Class 11 Physics Kinetic Theory Behaviour of Gases

Behaviour of Gases

• Gases at low pressures and high temperatures much above that at which they liquify (or solidify) approximately satisfy a relation between their pressure, temperature and volume:
• PV=KT (i)
• This is the universal relation which is satisfied by all gases.
• where P, V, T are pressure,volume and temperature resp. and
• K is the constant for a given volume of gas. It varies with volume of gas.
• K=NkB where
• N=number of molecules and
• kB = Boltzmann Constant and its value never change.
• From equation (i) PV= NkB
• Therefore PV/NT = constant=(kB)(Same for all gases).
• Consider there are 2 gases :- (P1,V1,T1) and (P2, V2,T2) where P, V and T are pressure, volume and temperature resp.
• Therefore P1,V1/(N1T1) = P2V2/(N2T2)
• Conclusion: - This relation is satisfied by all gases at low pressure and high temperature. Problem:- An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u).

Answer:- Volume of oxygen, V1 = 30 litres = 30 × 10–3 m3

Gauge pressure, P1 = 15 atm = 15 × 1.013 × 105 Pa

Temperature, T1 = 27°C = 300 K

Universal gas constant, R = 8.314 J mole–1 K–1

Let the initial number of moles of oxygen gas in the cylinder be n1.

The gas equation is given as:

P1V1 = n1RT1

n1 =P1V1/RT1 = (15.195x105x30x10-3)/ (8.314)x300= 18.276

But, n1=m/M Where, m1 =Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

m1 = n1M = 18.276 × 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, V2 = 30 litres = 30 × 10–3 m3

Gauge pressure, P2 = 11 atm = 11 × 1.013 × 105 Pa

Temperature, T2 = 17°C = 290 K

Let n2 be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

P2V2 = n2RT2

n2= P2V2/RT2

= (11.143x105x30x10-3)/ (8.314x290)

= 13.86

But, n2 = m2/M Where, m2 is the mass of oxygen

remaining in the cylinder m2 = n2M = 13.86 × 32

= 453.1 g

The mass of oxygen taken out of the cylinder is given by the relation:

Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder

= m1 – m2

= 584.84 g – 453.1 g

= 131.74 g

= 0.131 kg

Therefore, 0.131 kg of oxygen is taken out of the cylinder.

Problem:- An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

Answer:- Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m3

Bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T1 = 12°C = 285 K

Temperature at the surface of the lake, T2 = 35°C = 308 K

The pressure on the surface of the lake:

P2 = 1 atm = 1 ×1.013 × 105 Pa

The pressure at the depth of 40 m:

P1 = 1 atm + dρg Where, ρ is the density ofwater = 103 kg/m3 g is the acceleration due

to gravity = 9.8 m/s2

Therefore, P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa

We have: P1 V1/T1 = P2V2/T2

Where, V2 is the volume of the air bubble when it reaches the surface

V2 = P1 V1T2/T1P2

= (493300) (1.0x10-6)308/(285x1.013x105)

= 5.263 × 10–6 m3 or 5.263 cm3

Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.

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