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Enthalpy:- The heat content of a system at constant pressure is called enthalpy.

Let a system undergo a change from state A to state B by absorbing q calories of heat from the surroundings at a constant pressure P. In such a case change of volume will take place.

Let V_A be the volume of the system in state A and V_B is the volume of the system in state B. The work done by the system is given by equation.

W = p(V_B - V_A)

Substitute the value of W

ΔE = q - p(V_B - V_A)    ....(i)

But ΔE = E_B - E_A

∴ E_B - E_A = q - p(V_B - V_A)   ...(ii)

or, E_B - E_A = q - pV_B + pV_A

or, (E_B + pV_B) - (E_A - pV_A) = q   ...(iii)

The quantity E + pV is representing the total energy stored in a system and is called heat constant or enthalpy of a system. It is denoted by the symbol H

∴ H = E + pV   ...(iv)

E is definite quantity and is determined by the state of the system, p and V are also definite quantities. Therefore the quantity E + pV, i.e., H(enthalpy) is also a definite quantity and is determined by the state of the system.

It is neither possible nor it is required to determine the absolute value of enthalpy of a system. Only the change in enthalpy of the system which takes place due to change in any of the variable of the system is required and can be measured accurately.

For a system A, H_A = E_A + pV_A  ...(v)

For a system B, H_B = E_B + pV_B ...(vi)

Subtracting (vi) from (v),

H_B - H_A = (E_B - E_A) + p(V_B - V_A) ..(vii)

∴ ΔH = ΔE + PΔV  ....(viii)

ΔH represents the increase in that heat content of the system. Like ΔE it is also a definite property and depends on the initial and final state of the system. We can write Q = ΔE + pΔV

Enthalpy changes at constant value. When the volume is kept constant

ΔV = 0

pΔV = 0

∴ ΔH = ΔE

i.e., at constant volume the change in enthalpy is equal to the change in the internal energy.

Enthalpy changes at constant pressure. For a change at constant pressure, the enthalpy change equals the heat absorbed by the substance. Since for a monoatomic gas, all the internal energy is accounted for by the kinetic energy of its molecules, we can estimate enthalpy H, per mole of such a gas at a temperature T as

H = \(\frac{3}{2}\)RT + pV

= \(\frac{3}{2}\)RT + RT = \(\frac{5}{2}\)RT  [Since pV = RT]

γ = C_p/C_v 

C_p - C_v = R

C_p = γC_v 

(γ - 1)C_v = R; C_v = R/γ - 1

C_p = γR/γ - 1

Data : C_v = 12.47 J/mol.k, n = 2, T_f – T_i = 10 K 

The increase in the internal energy of the gas,

∆ U = _nC_v (T_f – T_i)

= (2) (12.47) (10) J 

= 249.4 J

Data : n = 4, T = 300 K, P_f = 0.8 P_i

\(\therefore\) \(\frac{P_i}{P_f}\) = \(\frac{10}8\), R = 8.314 j/mol.K

The work done by the gas on its surroundings, W = nRT ln \((\frac{P_i}{P_f})\)

= (4) (8.314) (300) 2.303 log₁₀ \((\frac{10}8)\)

= 2.3 × 10⁴ log₁₀ (1.25) = 2.3 × 10⁴ × 0.0969

= 2.229 × 10³ J

Data ; n = 2, T = 300 K, V, = 23 

L = 23 × 10^-3 m³ , V_f = 46 

L = 46 × 10^-3 m³ , R = 8.314 J/mol .K

The work done by the gas on its surroundings,

W = nRT ln \((\frac{V_f}{V_i})\) = 2.303 nRT log₁₀ \((\frac{V_f}{V_i})\)

= (2.303) (2) (8.314) (300) log₁₀ \((\frac{46\times10^{-3}}{23\times10^{-3}})\)

= (4.606) (8.314) (300) log₁₀²

= (4.606) (8.314) (300) (0.3010) 

= 3458J

Data : \(\gamma\)= 1.4, P_f = 2P_i, T_i = 300K

\(\frac{T_f}{T_i}\) = \((\frac{P_f}{P_i})^{(\gamma-1)/\gamma}\) = 2^{(1.4 - 1)/1.4} = 2^0.4/1.4 = 2^0.2857

∴ log \(\frac{T_f}{T_i}\) = = 0.2857 log 2 = 0.2857 (0.3010) = 0.086

∴ \(\frac{T_f}{T_i}\) = antilog 0.086 = 1.219

∴ T_f = 1219T 

= (1.219) (300) = 365.7 K 

This is the final temperature of the gas.

(b) A gas contained in a cylinder fitted with a piston

(c) isolated system