Partial Pressure

Let us consider a mixture of two non-reactive gases. If we remove one of the component gas from the mixture, the pressure inside the container is partial pressure. It is the pressure that the gas would exert if present alone in the container.

Dalton’s law of partial pressure

It states that “The total pressure exerted by the mixture of non-reacting gases is the sum of partial pressure of the individual gases under the identical condition of temperature”. If PA, PB, and PC are the partial pressure of three non reacting gases A, B and C enclosed in a cylinder. Then according to this law, the total pressure (PT ) is given by:

PT = PA + PB + PC

Mathematical derivation of Dalton’s law of partial pressure

Let nA, nB, and nC be the number of moles of non reacting gases A, B and C and n is the total number of moles of the gaseous mixture. Since the number of moles is an additive property,

nT = nA + nB + nC —–(i)

From ideal gas equation,

PV=nRT\\ n_{T}=\frac{P_{T}V}{RT} \;\;\;\; n_{A}=\frac{P_{A}V}{RT} \;\;\;\; n_{B}=\frac{P_{B}V}{RT} \;\;\;\; n_{C}=\frac{P_{C}V}{RT} \;\;\;\;

where PT is the total pressure of the gaseous mixture and PA, PB, and PC are the partial pressures. So equation (i) becomes:

\frac{P_{T}V}{RT}=\frac{P_{A}V}{RT} + \frac{P_{B}V}{RT} + \frac{P_{C}V}{RT}\\ or,\frac{P_{T}V}{RT}=\frac{V}{RT}\;(P_{A}+P_{B}+P_{C})\\ \therefore P_{T}=(P_{A}+P_{B}+P_{C})
Relation between partial pressure, total pressure and mole fraction

From Dalton’s law,

P_{A}=n_{A}\frac{RT}{V}\;\; P_{B}=n_{B}\frac{RT}{V}\;\; P_{C}=n_{C}\frac{RT}{V}\;\; P_{T}=n_{T}\frac{RT}{V}

Dividing PA, PB and PC by PT,

P_{A}=\left [\frac{n_{A}}{n_{T}} \right]P_{T},\;\; P_{B}=\left [\frac{n_{B}}{n_{T}}\right]P_{T},\;\; P_{C}=\left [\frac{n_{C}}{n_{T}} \right ]P_{T}\;\;

In general,

Partial\ pressure=\frac{Num.\ of\ mole\ of\ a\ gas}{Total\ num\ of\ moles} \times Total\ pressure\\ =Mole\ fraction \times Total\ pressure
Applications

1. Calculation of pressure of dry gas: Some gases are collected by downward displacement of water. The collected gas get saturated with water vapour. This is due to the pressure of dry gas and the pressure of water vapour. The pressure of water vapour is called aqueous tension(f). Applying Dalton’s law,

Pmoist gas = Pgas + Pwater vapour
Pgas = Pmoist gas – f


Solved numerical examples

1. 40 cc of oxygen at 650 mm Hg pressure, 30 cc of nitrogen at 700 mm Hg pressure and 10 cc of hydrogen at 760 mm Hg pressure are mixed together in a separate vessel. Find the total pressure and partial pressure of each gas.

For oxygen,
Initial volume (V1) = 40 cc
Final volume (V2) = 80 cc
Initial pressure (P1) = 650 mm Hg
Partial pressure (PO2) =?

From Boyle’s law,

P_{1}V_{1}=P_{O_{2}}V_{2}\\ P_{O_{2}}=\frac{P_{1}V_{1}}{V_{2}} = \frac{650 \times 40}{80}=325\ mm\ Hg
For Nitrogen,
Initial volume (V1) = 30 cc
Final volume (V2) = 80 cc
Initial pressure (P1) = 700 mm Hg
Partial pressure (PN2) = ?

From Boyle’s law,

P_{1}V_{1}=P_{N_{2}}V_{2}\\ P_{N_{2}}=\frac{P_{1}V_{1}}{V_{2}} =\frac{700 \times 30}{80}=262.5\ mm\ Hg
For hydrogen,
Initial volume (V1) = 10 cc
Final volume (V2) = 80 cc
Initial pressure (P1) = 760 mm Hg
Partial pressure (PH2) = ?

From Boyl’e law,

P_{1}V_{1}=P_{H_{2}}V_{2}\\ P_{H_{2}}=\frac{P_{1}V_{1}}{V_{2}} =\frac{760 \times 10}{80}=95\ mm\ Hg\\ Total\ pressure=P_{O_{2}}+P_{N_{2}}+P_{H_{2}}\\ =325 + 262.5 + 95 =682.5\ mm\ Hg

2. A 1L sample of dry gas at 25 °C has a composition of 0.894 gm of nitrogen gas and 0.274 gm of carbon dioxide gas. Calculate the total pressure.

Number of moles of N2 = 0.894/28 = 0.0319 mol
Number of moles of CO2 = 0. 274/44 = 0.0062 mol
Volume of gas = 1litre
Temperature = 25 + 273 = 298 K

From ideal gas equation,

PV = nRT
P = (n/V) x RT
Partial pressure of N2 = (0.0319/1) x 0.082 x 298 = 0.78 atm
Partial pressure of CO2 = (0.0062/1) x 0.082 x 298 = 0.15 atm
Total pressure = 0.78 + 0.15 = 0.93 atm.


Some Important Questions
  1. 1 g CO2, 2g N2 and 3g O2 are placed in an evacuated cylinder of 1-litre capacity at normal temperature. Find the total pressure of the mixture of gas. (4.21 atm)
  2. 1-litre sample of dry gas at 25°C has the following compositions: 0.8940 gm of N2, 0.2741 gm of O2, 0.0152 gm of Ar, 0.00107 gm of CO2. What is the partial pressure for each component in the mixture? What is the total pressure (0.997 atm)

References:
Mishra, AD, et al. Pioneer Chemistry. Dreamland Publication.
Mishra, AD et al. Pioneer Practical Chemistry. Dreamland Publication
Wagley, P. et al. Comprehensive Chemistry. Heritage Publisher & Distributors Pvt. Ltd.

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