capacity defined?
Ans. Heat capacity is defined as the amount of heat energy required to raise the temperature of a substance by one degree Celsius. It is a measure of the substance's ability to store thermal energy.
 3. What is the significance of heat capacity in thermodynamics?
Ans. Heat capacity plays a crucial role in thermodynamics as it helps determine the amount of heat energy required to raise the temperature of a substance. It allows us to calculate the change in internal energy of a system and understand its behavior in response to heat transfer.
 4. How is heat capacity different from specific heat capacity?
Ans. Heat capacity is the amount of heat energy required to raise the temperature of a substance, while specific heat capacity is the heat energy required to raise the temperature of a unit mass of the substance. Specific heat capacity is obtained by dividing the heat capacity by the mass of the substance.
 5. Can heat capacity be negative?
Ans. No, heat capacity cannot be negative. It is always a positive value as it represents the amount of heat energy required to raise the temperature of a substance. A negative value would imply that the substance releases heat instead of absorbing it, which is not possible as per the first law of thermodynamics.

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