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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, V_{1} = 15.5 L, V_{2} = 20 L, P_{ext} = 1 atm
Work done in isothermal process is given by W = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} ) = – 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 dm^{3} to the volume 2 dm^{3} against a constant external pressure of 1.01 × 10^{5} Nm^{2},. Calculate the pressure volume work done.

Solution:
Given: n = 2 moles, V_{1} = 10 dm^{3 }= 10 × 10^{3} m^{3} , V_{2} = 2 dm^{3 }= 2 × 10^{3} m^{3}, P_{ext} = 1 atm
Work done in isothermal process is given by W = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} ) = – 1.01 × 10^{5} Nm^{2} × ( 2 × 10^{3} m^{3} – 10 × 10^{3} m^{3})
∴ W = – P_{ext} × (V_{2} – V_{1} ) = – 1.01 × 10^{5} Nm^{2} × ( – 8 × 10^{3} m^{3})
∴ W = + 8.08 × 10^{2} 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 cm^{3} 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, V_{1} = 300 cm^{3}= 0.3 L, V_{2} = 2.5 L, P_{ext} = 1.9 atm
Work done in isothermal process is given by W = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} ) = – 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 cm^{3} against a constant external pressure of 1.216 × 10^{5} Pa. The pressure volume work involved in the process is 36.5 J. calculate the final volume.

Solution:
Given: n = 1 mole, V_{1} = 500 cm^{3}= 500 × 10^{6} m^{3}, P_{ext} = 1.216 × 10^{5} Pa = = 1.216 × 10^{5} Nm^{2}, Work of compression = + 36. 5 j, V_{2} = ?
Work done in isothermal process is given by W = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} )
∴ 36.5 J= – 1.216 × 10^{5} Nm^{2} × (V_{2} – 500 × 10^{6} m^{3})
∴ 36.5 J/ 1.216 × 10^{5} Nm^{2} = – (V_{2} – 500 × 10^{6} m^{3})
∴300 × 10^{6} m^{3} = (500 × 10^{6} m^{3} – V_{2})
∴ V_{2 } = (500 × 10^{6} m^{3} – 300 × 10^{6} m^{3} )
∴ V_{2 } = 200 × 10^{6} m^{3} = 200 cm^{3}
Hence final volume = 200 cm^{3}
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, V_{1} = 20 L, V_{2} = 8 L, Work of compression W = + 44.9 L atm, P_{ext} = ?
Work done in isothermal process is given by W = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} )
∴ 44.9 L atm = – P_{ext} × (8 L – 20 L)
∴ 44.9 L atm = – P_{ext} × ( 12 L)
∴ P_{ext }=44.9 L atm / 12 L)
∴ P_{ext }= 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) Latm b) joules and c) calories.

Solution:
Given: n = 1 mole, Δ V = 3 L, P_{ext} = 3 atm
Work done in isothermal process is given by W = – P_{ext} × 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.
C_{2}H_{4(g) } + HCl_{(g) }→ C_{2}H_{5}Cl_{(g)}

Solution: Given: P_{ext} = 2 atm and the given reaction is
C_{2}H_{4(g) } + HCl_{(g) }→ C_{2}H_{5}Cl_{(g)}
1 Vol 1Vol 1Vol
100 mL 100 mL 100 mL
Thus 100 ml of C_{2}H_{4(g)} reacts with 100 mL of HCl_{(g) } to give 100 mL of C_{2}H_{5}Cl_{(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 = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} ) = – 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: V_{1} = 5 L, mass of gas m = 4 kg = 4 × 10^{3} g, T = 27 °C = 27 +273 = 300 K, P_{ext} = 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 × 10^{3} g / 4 g = 10^{3}
By ideal gas equation, we have P V_{2} = nRT
∴ V_{2} = nRT / P = 10^{3} × 0.0821 × 300 / 1 = 2.463 × 10^{4} L = 24630 L
Work done in isothermal process is given by W = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} ) = – 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 × 10^{6} 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, P_{ext} = 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 = V_{1} = 18 x 1.6 x 10^{3} L = 0.0288 L
By ideal gas equation, we have P V_{2} = nRT
∴ V_{2} = nRT / P = 1.6 × 0.0821 × 373 / 1 = 48.93 L
Work done in isothermal process is given by W = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} ) = – 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, P_{ext} = 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 = V_{1} = 18 x 1.0 x 10^{3} L = 0.018 L
By ideal gas equation, we have P V_{2} = nRT
∴ V_{2} = nRT / P = 1.0 × 0.0821 × 373 / 0.987 = 31.03 L
Work done in isothermal process is given by W = – P_{ext} × ΔV
∴ W = – P_{ext} × (V_{2} – V_{1} ) = – 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 