Volume and Temperature Charles’s Law Volume and Temperature
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).
Temperature Scales Is 20o C twice 10o C ? Is 20 kg twice 10 kg? 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
Temperature vs. Volume Graph 5 10 15 20 25 30 Volume (mL) Fixed Volume in a cylinder 25 mL at 22C 31.6 mL, 23.1 mL Y=0.0847x + 23.137 Temperature (C) 100 – 273
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. –273C is known as “absolute zero” (no EK) Kelvin kept the convenient units of C, but started at absolute zero. Thus, K = C + 273. 62C = ? K: K=C+273 = 62 + 273 = 335 K Notice that Kelvin is represented as K not K.
Temperature MUST be in KELVIN 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 b = V1/T1 and b = V2/T2 V1/T1 = V2/T2 Temperature MUST be in KELVIN
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
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
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 25o C 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.