Gas laws:- Charles Law

Slide 1 : Charles's Law The relationship between temperature and volume

How Volume Varies With Temperature : If we place a balloon in liquid nitrogen it shrinks: How Volume Varies With Temperature So, gases shrink if cooled. Conversely, if we heat a gas it expands (as in a hot air balloon). Let’s take a closer look at temperature before we try to find the exact relationship of V vs. T.

Temperature scales : No. 68F (20C) is not double 50F (10C) Yes. 44 lb (20 kg) is double 22 lb (10 kg) What’s the difference? Weights (kg or lb) have a minimum value of 0. But the smallest temperature is not 0C. We saw that doubling P yields half the V. Yet, to investigate the effect of doubling temp-erature, we first have to know what that means. An experiment with a fixed volume of gas in a cylinder will reveal the relationship of V vs. T… Temperature scales Is 20C twice as hot as 10C? Is 20 kg twice as heavy as 10 kg?

Temperature vs. Volume Graph (fig.7,8 pg.430) : Temperature vs. Volume Graph (fig.7,8 pg.430) – 273

The Kelvin Temperature Scale : If a volume vs. temperature graph is plotted for gases, most lines can be interpolated so that when volume is 0 the temperature is -273 C. Naturally, gases don’t really reach a 0 volume, but the spaces between molecules approach 0. At this point all molecular movement stops. –273C is known as “absolute zero” (no EK) Lord Kelvin suggested that a reasonable temp-erature scale should start at a true zero value. He kept the convenient units of C, but started at absolute zero. Thus, K = C + 273. 62C = ? K: K=C+273 = 62 + 273 = 335 K Notice that kelvin is represented as K not K. The Kelvin Temperature Scale

Kelvin Practice : What is the approximate temperature for absolute zero in degrees Celsius and kelvin? Calculate the missing temperatures 0C = _______ K 100C = _______ K 100 K = _______ C – 30C = _______ K 300 K = _______ C 403 K = _______ C 25C = _______ K 0 K = _______ C Kelvin Practice 273 373 – 173 243 27 130 298 – 273 Absolute zero is – 273C or 0 K

Charles’s Law : Looking back at the temperature vs. volume graph, notice that there is a direct relationship. It can be shown that V/T = constant Read pages 432-3. Answer these questions: Give Charles’s law in words & as an equation. Charles’s Law: as the temperature of a gas increases, the volume increases proportionally, provided that the pressure and amount of gas remain constant, V1/T1 = V2/T2 Charles’s Law

Slide 8 : A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be? Do questions 16, 17, 19 on page 434 V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K Using Charles’ law: V1/T1 = V2/T2 3.5 L / 300 K = V2 / 200 K V2 = (3.5 L/300 K) x (200 K) = 2.3 L V1 = 1 L, T1 = 22°C = 295 K V2 = ?, T2 = 100 °C = 373 K V1/T1 = V2/T2, 1 L / 295 K = V2 / 373 K V2 = (1 L/295 K) x (373 K) = 1.26 L For more lessons, visit www.chalkbored.com