Pressure Volume Work (Work Done in Isothermal Process)

Physics Chemistry Biology Mathematics
Science > Chemistry > Chemical Thermodynamics and Energetics > You are Here

Problems involving Pressure Volume Work:

Example – 1:

  • 2 moles of an ideal gas are expanded isothermally from volume of 15.5 L to the volume 20 L against a constant external pressure of 1 atm. Calculate the pressure volume work in L atm and J.
  • Solution:

    Given: n = 2 moles, V1 = 15.5 L, V2 = 20 L, Pext = 1 atm

    Work done in isothermal process is given by   W  =  – Pext × ΔV

    ∴  W = – Pext × (V2 – V1 ) = – 1 atm × (20 L – 15.5  L)

    ∴  W  =  –  1 atm × (4.5 L) = – 4.5  L atm

    W = – 4.5 L atm  × 101.3  J L-1 atm-1 = – 455.8 J

    Hence work done  = – 4.5 L atm or – 455.8 J

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

Example – 2:

  • 2 moles of an ideal gas are compressed isothermally from volume of 10 dm3 to the volume 2 dm3 against a constant external pressure of 1.01 × 105 Nm-2,. Calculate the pressure volume work done.
  • Solution:

    Given: n = 2 moles, V1 = 10 dm3  = 10 × 10-3 m3 , V2 = 2 dm= 2 × 10-3 m3, Pext = 1 atm

    Work done in isothermal process is given by   W  =  – Pext × ΔV

    ∴  W = – Pext × (V2 – V1 ) = – 1.01 × 105 Nm-2 × ( 2 × 10-3 m3 – 10 × 10-3 m3)

    ∴  W = – Pext × (V2 – V1 ) = – 1.01 × 105 Nm-2 × ( – 8 × 10-3 m3)

    ∴  W = + 8.08 × 102 J = + 808 J

    Hence work done  =+  808 J

    Positive sign indicates the work is done by the surrounding on the system

Example – 3:

  • 3 moles of an ideal gas are expanded isothermally from volume of 300 cm3 to the volume 2.5 L against a constant external pressure of 1.9 atm at 300 K. Calculate the pressure volume work in L atm and J.
  • Solution:

    Given: n = 3 moles, V1 = 300 cm3= 0.3 L, V2 = 2.5 L, Pext = 1.9 atm

    Work done in isothermal process is given by   W  =  – Pext × ΔV

    ∴  W = – Pext × (V2 – V1 ) = – 1.9 atm × (2.5 L – 0.3  L)

    ∴  W  =  –  1.9 atm × (2.2 L) = – 4.18  L atm

    W = – 4.18 L atm  × 101.3  J L-1 atm-1 = – 423.4 J

    Hence work done  = – 4.5 L atm or – 423.4 J

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

Example -4:

  • 1 mole of an ideal gas is compressed isothermally from volume of 500 cm3 against a constant external pressure of 1.216 × 105 Pa. The pressure volume work involved in the process is 36.5 J. calculate the final volume.
  • Solution:

    Given: n = 1 mole, V1 = 500 cm3= 500 × 10-6 m3, Pext = 1.216 × 105 Pa = = 1.216 × 105 Nm-2, Work of compression = + 36. 5 j, V2 = ?

    Work done in isothermal process is given by   W  =  – Pext × ΔV

    ∴  W = – Pext × (V2 – V1 )

    ∴ 36.5 J= – 1.216 × 105 Nm-2 × (V2 – 500 × 10-6 m3)

    ∴ 36.5 J/  1.216 × 105 Nm-2 = – (V2 – 500 × 10-6 m3)

    ∴300 × 10-6 m3 = (500 × 10-6 m3  – V2)

    ∴    V2    = (500 × 10-6 m3  –  300 × 10-6 m3 )

    ∴    V2    = 200 × 10-6 m3 =  200 cm3

Hence final volume = 200 cm3

Example – 5:

  • 1 mole of an ideal gas is compressed isothermally from volume of 20 L to 8 L against constant externa pressure, when pressure volume work obtained is 44.9 L atm. Find the constant external pressure.
  • Solution:

    Given: n = 1 mol, V1 = 20 L, V2 = 8 L, Work of compression W = + 44.9 L atm, Pext = ?

    Work done in isothermal process is given by   W  =  – Pext × ΔV

    ∴  W = – Pext × (V2 – V1 )

    ∴ 44.9 L atm  = – Pext × (8 L  – 20 L)

∴ 44.9 L atm  = – Pext × (- 12 L)

∴ Pext   =44.9 L atm / 12 L)

∴ Pext   = 3.74 atm



Hence constant external pressure is 3.74 atm

Example – 6:

  • One mole of a gas expands by 3 L against a constant pressure of 3 atmosphere. Calculate the pressure volume work done in a) L-atm b) joules and c) calories.
  • Solution:

    Given: n = 1 mole, Δ V  = 3 L, Pext = 3 atm

    Work done in isothermal process is given by   W  =  – Pext × ΔV

    ∴  W = – 3 atm × 3 L = – 9 L atm

    ∴ W = – 9 L atm  × 101.3  J L-1 atm-1 = – 911.7 J

∴ W =  – 911.7 J / 4.184 J cal-1  = – 217.9 cal

Hence work done  = – 9  L atm or – 911.7 J  or -217.9 cal

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



Example – 7:

  • 100 mL of ethylene(g) and 100 mL of HCl (g) are allowed to react at  2 atm pressure as per the reaction given below. Calculate pressure volume type of work in it in joules.

C2H4(g)    +  HCl(g)    →     C2H5Cl(g)

  • Solution: Given: Pext = 2 atm and the given reaction is

C2H4(g)    +  HCl(g)    →     C2H5Cl(g)

1 Vol               1Vol                      1Vol

100 mL            100 mL                  100 mL

Thus 100 ml of C2H4(g) reacts with 100 mL of HCl(g)  to give 100 mL of C2H5Cl(g).



Initial volume = Volume of reactants = 100 mL + 100 mL = 200 mL = 0.2 L

Final volume = Volume of products = 100 mL = 0.1 L

Work done in isothermal process is given by   W  =  – Pext × ΔV

∴  W = – Pext × (V2 – V1 ) = – 2 atm × (0.1 L – 0.2  L)

∴  W  =  –  2 atm × (- 0.1 L) =  0.2  L atm



W = 0.2 L atm  × 101.3  J L-1 atm-1 = 20.26 J

Hence work done  = + 20.26 J

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



Example – 8:

  • A gas cylinder of 5 L capacity containing 4 kg of helium gas at 27 °C developed a leakage leading to the escape of the gas into atmosphere. If atmospheric pressure is 1.0 atm. calculate the pressure volume work done by the gas assuming ideal behaviour.
  • Solution:

    Given: V1 = 5 L, mass of gas m = 4 kg = 4 × 103 g, T = 27 °C = 27 +273 = 300 K,  Pext = P = 1.0 atm, R = 0.0821 L atm K-1 mol-1.

Number of moles = Given mass of He / Molecular mass of He = 4 × 103 g / 4 g =  103

By ideal gas equation, we have P V2  = nRT



∴   V2  = nRT / P = 103 × 0.0821 × 300 / 1   = 2.463 × 104 L = 24630 L

Work done in isothermal process is given by   W  =  – Pext × ΔV

∴  W = – Pext × (V2 – V1 ) = – 1 atm × (24630 L – 5  L)

∴  W  =  –  1 atm × (24625 L) = – 24625  L atm



W = – 24625 L atm  × 101.3  J L-1 atm-1 = – 2.494 × 106 J = – 2494 kJ

Hence work done  = – 2494 kJ

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

Example – 9:

  • 1.6 mol of water evaporates at 373 K against atmospheric pressure of 1 atm. Assuming ideal behaviour of water vapours calculate the work done.
  • Solution:

    Given: n = 1.6 mole, T = 373 K,  Pext = P = 1.0 atm, R = 0.0821 L atm K-1 mol-1.

    The molecular mass of water 18 g. and the density of water is 1 g per cc

Hence initial volume of water = V1 = 18 x 1.6 x 10-3 L = 0.0288 L

By ideal gas equation, we have P V2  = nRT

∴   V2  = nRT / P = 1.6 × 0.0821 × 373 / 1   = 48.93  L



Work done in isothermal process is given by   W  =  – Pext × ΔV

∴  W = – Pext × (V2 – V1 ) = – 1 atm × (48.93 L – 0.0288  L)

∴  W  =  –  1 atm × (48.90 L) = – 48.90  L atm

W = – 48.90 L atm  × 101.3  J L-1 atm-1 = – 4954.8 J



Hence work done  = – 4954.8 J

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

Example – 10:

  • 1.0 mol of water evaporates at 373 K against atmospheric pressure of  749.8 mm of Hg. Assuming ideal behaviour of water vapours calcul;ate the pressure volume work done.
  • Solution:

    Given: n = 1.0 mole, T = 373 K,  Pext = P = 749.8 mm of H = 749.8 / 760 = 0.987 atm, R = 0.0821 L atm K-1 mol-1.

    The molecular mass of water 18 g. and the density of water is 1 g per cc

Hence initial volume of water = V1 = 18 x 1.0 x 10-3 L = 0.018 L

By ideal gas equation, we have P V2  = nRT

∴   V2  = nRT / P = 1.0 × 0.0821 × 373 / 0.987   = 31.03  L

Work done in isothermal process is given by   W  =  – Pext × ΔV



∴  W = – Pext × (V2 – V1 ) = – 0.987 atm × (31.03 L – 0.018  L)

∴  W  =  –  0.987 atm × (31.01 L) = – 30.61  L atm

W = – 30.61 L atm  × 101.3  J L-1 atm-1 = – 3101 J

Hence work done  = – 3101 J



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

Science > Chemistry > Chemical Thermodynamics and Energetics > You are Here

Leave a Comment

Your email address will not be published. Required fields are marked *