 
					
					
						Charles’s law: Volume and temperature					
				 
				
					
						 المؤلف:  
						John T. Moore, EdD
						 المؤلف:  
						John T. Moore, EdD					
					
						 المصدر:  
						Chemistry Essentials For Dummies
						 المصدر:  
						Chemistry Essentials For Dummies					
					
						 الجزء والصفحة:  
						p 164
						 الجزء والصفحة:  
						p 164					
					
					
						 30-1-2017
						30-1-2017
					
					
						 2107
						2107					
				 
				
				
				
				
				
				
				
				
				
			 
			
			
				
				Charles’s law: Volume and temperature
Charles’s law (named after Jacques Charles, a 19th-century French chemist) has to do with the relationship between volume and temperature, keeping the pressure and amount of the gas constant. Ever leave a bunch of balloons in a hot car while running an errand? Did you notice that they expanded when you returned to the car?
Charles’s law says that the volume is directly proportional to Kelvin temperature. Mathematically, the law looks like this:
V = bT or V/T = b (where b is a constant) This is a direct relationship: As the temperature increases, the volume increases, and vice versa. For example, if you placed a balloon in the freezer, the balloon would get smaller. Inside the freezer, the external pressure, or atmospheric pressure, is the same, but the gas particles inside the balloon aren’t moving as fast, so the volume shrinks to keep the pressure constant. If you heat the balloon, the balloon expands and the volume increases.
If the temperature of a gas with a certain volume (V1) and Kelvin temperature (T1) is changed to a new Kelvin temperature
(T2), the volume also changes (V2):
V1/T1 = b V2/T2 = b
The constant, b, is the same, so
V1/T1 = V2/T2 (with the pressure and amount of gas held constant and temperature expressed in K) If you have three of the quantities, you can calculate the fourth.
For example, suppose you live in Alaska and are outside in the middle of winter, where the temperature is –23°C. You blow up a balloon so that it has a volume of 1.00 liter. You then take it inside your home, where the temperature is a toasty 27°C.
What’s the new volume of the balloon? First, convert your temperatures to Kelvin by adding 273 to the Celsius temperature:
Inside: –23°C + 273 = 250 K
Outside: 27°C + 273 = 300 K
Now you can solve for V2, using the following setup:
V1/T1 = V2/T2
Multiply both sides by T2 so that V2 is on one side of the equation by itself:
[V1T2]/T1 = V2
Then substitute the values to calculate the following answer:
[(1.00 L)(300 K)]/250 K = V2 = 1.20 L
It’s a reasonable answer, because Charles’s Law says that if you increase the Kelvin temperature, the volume increases.
				
				
					
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