Charles's Law
description
Transcript of Charles's Law
The relationship between temperature and
volume
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).
No. 68F (20C) is not double 50F (10C)
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 0C.• We saw that doubling P yields half the V.• An experiment with a fixed volume of gas in a
cylinder will reveal the relationship of V vs. T…
Temperature scalesIs 20C twice as hot as 10C?
Is 20 kg twice as heavy as 10 kg?
CHARLES' LAW
If a gas has a constant mass and is held at a constant pressure then
the volume divided by the temperature (in kelvins) is a
constant value.
V = kT or = k
VT
Mathematically,
Graphically
Using a glass syringe, a scientist draws exactly 25.5 cm3 of dry oxygen at 20°C
from a metal cylinder. She heats the syringe to 65 °C. What volume will the
oxygen occupy?1. What are you trying to
determine?
What volume will the oxygen occupy?
Vf or V2
Step 2: Identify what is given:Vi = 25.5 cm3
Ti = 20.0 °C
Tf = 65.0 °C
Step 3: Convert temperatures from °C to KTi = 20.0 °C Tf = 65.0 °C
= 20.0 °C + 273 = 65.0 °C + 273
= 293 K = 338 K
Step 4: Use equation to determine final volume:
=
=
x 338 K = Vf
Vi
Ti
Vf
Tf
25.5293
Vf
33825.5
293
Vf = 29.42 cm3
A balloon is filled with 2.50 L of dry helium at 23.5°C. the balloon is placed in a
freezer overnight. The resulting volume is found to be 2.15L. What was the
temperature (in °C) in the freezer.
1. What are you trying to determine?
What was the temperature (in °C) in the freezer.
Tf
Step 2: Identify what is given:Vi = 2.50 L
Ti = 23.5 °C
Vf = 2.15 L
Step 3: Convert temperature from °C to KTi = 23.5 °C + 273
= 297 K
Step 4: Use equation to determine final temperature:
=
= 2.50 L (Tf) = 297K (2.15L)
Tf = 255.42K
Vi
Ti
Vf
Tf
2.50297
2.15 Tf
Cross multiply
Divide by 2.50L to isolate Tf
Step 5: Convert the temperature back to °C
Tf = 255.42K – 273
= -17.6 °C
3. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K?
4. If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?
V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200KUsing Charles’ law: V1/T1 = V2/T2
3.5 L / 300 K = V2 / 200 KV2 = (3.5 L/300 K) x (200 K) = 2.3 L
V1 = 1 L, T1 = 22°C = 295 KV2 = ?, T2 = 100 °C = 373 KV1/T1 = V2/T2, 1 L / 295 K = V2 / 373 KV2 = (1 L/295 K) x (373 K) = 1.26 L