Class 11 Physics Kinetic Theory Deducing Daltons Law of partial pressures

Deducing Dalton’s Law of partial pressures

• Dalton’s law of partial pressure states that the total pressure of a mixture of ideal gases is the sum of partial pressures.
• Consider if there are several ideal gases mixed together in a vessel,then the total pressure of that vessel is equal to sum of partial pressure.
• Partial pressure is the pressure exerted by a particular gas if only that gas is present in the vessel.
• For example: -
• Consider if in a vessel there is a mixture of 3 gases, A,B and C.So the partial pressure of A is equal to pressure exerted only by Aand considering B and C are not present.
• Similarly partial pressure of B is equal to the pressure exerted only by B and considering A and C are not there.
• Similarly for C.
• According to Dalton’s law the total pressure of mixture is sum of partial pressure of A, partial pressure of B and partial pressure of C.

To show how perfect gas equation concludes Dalton’s law of partial pressure:-

• Suppose there is a mixture of ideal gases which means these gases do not interact with each other.
• By perfect gas equation PV=μRT
• Where V=volume of vessel,P=Pressure and T=temperature andμ (no. of moles).
• As there are mixture of gases therefore μ=μ12+ ---- so on.
• PV=( μ12)RT
• P=( μ12)RT/V =>μ1RT/V + μ2RT/V + ---
• P=P1+P2 +----
• Where P1=partial pressure of gas 1 and P2=partial pressure of gas 2.
• Therefore P= P1+P2+---total pressure due to the mixture of gases is equal to the sum of the partial pressure of the gas.

Problem: -Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of the oxygen to be 3Å.

Answer:- At STP, Temperature = 273K and Pressure=1atm, diameter=3Å =3x10-8 cm.

Actual volume by 1mole O2 gas at STP =22.4l = 22400cm3.

Molecular volume of O2 = (4/3) πr3xNA

= (4/3) 3.14x (1.5x10-8)3x6.02x1023 =8.51cm3

Therefore, Ratio =8.51/22400 = 3.8x10-4

Problem:- The density of water is 1000kg m–3. The density of water vapour at 100 °Cand 1 atm pressure is 0.6 kg m–3. Thevolume of a molecule multiplied by the totalnumber gives, what is called, molecularvolume. Estimate the ratio (or fraction) ofthe molecular volume to the total volumeoccupied by the water vapour under theabove conditions of temperature andpressure.

Answer:- For a given mass of water molecules,the density is less if volume is large. So the volume of the vapour is 1000/0.6 = /(6 ×10 -4 )times larger. If densities of bulk water and watermolecules are same, then the fraction ofmolecular volume to the total volume in liquidstate is 1. As volume in vapour state hasincreased, the fractional volume is less by thesame amount, i.e. 6×10-4.

Problem: -  Estimate the volume of awater molecule if the density of water is 1000kg m–3. The density of water vapour at 100 °Cand 1 atm pressure is 0.6 kg m–3?

Answer: - In the liquid (or solid) phase, themolecules of water are quite closely packed. Thedensity of water molecule may therefore, beregarded as roughly equal to the density of bulkwater = 1000 kg m–3. To estimate the volume ofa water molecule, we need to know the mass ofa single water molecule. We know that 1 moleof water has a mass approximately equal to(2 + 16)g = 18 g = 0.018 kg.

Since 1 mole contains about 6 × 1023molecules (Avogadro’s number), the mass ofa molecule of water is (0.018)/(6 × 1023) kg =3 × 10–26 kg. Therefore, a rough estimate of thevolume of a water molecule is as follows:

Volume of a water molecule

= (3 × 10–26 kg)/ (1000 kg m–3)

= 3 × 10–29 m3

Hence, Radius ≈ 2 ×10-10 m = 2 Å.

Problem:- What is the averagedistance between atoms (interatomicdistance) in water?

Answer: A given mass of water in vapour statehas 1.67×103 times the volume of the same massof water in liquid state. This is alsothe increase in the amount of volume availablefor each molecule of water. When volumeincreases by 103 times the radius increases byV1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So theaverage distance is

= 2 × 20 = 40 Å.

Problem:-  A vessel contains two nonreactivegases: neon (monatomic) andoxygen (diatomic). The ratio of their partialpressures is 3:2. Estimate the ratio of (i)number of molecules and (ii) mass densityof neon and oxygen in the vessel. Atomicmass of Ne = 20.2 u, molecular mass of O2= 32.0 u.

Answer: - Partial pressure of a gas in a mixture isthe pressure it would have for the same volumeand temperature if it alone occupied the vessel.

(The total pressure of a mixture of non-reactivegases is the sum of partial pressures due to itsconstituent gases.) Each gas (assumed ideal)obeys the gas law.

Since V and T are common tothe two gases, we have

P1V = μ1 RT and P2V =μ2RT, i.e. (P1/P2) = (μ1 / μ2).

Here 1 and 2 referto neon and oxygen respectively. Since (P1/P2) =(3/2) (given), (μ1/ μ2) = 3/2

(i) By definition μ1 = (N1/NA ) and μ2 = (N2/NA)where N1 and N2 are the number of moleculesof 1 and 2, and NA is the Avogadro’s number.

Therefore, (N1/N2) = (μ1 / μ2) = (3/2).

(ii) We can also write μ1 = (m1/M1) and μ2 =(m2/M2) where m1 and m2 are the masses of 1 and 2; and M1 and M2 are their molecularmasses. (Both m1 and M1; as well as m2 and M2 should be expressed in the same units).If ρ1 and ρ2 are the mass densities of 1 and 2 respectively, we have,

= (½) (m1/V)/m2/V = (m1/m2) (1/2) (M1/M2)

= (3/2) (20.2/32.0)

=0.947

Problem:- An air bubble of volume 1.0cm3 rises from the bottom of a lake 40m deep at a temperature of 120C. To what volume does it grow when it reaches the surface, which is at temperatures of 350C?

Answer:- Let the depth (d) of the lake = 40m. Temperature (T1) at the bottom = 120C = 285K and (V1) Volume of air bubble= 1.0cm3 =1x10-6m3.Temperature at the surface (T2) =350C = 308K, Pressure at the surface (P2) = 1 atm= 1.013x105 Pa.

Pressure at the bottom P1 = P2+ρdg

=1+103x9.8x40

=493300 Pa

Also (P1V1)/T1= (P2V2)/T2

V2 = (P1V1T2)/P2T1

After calculating,

V2=5.263 cm3

.