2. The relationship between temperature and volume

3. 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). Lets take a closer look at temperature before we try to find the exact relationship of V vs. T.

4. No. 68 F (20 C) is not double 50 F (10 C) Yes. 44 lb (20 kg) is double 22 lb (10 kg) Whats 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?

5. Temperature vs. Volume Graph (fig.7,8 pg.430) 5 10 15 20 25 30 Volume (mL) Temperature ( C) 0 100 – 273 25 mL at 22C 31.6 mL, 23.1 mL Y=0.0847x + 23.137

6. 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 dont 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 E K ) 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

7. What is the approximate temperature for absolute zero in degrees Celsius and kelvin? Calculate the missing temperatures 0 C = _______ K100 C = _______ K 100 K = _______ C – 30 C= _______ K 300 K = _______ C 403 K = _______ C 25 C = _______ K0 K = _______ C Kelvin Practice 273373 – 173243 27130 298 – 273 Absolute zero is – 273 C or 0 K

8. 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: 1.Give Charless law in words & as an equation. Charless Law: as the temperature of a gas increases, the volume increases proportionally, provided that the pressure and amount of gas remain constant, V 1 /T 1 = V 2 /T 2 Charless Law

9. 2.A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? 3.If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be? 4.Do questions 16, 17, 19 on page 434 V 1 = 3.5 L, T 1 = 300K, V 2 = ?, T 2 = 200K Using Charles law: V 1 /T 1 = V 2 /T 2 3.5 L / 300 K = V 2 / 200 K V 2 = (3.5 L/300 K) x (200 K) = 2.3 L V 1 = 1 L, T 1 = 22°C = 295 K V 2 = ?, T 2 = 100 °C = 373 K V 1 /T 1 = V 2 /T 2, 1 L / 295 K = V 2 / 373 K V 2 = (1 L/295 K) x (373 K) = 1.26 L For more lessons, visit www.chalkbored.com www.chalkbored.com