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=Nk
_{B}where- N=number of molecules and
- k
_{B}= Boltzmann Constant and its value never change.

- From equation (i) PV= Nk
_{B} - Therefore PV/NT = constant=(k
_{B})(Same for all gases). - Consider there are 2 gases :- (P
_{1},V_{1},T_{1}) and (P_{2}, V_{2},T_{2}) where P, V and T are pressure, volume and temperature resp. - Therefore
**P**_{1},V_{1}/(N_{1}T_{1}) = P_{2}V_{2}/(N_{2}T_{2}) - 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cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J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 O_{2} = 32 u).

**Answer:- **Volume of oxygen, V_{1} = 30 litres = 30 × 10^{–3} m^{3}

Gauge pressure, P_{1} = 15 atm = 15 × 1.013 × 10^{5} Pa

Temperature, T_{1} = 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 n_{1}.

The gas equation is given as:

P_{1}V_{1} = n_{1}RT_{1}

n_{1} =P_{1}V_{1}/RT_{1} = (15.195x10^{5}x30x10^{-3)}/ (8.314)x300= 18.276

But, n_{1}=m/M Where, m_{1} =Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

m_{1} = n_{1}M = 18.276 × 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, V_{2} = 30 litres = 30 × 10^{–3} m^{3}

Gauge pressure, P_{2} = 11 atm = 11 × 1.013 × 10^{5} Pa

Temperature, T_{2} = 17°C = 290 K

Let n_{2} be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

P_{2}V_{2} = n_{2}RT_{2}

n_{2}= P_{2}V_{2}/RT_{2}

= (11.143x10^{5}x30x10^{-3})/ (8.314x290)

= 13.86

But, n_{2} = m_{2}/M Where, m_{2} is the mass of oxygen

remaining in the cylinder m_{2} = n_{2}M = 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

= m_{1} – m_{2}

= 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 cm^{3} 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, V_{1} = 1.0 cm^{3} = 1.0 × 10^{–6} m^{3}

Bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T_{1} = 12°C = 285 K

Temperature at the surface of the lake, T_{2} = 35°C = 308 K

The pressure on the surface of the lake:

P_{2} = 1 atm = 1 ×1.013 × 10^{5} Pa

The pressure at the depth of 40 m:

P_{1} = 1 atm + dρg Where, ρ is the density ofwater = 10^{3} kg/m^{3} g is the acceleration due

to gravity = 9.8 m/s^{2}

Therefore, P_{1} = 1.013 × 10^{5} + 40 × 10^{3} × 9.8 = 493300 Pa

We have: P_{1} V_{1}/T_{1} = P_{2}V_{2}/T_{2}

Where, V_{2} is the volume of the air bubble when it reaches the surface

V_{2} = P_{1} V_{1}T_{2}/T_{1}P_{2}

= (493300) (1.0x10^{-6})308/(285x1.013x10^{5})

= 5.263 × 10^{–6} m^{3} or 5.263 cm^{3}

Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm^{3}.

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