0:02 this is msj chem in this video i'll be
0:03 looking at
0:06 examples involving the gas laws in our
0:08 first example we have a sample of gas
0:11 with a volume of 15 cubic centimeters
0:14 at a pressure of 575 kilopascals
0:16 assuming that temperature remains
0:19 constant what volume will the gas occupy
0:22 at a pressure of 968 kilopascals
0:24 to solve this problem we'll use the
0:26 equation p1
0:30 times v1 is equal to p2 times v2
0:33 which is boyle's law in the question we
0:33 are given
0:39 p1 v1 and p2 so we need to solve for v2
0:41 so here we have the equation rearranged
0:43 to solve for v2
0:45 next we substitute the values into the equation
0:46 equation
0:50 p1 is 575 kilopascals
0:54 and v1 is 15.0 cubic centimeters
0:58 p2 is 968 kilopascals
1:01 and this gives us a value for v2 of 8.91
1:04 cubic centimeters in our next example we
1:06 have a sample of gas with a volume of
1:09 32.0 cubic decimeters
1:12 at a temperature of 256 kelvin
1:14 assuming the pressure remains constant
1:15 what volume
1:18 will the gas occupy at a temperature of
1:20 391 kelvin
1:22 to solve this we'll use this equation
1:24 which is charles's law
1:27 in the question we're given v1 t1
1:30 and t2 so we need to rearrange this
1:32 equation to solve for v2
1:36 and v2 is equal to v1 times t2
1:39 divided by t1 so next we substitute the values
1:40 values
1:43 into the equation v1 is 32.0
1:48 cubic decimeters and t2 is 391 kelvin
1:52 t1 is 256 kelvin and this gives us a
1:53 value for v2
1:57 of 48.9 cubic decimeters
1:59 in our next example we have a sample of
2:03 gas with a pressure of 73.9 kilopascals
2:06 at a temperature of 347 kelvin
2:08 assuming that volume remains constant
2:10 what will be the pressure of the gas
2:14 at a temperature of 602 kelvin so here
2:16 we have the equation for gaylor sac's law
2:16 law
2:20 in the question we're given p1 t1
2:24 and t2 so we need to solve for p2
2:26 and here we have the equation rearranged
2:28 to solve for p2
2:30 next we substitute the values into the equation
2:31 equation
2:34 p1 is 73.9 kilopascals
2:38 t2 is 602 kelvin and t1
2:42 is 347 kelvin and this gives us a value
2:46 for p2 of 128 kilopascals
2:48 in the next example a sample contains
2:50 5.13 moles of gas
2:54 with a volume of 1.28 cubic meters
2:55 assuming that temperature and pressure
2:58 remain constant what volume will the gas occupy
2:59 occupy
3:02 if 3.49 moles of gas are added
3:04 so here we have the equation for
3:06 avogadro's law
3:10 in the question we're given v1 and n1
3:12 to find n2 we need to add the two amount
3:14 in moles of gas
3:18 which is 5.13 plus 3.49
3:20 to find v2 we need to rearrange the equation
3:21 equation
3:25 so v2 equals v1 times n2
3:31 divided by n1 v1 is 1.28 cubic meters
3:34 n2 is 8.62 moles and
3:38 n1 is 5.13 moles and this gives us
3:43 a value for v2 of 2.15 cubic meters
3:45 in our last example a sample of gas has
3:48 a volume of 1.54 cubic meters
3:51 at a temperature of 447 kelvin
3:55 and a pressure of 12.0 kilopascals
3:57 if the temperature and pressure are
4:00 changed to 561 kelvin
4:03 and 15.7 kilopascals respectively
4:06 what volume will the gas occupy to solve
4:09 this we'll use the combined gas law
4:12 in the question we're given p1 v1
4:16 t1 p2 and t2
4:18 so we need to rearrange the equation to
4:20 solve for v2
4:23 and v2 equals v1 times p1
4:26 times t2 divided by t1
4:31 times p2 v1 is 1.54 cubic meters
4:34 p1 is 12.0 kilopascals
4:38 t2 is 561 kelvin t1
4:41 is 447 kelvin and p2
4:45 is 15.7 kilopascals so this gives us a
4:46 value for v2
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