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Formulae Required for Calculation of Maximum Work:
Number of moles = Given mass / molecular mass
W_{max} = 2.303 nRT log_{10}(V_{2}/V_{1})
W_{max} = 2.303 nRT log_{10}(P_{1}/P_{2})
By Boyle’s Law for isothermal Process, P_{1}V_{1 }= P_{2}V_{2}
Example – 1:
 3 moles of an ideal gas are expanded isothermally and reversibly from volume of 10 m^{3} to the volume 20 m^{3} at 300 K. Calculate the work done.

Solution:
Given: n = 3 moles, V_{1} = 10 m^{3}, V_{2} = 20 m^{3}, T = 300 K, R = 8.314 J K^{1} mol^{1}.
Work done in isothermal reversible process is given by
W_{max} = 2.303 nRT log_{10}(V_{2}/V_{1})
∴ W_{max} = 2.303 × 3 mol × 8.314 J K^{1} mol^{1 }× 300 K × log_{10}( 20 m^{3}/ 10 m^{3})
∴ W_{max} = 2.303 × 3 × 8.314 × 300 × log_{10}(2)
∴ W_{max} = 2.303 × 3 × 8.314 × 300 × 0.3010
∴ W_{max} = – 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 × 10^{5} Pa pressure to 100 kPa at 298 K. Calculate the work done.

Solution:
Given: m = 24 g, P_{1} = 1.6 × 10^{5} Pa, P_{2} = 100 kPa = 100 × 10^{3} Pa = 1 × 10^{5} 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
W_{max} = 2.303 nRT log_{10}(P_{1}/P_{2})
∴ W_{max} = 2.303 × 0.75 mol × 8.314 J K^{1} mol^{1 }× 298 K × log_{10}( 1.6 × 10^{5} Pa / 1 × 10^{5} Pa)
∴ W_{max} = 2.303 × 0.75 × 8.314 ^{ }× 298 × log_{10}( 1.6)
∴ W_{max} = 2.303 × 0.75 × 8.314 ^{ }× 298 × 0.2041
∴ W_{max} = – 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 CO_{2} 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, P_{1} = 150 kPa, Maximum work of compression = W = + 1.245 kJ = 1245 J, T = 293 K, R = 8.314 J K^{1} mol^{1}, P_{2} = ?,
Number of moles of CO_{2} = Given mass of CO_{2} / molecular mass of CO_{2}
Number of moles of CO_{2} = n = 4.4 × 10^{2} kg / 44 × 10^{3} kg mol^{1} = 1 mol
Work done in isothermal reversible process is given by
W_{max} = 2.303 nRT log_{10}(P_{1}/P_{2})
∴ 1245 J = 2.303 × 1 mol × 8.314 J K^{1} mol^{1 }× 293 K × log_{10}(P_{1}/P_{2})
∴ 1245 J = 2.303 × 1 mol × 8.314 J K^{1} mol^{1 }× 293 K × log_{10}(P_{2}/P_{1})
∴ log_{10}(P_{2}/P_{1}) = (1245) J / (2.303 × 1 × 8.314 × 293) J
∴ log_{10}(P_{2}/P_{1}) = 0.2219
∴ (P_{2}/P_{1}) = Antilog (0.2219) = 1.667
∴ P_{2} = 1.667 × P_{1} = 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 × 10^{5} Nm^{2} when work obtained is 17.33 kJ. Find the final pressure.

Solution:
Given: m = 42.8 × 10^{2} kg, P_{1} = 15.15 × 10^{5} Nm^{2}, Maximum work of expansion = W = – 17.33 kJ = – 17.33 × 10^{3} J, T = 293 K, R = 8.314 J K^{1} mol^{1}, P_{2} = ?,
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
W_{max} = 2.303 nRT log_{10}(P_{1}/P_{2})
∴ – 17.33 × 10^{3} J = 2.303 × 1 mol × 8.314 J K^{1} mol^{1 }× 300 K × log_{10}(P_{1}/P_{2})
∴ 17.33 × 10^{3} J = 2.303 × 1 mol × 8.314 J K^{1} mol^{1 }× 300 K × log_{10}(P_{1}/P_{2})
∴ log_{10}(P_{1}/P_{2}) = (17.33 × 10^{3}) J / (2.303 × 1 × 8.314 × 300) J
∴ log_{10}(P_{1}/P_{2}) = 3.0170
∴ (P_{1}/P_{2}) = Antilog (3.0170) = 1040
∴ P_{2} = P_{1} / 1040 = 15.15 × 10^{5} 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, V_{2} = 2 L, Maximum work of compression = W = + 2.983 kJ = 2.983 × 10^{3} J , T = 22° C = 22 + 273 = 295 K, R = 8.314 J K^{1} mol^{1}, V_{1} = ?,
Work done in isothermal reversible process is given by
W_{max} = 2.303 nRT log_{10}(V_{2}/V_{1})
∴ 2.983 × 10^{3} J = 2.303 ×3 mol × 8.314 J K^{1} mol^{1 }× 295 K × log_{10}(V_{2}/V_{1})
∴ 2.983 × 10^{3} J = 2.303 ×3 mol × 8.314 J K^{1} mol^{1 }× 295 K × log_{10}(V_{1}/V_{2})
∴ log_{10}(V_{1}/V_{2}) = 2.983 × 10^{3} J / (2.303 × 3 × 8.314 × 295) J
∴ log_{10}(V_{1}/V_{2}) = 0.1760
∴ (V_{1}/V_{2}) = Antilog (0.1760) = 1.5
∴ V_{1}= 1.5 × V_{2 }= 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, V_{1} = 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 = – P_{ext} × 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 = (V_{2} – V_{1}) = (V_{2} – 12.7 L) = 3.3 L
V_{2} = 3.3 L + 12.7 L = 16 L
Work done in isothermal reversible process is given by
W_{max} = 2.303 nRT log_{10}(V_{2}/V_{1})
∴ W_{max} = 2.303 × 0.280 mol × 8.314 J K^{1} mol^{1 }× 310 K × log_{10}( 16 L/ 12.7 L)
∴ W_{max} = 2.303 × 0.280 × 8.314 × 310 × log_{10}(1.260)
∴ W_{max} = 2.303 × 0.280 × 8.314 × 310 × 0.1004
∴ W_{max} = – 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 P_{ext} = 0
W = – P_{ext} × 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, V_{1} = 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 = – P_{ext} × 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 = (V_{2} – V_{1}) = (V_{2} – 13 L) = 3 L
V_{2} = 3 L + 13 L = 16 L
Work done in isothermal reversible process is given by
W_{max} = 2.303 nRT log_{10}(V_{2}/V_{1})
∴ W_{max} = 2.303 × 0.3 mol × 8.314 J K^{1} mol^{1 }× 320 K × log_{10}( 16 L/ 13 L)
∴ W_{max} = 2.303 × 0.3 × 8.314 × 320 × log_{10}(1.231)
∴ W_{max} = 2.303 × 0.3 × 8.314 × 320 × 0.0902
∴ W_{max} = – 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 P_{ext} = 0
W = – P_{ext} × ΔV = 0 atm × 3 L = 0
Work done = 0
Science > Chemistry > Chemical Thermodynamics and Energetics > You are Here 
Why carbon dioxide known as dry ice
Dry ice is solidified carbon dioxide, and is called so because the gas on solidifying gives an appearance similar to that of ice. At (78.5 degrees C) it undergoes sublimation i.e. it changes directly into gas.