Charles’s Law

1. Charles’s Law Volume and Temperature

2. How Volume Varies With Temperature If a balloon is placed in liquid nitrogen it shrinks: So, gases shrink if cooled. Conversely, if we heat a gas it expands (as in a hot air balloon).

3. Temperature Scales • Is 20o C twice 10o C ? NO. 20o C = 68 oF or 293 Kelvin 10oC = 50 oF or 283 Kelvin • Is 20 kg twice 10 kg? YES. 20 kg = 44 lbs; 10 kg = 22 lbs What’s the Difference? • Weight has a minimum of zero, but oC doesn’t • We use Kelvin for Charles’s law because 0 Kelvin equates to zero molecular movement

4. 30 25 20 Volume (mL) 15 10 5 0 100 Temperature (C) Temperature vs. Volume Graph Fixed Volume in a cylinder 25 mL at 22C 31.6 mL, 23.1 mL Y=0.0847x + 23.137 – 273

5. The Kelvin Temperature Scale • If a volume vs. temperature graph is plotted for gases, most lines can be interpreted 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) • Kelvin 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.

6. Charles’s Law • Looking at the temperature vs volume graph, we can see that: • If temperature increases, volume increases • If temperature decreases, volume decreases • Charles’s Law: V = bT (b is a proportionality constant) So… b/ = V/T b = V1/T1 and b = V2/T2 V1/T1 = V2/T2 Temperature MUST be in KELVIN

7. Calculating Volume using Charles’s Law A 2.0-L sample of air is collected at 298 K and then cooled to 278 K. The pressure remains constant at 1.0 atm. • Does the volume increase or decrease? • Since the ratio of V/T = b, if T is decreased then V must also decrease so that the constant remains the same • Calculate the volume of air at 278 K Initial Conditions Final Condiitions V1 = 2.0 L V2 = ? T1 = 298 K T2= 278 K 2.0/298 = V2/278 V2 = (278)(2.0/298) V2 = 1.9 L

8. Calculating Temperature In former times, gas volume was used as a way to measure temperature using gas thermometers. Consider a gas with a volume of 0.675-L at 35o C and constant pressure of 1.00 atm. What is the temperature (in oC) of a room where this gas has a volume of 0.535-L ? Initial Conditions Final Condiitions V1 = 0.675 L V2 = 0.535 L T1 = 35o C T2= ? (35 + 273 = 308 K) Then divide by to get: T2 = 0.535/(0.675/308) So T2 = 244 K 0.675/308 = 0.535/T2 Divide both sides by T2 to get: T2(0.675/308) = 0.535 244 – 273 = -29o C

9. Charles’s Law Practice: • V = 1.14 L at 21o C; what is the volume at 42o C • V = 257 mL at 45o C; V = 300 mL at ? o C • V = 2.78 L at -50o C; V = 5.00 L at ? o C • V = 73.5 mL at 0o C; V = ? at 25oC • Suppose 1.25 L of argon is cooled from 291 K to 78 K. What will the new volume of argon be? • Aerosol cans should be stored within a specific temperature range because they are pressurized. Calculate the potential volume of the gas contained in a 500mL can when it is heated from 25o C to 54o C. • Suppose a 25.2 mL sample of helium gas at 29o C is heated to 151o C. What will the new volume of the helium sample be? • Suppose a gas sample is cooled from 600 K to 300 K. Describe how the new volume of gas would relate to the original volume of gas.