Enthalpy, Heat, Internal Energy & Work Done Calculations in Different Processes | Chemistry for JEE Main & Advanced PDF Download

Enthalpy, Heat, Internal Energy & Work Done Calculations in Different Processes

(1) Isochoric process

V = constant  

dV = 0

dU = dq_V
ΔU = q_V  = nC_VΔT

Calculation of q, w, ∆H, ∆U for an IRREVERSIBLE ISOCHORIC process involving an ideal gas:

w= 0   

∵ dV = 0

(2) Isobaric process

Reversible & isobaric process

W = - P (V₂-V₁)
= - nR (T₂-T₁)
Irreversible & isobaric process
P₁ = P₂ = P_ext
For reversible & irreversible isobaric or isochoric process, workdone is same.

(3) Isothermal process

W = -P∫ dv
= -∫P_gas dV

In Expansion W = - ve
ΔE = 0

Calculation of q, w , ∆H, ∆U for a reversible isothermal process involving an ideal gas :
ΔU = q + w = 0 ⇒ -w

(b) Single stage irreversible expansion

W = - P_ext(V₂ - V₁)

|W_rev| > |W_irr| (in case of expansion)

(c) Two Stage irreversible Expansion

Stage I. P_ext  = 3 atm, P_i = 5 atm
Stage II. P_ext  = 2 atm , P_f = 2 atm

Work done in 2^nd stage > Work done in I^st stage

(d)  n- stage expansion 

 Compression - (One stage Compression ) 

| W_irr | = P_ext DV
P₁ = 1 atm , P₂ = 5 atm , P_ext = 5 atm

 | W_irr | > | W_rev | For compression

Calculation of q, w, ∆H, ∆U for an Irreversible isothermal process involving an ideal gas:

ΔH = ΔU = 0 ∵ ΔT = 0

Also,

 
For isobaric process

= -nR(T₂ - T₁) (∵ P_ext = P)
= -nRΔT

Example 1. 2 moles of an ideal gas initially present in a piston fitted cylinder at 300 K, and 10 atm are allowed to expand against 1 atm but the piston was stopped before it stablished the mechanical equilibrium. If temperature were maintained constant through out the change and system delivers 748.26 J of work, determine the final gas pressure and describe the process on PV diagram. 
Solution. 
W_irrv = - 748.26
W_irr = - P_ext [1/P₂- 1/P₁]nRT
P₂ = 4atm

Example 2. 1150 Kcal heat is released when following reaction is carried out at constant volume. 
C₇H_16(l) 11O_2(g) → 7CO_2(g) 8H₂O_(l) 
Find the heat change at constant pressure. 
The pressure of liquid is a linear function of volume (P = a + bV) and the internal energy of the liquid is U = 34 + 3PV find a, b, w, ΔE & ΔH for change in state from 100 Pa, 3m³ to 400 Pa, 6m³ 

Solution. 100 = a + bV
⇒ 100 = a + 3b
Also, 400 = a + 6b
⇒ a = -200, b = 100
ΔU = 34 + 3(P₂V₂ - P₁V₁)
= 6300 J
ΔH = ΔU + P₂V₂ - P₁V₁
= 6300 + 2100 = 8400 J
P is a linear function
P_ext = (400 + 100)/2 = 250
W = - P_ext(dV)
= - 250(6 - 3) = - 750 J

Example 3. 4 moles of an ideal gas (C_v = 15 J) is subjected to the following process represented on P - T graph. From the given data find out whether the process is isochoric or not ? also calculate q, w, ΔU, ΔH,

Solution. PV = nRT
4V = 4R × 400
V = 400 R ........(1)
3V = 4R × 300
⇒ V = 400 R ........(2)
i.e., V is constant
w = 0
ΔU = nC_V  + ΔT ⇒ 4 ×15 ×100 = 6000 J
ΔH = nC_P + ΔT ⇒ n (C_V + R ) ΔT
⇒ 4 ×(15 8.3)×100
⇒ 9320 J

Example 4. 2 mole of a gas at 1 bar and 300 K are compressed at constant temperature by use of a constant pressure of 5 bar. How much work is done on the gas ?  
Solution. 
 = 19953.6 J

Example 5. 2 moles of an ideal diatomic gas (C_V  = 5/2 R) at 300 K, 5 atm expanded irreversibly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. 
(1) Calculate final temperature q, w, ΔH & ΔU 
(2) Calculate corresponding values if the above process is carried out reversibly. 
Solution. 
w = C_V (T₂ - T₁) 

Given T₂ = 270K
P_ext = 1,P₂ = 2,P₁ = 5
q = 0
w = ΔU = nC_VΔT
= -1247.1 J
ΔH = nC_PΔT
= 1745.94 J
If process is reversible Pv^γ = Constant
P^{1- γ}T^γ  = Constant

Example 6. 10 gm of Helium at 127°C is expanded isothermally from 100 atm to 1 atm Calculate the work done when the expansion is carried out
(i) In single step
(ii) In three steps the intermediate pressure being 60 and 30 atm respectively and
(iii) Reversibly.
Solution. 
(i) Work done = V.ΔP

(ii) In three steps
V_I = 83.14 × 10^-5 m³
W_I = (83.14 × 10^-5) × (100 - 60) × 10⁵
= 3325.6 Jules

W_II = V. DP
W_II = 138.56 × 10^-5 (60 - 30) × 10⁵
= 4156.99 ≈ 4157 J.

W_III = 277.13 × 10^-5 (30 - 1) × 10⁵
W_III = 8036.86 J.
W _total  =  W_I + W_II + W_III 
= 3325.6 + 4156.909 + 8036.86 = 15519.45 J.
(iii) For reversible process

Example 7. Calculate the amount of work done by 2 mole of an ideal gas at 298 K in reversible isothermal expansion from 10 litre to 20 litre.

Solution. Amount of work, done in reversible isothermal expansion,

Given, n = 2, R = 8.314 JK^-1 mol^-1, T = 298 K, V₂ = 20 L and V₁ = 10 L.
Substituting the values in above equation,

= -2.303 × 8.314 × 298 × 0.3010 = -3434.9 J
i.e., work is done by the system.

Example 8. 5 moles of an ideal gas expand isothermally and reversibly film a pressure of 10 atm to 2atm at 300K. What is the largest mass which can be lifted through a height of l metre in this expansion?

Solution. Work done by the system

Let M be the mas which can be lifted through a height of 1 m.

Work done in lifting the mass = Mgh = M × 9.8 × 11
So, M × 9.8 = 20.075 × 10³ M
= 2048.469 kg

(4) Adiabatic Ideal Gas Expansion & Compression

dq = 0
dU = dW 

For an ideal gas C_P - C_V = R

Reversible Adiabatic Ideal Gas Expansion & Compression

nC_VdT = - P_ext dV
P_int = dP = P_ext

⇒ T₂(V₂)^{γ - 1} = V₁^{γ - 1}T₁

TV^{γ - 1} = Constant

Calculation of q, w, ∆H, ∆U for reversible adiabatic process:

dq = 0 ⇒ dU = dw

for a reversible change

Now substituting V = nRT/P in equation

substituting T = PV/nR in eq...

⇒ Equation (i), (ii) and (iii) is valid only for reversible adiabatic process, for irreversible adiabatic process these equations are not applicable.

• Expression for work:

• Expression for ΔH and ΔU

Irreversible Adiabtic Ideal Gas Expansion & Compression

dU = dW

⇒ nC_V(T₂ - T₁) = - P_ext dV

= - P_ext [V₂ - V₁]

Calculation of Q, W, ∆H, ∆U For Irreversible Adiabatic Process involving An Ideal Gas:

If large amount of dust particles are removed abruptly an irreversible adiabatic expansion take place.

In an irreversible adiabatic process, an ideal gas is subjected to compression or expansion in a thermally insulated vessel.

The heat absorbed in the process =0
⇒ dU = w_irr ....(i)

If P_ext, = P₂ = P_final
Then

eq. (ii) or (iii) can be solved for T₂
Expression for w

Note: If two states A and B are connected by a reversible path then they can never be connected by an irreversible path.
If the two states are linked by an adiabatic reversible and irreversible path then w_rev = ∆U_rev
But as U is a state function
Therefore, ∆U _irrev = ∆U_rev
w_irrev = w_rev
as work is a path function.

If we assume that
w_irrev = w_rev

It implies that   which again is a contradiction as U is a state function.
∆U _irrev ≠ ∆U_rev
Two states A and B can never lie both on a reversible as well as irreversible adiabatic path.

There lies only one unique adiabatic path linkage between two states A and B.

Example 1. Two moles of an ideal monoatomic gas at NTP are compressed adiabatically and reversibly to occupy a volume of 4.48 dm³. Calculate the amount of work done, ∆U, final temperature and pressure of the gas. Cv for ideal gas 12.45J K ^-1 mol^-1.
Solution. For an ideal gas,

Initial volume, V₁ = 2 × 22.4 = 44.8 dm³ 
Initial pressure, P₁ = 1 atm
Initial temperature, T₁ = 273 K
Final volume, V₂ = 4.48 dm³
Let the final pressure be P₂ and temperature be T₂ 
Applying

or

or

P₂ = (10)^1.667(P₁ = 1 given)
log P₂ = 1.667 log 10= 1.667
P₂ = antilog 1.667= 46.45 atm
Final temperature 

= 1268 K

Work done on the system = n.C_v.ΔT
= 2 × 12.45 × (1268 - 273)
= 2 × 12.45 × 995 = 24775.5 J
From the first law of thermodynamics,
ΔE = q + w = 0 + 24775.5 = 24775.5 J

Example 2. A certain volume of dry air at NTP is expanded reversibiy to four times its volume (a) isothermally (b) adiabatically. Calculate thefinal pressure and temperature in each case, assuming ideal behaviour.
(C_P \ C_V for air = 1.4)
Solution. 
Let V₁ be the initial volume of dry air at NTP.
(a) Isothermal expansion: During isothermal expansion, the temperature remains the same throughout. Hence, final temperature will be 273 K.
Since P₁V₁ = P₂V₂

(b) Adiabatic expansion:

Comparison of Reversible Isothermal And Reversible Adiabatic Ideal Gas Expansion 

Expansion

(i) If final volumes are same
Isothermal process.
P₁v₁ = P_iso V₂ 
⇒  

Adiabatic process.

P₁v₁^γ  = P_adia V₂^γ 

(ii) If final pressures are same
Isothermal process.

P₁V₁ = P₂ V_iso 

 .......(1)

P₁V₁^γ  = P₂V^γ _adia

In ideal gas expansion,
| W_iso | > | W_adia |

Hence

Compression 

(i) If final volumes are same

For isothermal process

P₁V₁ = P_isoV₂

 ..........(1)

Adiabatic process.

P₁V₁^γ  = P_adiaV₂^γ 

 ..........(2)

(ii) If final pressures are same

P₁V₁ = P₂ V_iso ..........(1)

P₁V₁^γ  = P₂ V^γ _adia ..........(2)

 ⇒  

⇒