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The candidates have to first appear in Paper-1 of JEE(Main)-2014. Only 1,50,000 of the top scorers of Paper-1 of JEE(Main)-2014, including all categories, will be eligible to write JEE(Advanced) – 2014. Candidates who appeared in their qualifying examination (QE) (10+2 or equivalent) earlier than 2013 are not eligible.

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Specific Heat Capacity Of Gases

Specific Heat Capacity Of Gases

Specific heat of a gas is numerically equal to the amount of heat necessary to raise the temperature of unit mass of gas by 1°C. In order to raise the temperature of unit mass of a gas through 1°C more heat will be required if the gas was kept at constant pressure than when it is at constant volume.

(i) Molar Specific heat capacity at constant Volume: The amount of heat required to raise the temperature of 1 mole of gas by 1 °C at constant volume is called the molar specific heat and it is represented by C_v.

            C_v = (ΔQ/mΔt)_v = constant

By first law of thermodynamics ΔQ = ΔU + W

But W = 0 for isochoric process, then ΔQ = ΔU

by definition of specific heat ΔQ = nC_vΔT

Where CV is the specific heat (for 1 mole of gas), then ΔU = nCvΔT The relation ΔU = nCvΔT, is used to find the change in internal energy of the system and is valid for any process where a temperature change has taken place. Consider two isotherms on the P-V diagram: Process 1 –> 2 represents an isochoric process Process 1 –> 3 represents an isobaric process

In both processes the temperature has changes from T₁ to T₂ as 2 and 3 lie on the same isotherm. Thus change in internal energ

             U_1–>2 = U_1–>3 = nC_vΔT

(ii) Molar Specific heat capacity at constant Pressure: The amount of heat required to raise the temperature of 1 mol of gas by 1°C keeping its pressure constant, is called molar specific heat at constant pressure and it is represented by C_p,

             C_p = (ΔQ/mΔT)_P  = constant

(iii) Relation between C_p and C_v: For an isobaric process by definition

From first law ΔQ = ΔU + W

=> nC_PΔT = nC_VΔT + W      [∴ ΔU = nC_VΔT for any process]

For isobaric process

W = PΔV = nRΔT            nC_PΔT = nC_VΔT + nRΔT

=> C_P = C_V + R      or     C_P ­- C_V = R     (Meyer’s relation)

(iv) Relation between specific heat (CP and CV) and degrees of freedom: If f is the number of degree of freedom of a gas molecule then the internal energy of n moles of that gas is given by

U = f × 1/2 nRT            (from law of equipartition of energy)

(∴ U = f/2 kT = f/2 R/N T (for n moles) = nN × f/2 R/N T = n f/2 RT)

=> dU = f × 1/2 nRdT

Also dU = nCVdT, 

So, nCVdT = f × 1/2 nRdTf/2 R

but CP = CV + R

∴ C_P = (f/2 f/2 + 1)R. This is the relation between specific heat ratio and degree of freedom.

(v) Adiabatic Expansion of an Ideal gas: As defined earlier in adiabatic process ΔQ = 0

For a system containing an ideal gas adiabatic process can happen in two ways

(i) if the system boundary is adiabatic

(ii) if the system boundary is diathermic but the process takes place so fast that their is no time for the transfer of the heat. For example, Propagation of sound in air.  

For an adiabatic process as shall be proved

PV? = a constant, here g = C_P/ C_V. ? is known as adiabatic constant

Since PV = nRT for an ideal gas

P = nRT/V so,

(nRT/V)V^g = Constant           ∴ TV^g–1 = a constant

Proof: Let n moles of an ideal gas expand adiabatically by a small amount ΔV.

By first law   ΔQ = ΔU + PΔV

As ΔQ = 0    we have ΔU = –PΔV

But       U = nC_vΔT

So,       nΔT = –(P/C_v)ΔV                                                                               …(i)

From the ideal gas law (after differentiating) ,          PΔV = VΔP = nRΔT

Replacing R by its equal, C_P ­- C_V leads to this,         nΔT = PΔV + VΔP/C_p – C_v        …(ii)

equating (1) and (2) with a little algebra leads to

            ΔP/P + (C_p/C_v) ΔV/V = 0            Cp/Cv = g

and integrating we have, ln P + g In V = a constant,

or PV^g = a constant, hence proved.

Again g = C_p/C_v = C_v + R/C_V = 1 + R/C_v = 1 + R/(1/2 fR) = 1 + 2/f

(vi) Work done in an adiabatic process (P₁V₁T₁) to (P₂V₂T₂)

For adiabatic process, W = –ΔU = –n C_vΔT

From ideal gas equation P₁V₁ – P₂V₂ = nR (T₁ – T₂) = – nRT

     W = C_v(P₁V₁ – P₂V₂)/R(P₁V₁ – P₂V₂)/?–1 = nR(T₁ – T₂)/–1

(vii) The table shows the values of f, Cv, Cp and g for different gasses:

Nature of gas	Degree of freedom f = (T+R+V)	Cv	Cp	g
Monatomic	3+0+0 = 3	3	5	5/3
Diatomic	3+2+0 = 5	5	7	7/5
Polyatomic (linear)	3+2+0 = 5	5	7	7/5
Polyatomic (non-linear)	3+2+1 = 6	3R	4R	4/3

Note: At room temperature the energy associated with vibrational motion is negligible in comparison to translational and rotational KE.

Expressions For ΔU, W, And ΔQ For Different Process

Processes	Relation between thermodynamic variables	Work Done (W)	Heat Exchange (ΔQ)
Isothermal Process (T constant)	P∝1/V	W = 2.303 nRT log10 (V2/V1)	ΔQ = 2.303 nRT log10(V2/V1)
Adiabatic Process (No heat exchange)	PVg = constant	W = (P1V1 – P2V2)/(g–1) = nR(T1– T2)/(g-1)	ΔQ = 0
Isochoric   Process (V = constant)	P ∝T	W = 0	ΔQ = n CV ΔT* (Use definition of Cv)
Isobaric Process (P = Constant)	V ∝T	W = P Δ V = P(V2 –V2) W = nR (T2 –T1)	ΔQ = n CP ΔT* (Use definition of Cp)