Pressure-Volume Work & Work Done Calculations | Physical Chemistry for NEET PDF Download

EXPRESSION FOR PRESSURE – VOLUME WORK
1. Work Done in Reversible Isothermal Expansion: Consider an ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The cylinder is not insulated.
The external pressure, P_ext, is equal to pressure of the gas, P_gas Let it be P.
P_ext = P_gas = P
If the external pressure is decreased by an infinitesimal amount dP, the gas will expand by an infinitesimal volume, dV.

As a result of expansion, the pressure of the gas within the cylinder falls to P_gas - dP, i.e., it becomes again equal to the external pressure and, thus, the piston comes to rest.

Such a process is repeated for a number of times, i.e., in each step the gas expands by a volume dV.

Isothernal reversible expansion work

Since, the system is in thermal equilibrium with the surroundings, the infinitesimally small cooling produced due to expansion is balanced by the absorption of heat from the surroundings and the temperature remains constant throughout the expansion.The work done by the gas in each step of expansion can be given as,
dw = -(P_ext - dP)dV
= -P_ext.dV
= -PdV
dP. dV, the product of two infinitesimal quantities, is neglected. 

• The negative sign of this expression is required to obtain conventional sign for w, which will be positive. It indicates that in case of compression work is done on the system.
• Here (V_f – V_i ) will be negative and negative multiplied by negative will be positive. Hence the sign obtained for the work will be positive.

If the pressure is not constant at every stage of compression, but changes in number of finite steps, work done on the gas will be summed over all the steps and will be equal to -Σp∆V
pV-plot when pressure is not constant and changes in finite steps during compression from initial volume, V_i to final volume, V_f . Work done on the gas is represented by the shaded area.

• The total amount of work done by the isothermal reversible expansion of the ideal· gas from volume V₁ to volume V₂ is, therefore
  
  For an ideal gas, p = nRT/V
  So,
  
  Integrating,
  
  At constant temperature, according to Boyle's law,
  P₁V₁ = P₂V₂
  or V₂/V₁ = P₁/P₂
  So,
  
  Isothermal compression work of an ideal gas may be derived similarly and it has exactly the same value with positive sign.
  
  
  WORK DONE IN IRREVERSIBLE ISOTHERMAL EXPANSION

Two types of irreversible isothermal expansions are observed, i.e.,
(i) Free expansion and
(ii) Intermediate expansion.

• In free expansion, the external pressure is zero, i.e. , work done is zero when gas expands in vacuum.
• In intermediate expansion, the external pressure is less than gas pressure. So, the work done when volume changes from V₁ to V₂ is given by

Since, P_ext is less than the pressure of the gas, the work done during intermediate expansion is numerically less than the work done during reversible isothermal expansion in which P_ext is almost equal to P_gas .

ISOTHERMAL AND FREE EXPANSION OF AN IDEAL GAS

For isothermal (T = constant) expansion of an ideal gas into vacuum ; w = 0 since p_ext = 0.

Also, Joule determined experimentally that q = 0; therefore, ∆U = 0
∆U = q + w can be expressed for isothermal irreversible and reversible changes as follows:
1. For isothermal irreversible change
q = – w = P_ext (V_f – V_i )

3. For adiabatic change, q = 0,
∆U = w_ad

REVERSIBLE ISOTHERMAL COMPRESSION OF AN IDEAL GAS

This can be achieved by placing particles of sand one by one at a very slow take in the assembly which keeps the temperature of gas constant in this case the expression of work done will be exactly similar to as obtained in case of reversible expansion of gas
W= -nRT ln (V_f /V_i)
This will automatically come out to be +ve as V_f < V_i

Example 1. 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 2. 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 3. 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

WORKDONE CALCULATION

(1) Isochoric process: 

V = constant  

dV = 0

dU = dq_V
ΔU = q_V  = nC_VΔT

(2) Isobaric process:
W = - P_ext (V₂-V₁)
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. 

(a) Reversible Expansion or compression

W = -P∫ dv
= -∫P_gas dV

In Expansion W = - ve
ΔE = 0

(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

Workdone in 2^nd stage > Workdone 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

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

ADIABATIC IDEAL GAS EXPANSION AND COMPRESSION

dq = 0
dU = dW 

For an ideal gas C_P - C_V = R

REVERSIBLE ADIABATIC EXPANSION OR COMPRESSION

nC_VdT = - P_ext dV
P_int = dP = P_ext

⇒ T₂(V₂)^{γ - 1} = V₁^{γ - 1}T₁

TV^{γ - 1} = Constant

IRREVERSIBLE ADIABATIC EXPANSION OR COMPRESSION

dU = dW

⇒ nC_V(T₂ - T₁) = - P_ext dV

= - P_ext [V₂ - V₁]

COMPARISON OF REVERSIBLE ISOTHERMAL AND REVERSIBLE ADIABATIC IDEAL GAS 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)

 ⇒  

⇒