Ch10 Ideal Gas and Kinetic Theory of Gas

PHYSICS 1ST PAPER CHAPTER 10 Ideal Gas and Kinetic Theory of Gas Question 1 Saima heated a gas of volume 5 litre at constant pressure from 0 ◦C to 35 ◦C. As a result its volume increases by 0.641 litre. (a) What is an ideal gas? (b) What do you mean by STP? (c) Find absolute zero temperature from the information of the stimulus. (d) The stated information supports which of the gas laws? Deduce its mathematical equation. Solution (a) A gas that obeys the gas laws under all conditions is called an ideal gas. (b) STP means standard temperature and pressure. The standard temperature is 0 ◦C or 273 K. The standard pressure is the pressure of a 76 cm column of mercury. It equals 1.013 × 105 N m−2. The volume of one mole of any gas at STP is 22.4 litre. (c) Here, Volume at 0 ◦C 𝑉0 = 5 litre Rise of temperature 𝜃= 35 ◦C Increase of volume 𝑉−𝑉0 = 0.641 litre At constant pressure, the volume at 𝜃◦C is 𝑉= 𝑉0(1+𝛾𝜃). Here 𝛾is the coefficient of volume expansion. 𝑉−𝑉0 = 𝑉0𝛾𝜃 𝛾= 𝑉−𝑉0 𝑉0 𝜃 = 0.641 5 × 35 ◦C−1 = 3.663 × 10−3 ◦C−1 At absolute zero temperature the volume of the gas becomes zero. 𝑉0(1 + 𝛾𝜃) = 0 𝜃= −1 𝛾= − 1 3.663 × 10−3 ◦C = −273 ◦C Absolute zero temperature = −273 ◦C ahbab.dev 1

−273 −200 −100 35 5 5.641 𝜃(◦C) 𝑉(litre) (d) The pressure was constant and the volume increased with temperature. Hence the information supports Charles’ law. For each degree, the volume increased by a fixed fraction 𝛾of 𝑉0. 𝛾= 3.663 × 10−3 ◦C−1 = 1 273 ◦C−1 Let 𝑉1 and 𝑉2 be the volumes at 𝜃1 and 𝜃2. 𝑉1 = 𝑉0  1 + 𝜃1 273  = 𝑉0 273 + 𝜃1 273 𝑉2 = 𝑉0  1 + 𝜃2 273  = 𝑉0 273 + 𝜃2 273 The absolute temperatures are 𝑇1 = 273 + 𝜃1 and 𝑇2 = 273 + 𝜃2. Dividing the first equation by the second, 𝑉1 𝑉2 = 𝑇1 𝑇2 ⇒ 𝑉1 𝑇1 = 𝑉2 𝑇2 𝑉∝𝑇 at constant pressure At constant pressure, the volume of a given gas varies as its absolute temperature. This is the mathematical form of Charles’ law. ahbab.dev 2

Question 2 A group of students in the physics laboratory observed that while bubbles are coming from the bottom to the surface of a water-filled vessel, their volume becomes 1.1 times. At one stage of the experiment, when one student mixed another liquid with the water, the density of the water became double. (Atmospheric pressure is 1 × 105 N m−2.) (a) State Boyle’s law. (b) Why does the pressure inside the wheel of a moving car increase? (c) If the temperature of the water remains constant, then what is the height of the vessel? (d) Will there be any change of the volume of the bubbles when they come to the surface after mixing the liquid with water? Solution (a) At constant temperature, the volume of a given mass of gas varies inversely as pressure. (b) The tyre of a moving car rubs against the road. Friction produces heat and the air inside becomes hot. The volume of the tyre stays almost constant. At constant volume, the pressure of a gas is proportional to its absolute temperature. 𝑃1 𝑇1 = 𝑃2 𝑇2 Hence the pressure inside the wheel increases as the temperature rises. (c) Here, Atmospheric pressure 𝑃= 1 × 105 N m−2 Density of water 𝜌= 1000 kg m−3 Volume at the bottom 𝑉1 = 𝑉 Vol