00:01
So here they have taken one of the bucket and they submerged the graduated cylinder.
00:10
Submerge the graduated cylinder in the trough or bucket and filled with water.
00:17
So like this it has to submerge and it is filled with the water like that.
00:23
So here no air bubbles are the start of the experiment.
00:27
No air bubbles.
00:29
No air bubbles.
00:33
Now what they are doing here? we have to here we are placed the butane lighter so underneath the opening of the graduated cylinder so here we have placed the the butane lighter under the opening of graduated cylinder so where the graduated cylinder then there we have to place the butane lighter opening of graduated cylinder that is the given condition so now here we have to fill the particular cylinder with butane with butane how by holding the trigger on the butane lighter okay now we have filled with the butane now we have to see here there is no leakage of gas no leakage of gas that is escaped from the around the graduated cylinder then now we are displacing 90 m l of water from graduated cylinder now adjust the so from the graduated cylinder we have displaced the 90 m l of water now we have to adjust the particular cylinder adjust so that 100 m l mark this is lines up to the level of water level of water so here we have to fill the graduated cylinder to 100 m l mark with butin so here we have to fill the graduated cylinder to the 100 m l mark to 100 ml mark with the butin 100m mark with butin that is the question so coming to the solution so here now we are here we are using ideal gas equation that is p1 v1 b1 t1 that is equal to p2 v2 by t 2 p2 means at a s tp condition and a volume at stp condition now temperature at stp condition so here the p1 value that is partial pressure of butin what is p1 that is partial pressure of butane so the partial pressure of butane so value it is 0 .96768 atmosphere and the value of volume of that is v1 that is what is the v1 volume of gas collected the volume of gas collected it will be 100 ml it will be what is the value here it will be 100m that is whenever we convert into liters that is 0 .1 liters now what is the of t1 now the value of t1 so that even temperature it will be water water water water temperature it is a 273 plus 22 so that is equal to 295 kelvin's now the value of pstp we know the pressure at stp 1 atmosphere and temperature at stp what is the temperature at stp that will be 25 degrees centigrade or it is a 273 kelvin's in terms of the kelvin's so now we have to find out the volume of butane gas volume of butane gas at stp volume of butane gas at stp so how to find out that is vstp so simply here we are using p1 v1 is equal to p2 v2 so that is equal to p2 v1 divided by t1 into t at stp divided by pressure at stp condition.
04:57
So simply substituting the values in this particular equation.
05:02
So 0 .967368 into 0 .1 divided by 295 into 273 into 273 divided by 1.
05:13
Whenever we do the calculation will get 0 .08 95 liters.
05:17
What will be the volume of butane gas at stp condition the volume of butane gas at stp condition that is equal to 89 .5 ml 89.
05:33
82m.
05:34
Now now so we are going to solve here that is the 10th question.
05:46
So that is the what will be the so what will be the most of butane gas...
