The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

: The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The document The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev is a part of the Course I. E. Irodov Solutions for Physics Class 11 & Class 12.

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,    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev                             (1)

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

pdV + Vdp = v RdT                      (2)

Thus from (1) and (2) 

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Dividing both the sides by pV

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

or, The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

and    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

By the first law of thermodynamics Q = ΔU + A 

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

Here,    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

In the polytropic process molar heat capacity is given by

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

From first law of thermodynamics

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b) Sought work done,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

 
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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b) Work done by the gas is given by

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(c) Molar heat capacity is given by

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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.  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

where Cn is the molar heat capacity in the process. It is given that  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

or,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(c)   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

But from part (a), we have  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

From part (b); we know  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So, The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev (where T is the final temperature)

Work done by the gas for one mole is given by

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Here polytropic exponent  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b) Molar heat capacity is given by

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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.

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So polytropric index n = 1 - α

(a) Work done by the gas is given by

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Hence   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b) Molar heat capacity is given by

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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: 

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Solution. 52. By the first law .of thermodynamics

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Molar specific heat according to definition

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

We have   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

After differentiating, we get  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Hence  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b)   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev (for one mole of gas)

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRevThe First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Therefore   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Hence  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b) Work done is given by

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

By the first law of thermodynamics Q = ΔU +A

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev   (see solution of 2.52)

We have  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

After differentiating, we get,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Hence   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b)  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev for one mole of gas

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Now  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

By the first law of thermodynamics Q = ΔU + A

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Integrating both sides, we get  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev is a constant.

Or,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

and  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Integrating both sides, we get   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRevThe First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or     The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev  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,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Given    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b) The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Given   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Integrating both sides, we get

or, The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

But from second law

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

On the other hand   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,    The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b) From the first law  

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b)  We use  

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Now,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So along constant p, The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

On differentiating,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

and   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRevas the vessels are themally insulated. 

As this is free expansion,   The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

But  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,  The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

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. The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev (as A = 0 in free expansion).

So at constant temperature.

The First Law Of Thermodynamics Heat Capacity (Part - 2) - Heat, Irodov JEE Notes | EduRev

= 0.33 kJ from the given data.

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