Maximum Work (Work Done in Isothermal Reversible Process)

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Formulae Required for Calculation of Maximum Work:

Number of moles  = Given mass   / molecular mass

Wmax  =  -2.303 nRT log10(V2/V1)

Wmax  =  -2.303 nRT log10(P1/P2)

By Boyle’s Law for isothermal Process,  P1V1  =   P2V2



Example – 1:

  • 3 moles of an ideal gas are expanded isothermally and reversibly from volume of 10 m3 to the volume 20 m3 at 300 K. Calculate the work done.
  • Solution:

    Given: n = 3 moles, V1 = 10 m3, V2 = 20 m3, T = 300 K, R = 8.314 J K-1 mol-1.

    Work done in isothermal reversible process is given by

    Wmax  =  -2.303 nRT log10(V2/V1)

    ∴ Wmax  =  -2.303 × 3 mol × 8.314 J K-1 mol-1 × 300 K × log10( 20 m3/ 10 m3)

∴ Wmax  =  -2.303 × 3 × 8.314 × 300 × log10(2)

∴ Wmax  =  -2.303 × 3 × 8.314 × 300 × 0.3010

∴ Wmax  =  – 5187 J = – 5.187 kJ

Hence maximum work done  =  – 5187 J = – 5.187 kJ

Negative sign indicates the work is done by the system on the surroundings

Example – 2:

  • 24 g of oxygen are expanded isothermally and reversibly from 1.6 × 105 Pa pressure to 100 kPa  at 298 K. Calculate the work done.
  • Solution:

    Given: m = 24 g, P1 = 1.6 × 105 Pa, P2 = 100 kPa = 100 × 103 Pa = 1 × 105 Pa, T = 298 K, R = 8.314 J K-1 mol-1.

    Number of moles of oxygen = Given mass of oxygen  / molecular mass of oxygen

Number of moles of oxygen =  24 g / 32 g mol-1 = 0.75 mol

Work done in isothermal reversible process is given by

Wmax  =  -2.303 nRT log10(P1/P2)

∴ Wmax  =  -2.303 × 0.75 mol × 8.314 J K-1 mol-1 × 298 K × log10( 1.6 × 105 Pa / 1 × 105 Pa)

∴ Wmax  =  -2.303 × 0.75 × 8.314  × 298  × log10( 1.6)

∴ Wmax  =  -2.303 × 0.75 × 8.314  × 298  × 0.2041

∴ Wmax  =  – 873.4 J

Hence maximum work done  =  – 873.4 J

Negative sign indicates the work is done by the system on the surroundings

Example – 3:

  • 4.4 × 10-2 kg of CO2 is compressed isothermally and reversibly at 293 K from the initial pressure of 150 kPa when work obtained is 1.245 kJ. Find the final pressure.
  • Solution:

    Given: m = 4.4 × 10-2 kg, P1 = 150 kPa, Maximum work of compression = W = + 1.245 kJ = 1245 J, T = 293 K, R = 8.314 J K-1 mol-1, P2 = ?,

Number of moles of CO2 = Given mass of CO2 / molecular mass of CO2

Number of moles of CO2 = n = 4.4 × 10-2 kg / 44 × 10-3 kg mol-1  = 1 mol

Work done in isothermal reversible process is given by

Wmax  =  -2.303 nRT log10(P1/P2)

∴ 1245  J =  -2.303 × 1  mol × 8.314 J K-1 mol-1 × 293 K × log10(P1/P2)

∴ 1245  J =  -2.303 × 1  mol × 8.314 J K-1 mol-1 × 293 K × log10(P2/P1)

∴  log10(P2/P1) = (1245) J / (2.303 × 1 × 8.314 × 293)  J

∴  log10(P2/P1) = 0.2219

∴  (P2/P1) = Antilog (0.2219) = 1.667

∴  P2 =  1.667 ×  P1  = 1.667 × 150 kPa = 250 kPa

Hence Final Pressure is 250 kPa



Example – 4:

  • 2.8 × 10-2 kg of nitrogen is expanded isothermally and reversibly at 300 K from the initial pressure of 15.15 × 105 Nm-2 when work obtained is 17.33 kJ. Find the final pressure.
  • Solution:

    Given: m = 42.8 × 10-2 kg, P1 = 15.15 × 105 Nm-2, Maximum work of expansion = W = – 17.33 kJ = – 17.33 × 103 J, T = 293 K, R = 8.314 J K-1 mol-1, P2 = ?,

    Number of moles of Nitrogen = Given mass of Nitrogen / molecular mass of Nitrogen

Number of moles of Nitrogen = n = 2.8 × 10-2 kg / 28 kg mol-1  = 1  mol

Work done in isothermal reversible process is given by

Wmax  =  -2.303 nRT log10(P1/P2)

∴ – 17.33 × 103 J =  -2.303 × 1 mol × 8.314 J K-1 mol-1 × 300 K × log10(P1/P2)

∴ 17.33 × 103 J =  2.303 × 1  mol × 8.314 J K-1 mol-1 × 300 K × log10(P1/P2)

∴  log10(P1/P2) = (17.33 × 103) J / (2.303 × 1 × 8.314 × 300)  J

∴  log10(P1/P2) = 3.0170

∴  (P1/P2) = Antilog (3.0170) = 1040

∴  P2 =  P1 / 1040 = 15.15 × 105 Nm-2 / 1040 =  14056.7 Nm-2

Hence Final Pressure is 14056.7 Nm-2

Example – 5:

  • 3 moles of an ideal gas are compressed isothermally and reversibly at 22° C to a volume 2L when work done is 2.983 kJ. Find the initial volume.
  • Solution:

    Given: n = 3 mol, V2 = 2 L, Maximum work of compression = W = + 2.983 kJ = 2.983 × 103 J , T = 22° C = 22 + 273 = 295 K, R = 8.314 J K-1 mol-1, V1 = ?,

Work done in isothermal reversible process is given by

Wmax  =  -2.303 nRT log10(V2/V1)

∴  2.983 × 103 J =  -2.303 ×3 mol × 8.314 J K-1 mol-1 × 295 K × log10(V2/V1)

∴ 2.983 × 103 J =  2.303 ×3 mol × 8.314 J K-1 mol-1 × 295 K × log10(V1/V2)

∴  log10(V1/V2) = 2.983 × 103 J / (2.303 × 3 × 8.314 × 295)  J

∴  log10(V1/V2) = 0.1760

∴  (V1/V2) = Antilog (0.1760) = 1.5

∴  V1=  1.5  × V2  = 1.5  ×  2 L  = 4 L

Problems Based on Pressure Volume Work, Maximum Work and Work in Free Expansion

Example – 6:

  • 280 mmol of an ideal gas occupy 12.7 L at 310 K. Calculate the work done when gas expands a) isothermally against constant external pressure 0.25 atm, b) isothermally and reversibly c) in a vacuum until its volume becomes 3.3 L.
  • Solution:

    Given: n = 280 mmoles = 0.280 mol, V1 = 12.7 L, ΔV = 3.3 L, T = 310 K, R = 8.314 J K-1 mol-1. External Pressure = 0.25 atm

  • a)  The gas expands isothermally against constant external pressure 0.25 atm

W = – Pext × ΔV = – 0.25 atm × 3.3 L  = – 0.825 L atm

W = – 0.825 L atm  × 101.3  J L-1 atm-1 = – 83.6 J

Hence work done  = – 83.6 J

Negative sign indicates the work is done by the system on the surroundings

  • b) Gas expands isothermally and reversibly

ΔV =  (V2 –  V1)  =  (V2 –  12.7 L) =  3.3 L

V2 =  3.3 L  +  12.7 L = 16 L

Work done in isothermal reversible process is given by

Wmax  =  -2.303 nRT log10(V2/V1)

∴ Wmax  =  -2.303 × 0.280 mol × 8.314 J K-1 mol-1 × 310 K × log10( 16 L/ 12.7 L)

∴ Wmax  =  -2.303 × 0.280 × 8.314 × 310 × log10(1.260)

∴ Wmax  =  -2.303 × 0.280 × 8.314 × 310 × 0.1004

∴ Wmax  =  – 166.9  J

Maximum work = 166,9 J

Negative sign indicates the work is done by the system on the surroundings



  • Gas expands in vacuum

In this case, it is a free expansion, there is no opposing pressure. Hence Pext = 0

W = – Pext × ΔV = 0  atm × 3.3 L  = 0

Work done = 0

Example – 7:

  •  300 mmol of a perfect gas occupies 13 L at 320 K. Calculate the work done in joules when the gas expands a) isothermally against the constant external pressure of 0.20 atm. b) Isothermal and reversible process c) Into vacuum until the volume of the gas is increased by 3 L.
  • Solution:

    Given: n = 300 mmoles = 0.3 mol, V1 = 13 L, ΔV = 3  L, T = 320 K, R = 8.314 J K-1 mol-1. External Pressure = 0.20 atm

  • a)  The gas expands isothermally against constant external pressure 0.20 atm

W = – Pext × ΔV = – 0.20 atm × 3 L  = – 0.6 L atm

W = – 0.6 L atm  × 101.3  J L-1 atm-1 = – 60.78 J

Hence work done  = – 60.78 J

Negative sign indicates the work is done by the system on the surroundings

  • b) Gas expands isothermally and reversibly

ΔV =  (V2 –  V1)  =  (V2 –  13 L) =  3 L

V2 =  3 L  +  13 L = 16 L

Work done in isothermal reversible process is given by

Wmax  =  -2.303 nRT log10(V2/V1)

∴ Wmax  =  -2.303 × 0.3 mol × 8.314 J K-1 mol-1 × 320 K × log10( 16 L/ 13 L)

∴ Wmax  =  -2.303 × 0.3 × 8.314 × 320 × log10(1.231)

∴ Wmax  =  -2.303 × 0.3 × 8.314 × 320 × 0.0902

∴ Wmax  =  – 165.8  J

Maximum work = – 165.8  J

Negative sign indicates the work is done by the system on the surroundings

  • Gas expands in vacuum

In this case, it is the free expansion, there is no opposing pressure. Hence Pext = 0

W = – Pext × ΔV = 0  atm × 3 L  = 0

Work done = 0

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