The pressure in a car tire is 198 kpa at 27c after a long...

The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant. A sample of nitrogen gas has a pressure of 6.58 kPa at 539 K. If the volume does not change, what will the pressure be at -62°C. 3. Find the volume of a gas at STP, if its volume is 80.0 mL at 109 kPa and -12.5°C.

Transcript

00:01 Hello everyone so here we are going to be writing the question first of all the given part whatever the thing is given in the question that is a pressure is given that is we are going to be taking as a p1 -198 kpa and even it would be a 27 degree shelter so we need to convert it into the calvin so how it's going to be converted to 73 270 sorry to 7 27 plus 273 we are going to get here 264 calvin right and when we are going to see the pressure two that is a p2 is given 2 to 5 pka is given and t2 that is actually we need to be calculate over here and here the v is remain a what it's remained the constant so according to the question let us see what the question is going to be safe so these are the given value we need to find out what is the temperature of the air in the tire okay the pressure of the car tire that they have been provided along with the temperature after a long drive also the pressure is going to be given volume it's remain the constant what will be the temperature inside the tire of that particular air and for the second questions we need to see that the sample of the nitrogen gas is going to be is having the pressure this much at that this much temperature we are need to be find out what will be the pressure at minus 62 degrees celsius and according to the third questions we need to be find out over here we need to find out the volumes of the gas at the standard as cp okay so let us write it over here that according to the gailusak law i'm going to be right that of particular equation p relation b1 it's going to become a t1 is equal to p2 divided by t2 i would like to write this as in t1.

02:03 So what the thing is going to be finding out over here by us? that is 198 kpa.

02:09 The thing is given in the question.

02:11 The given thing we are going to be putting this in the formula.

02:14 Okay, is equal to 225 kpa divided by t2.

02:21 So our t2 is equal to 2 to 5 kpa multiplied by 264.

02:28 K divided by 198 kpa so this b a is going to be cancelled our answer would be 59400 0 divided by 198 the temperature at the t2 is equal to 300 kelvin so this we are getting from the first answer we are getting right and next we are going to see the second question in the second questions we need to again are asking about the pressure only right so going to be writing the p1 is equal to 6 .5 8 kb a p2 we are no need to be find out over here t1 is given 539 calvin t2 is given minus 62 degrees celsius when we are going to be converted into the calvin minus 62 plus 273 how much we are getting 2 -11 calvin so according to the galusac law again we are going to be using galozyc law p1 t1 is equal to t2 the values which is given we are going to be putting the value we are going to be get for the p2 is equal to p1 t2 divided by t1 here we are going to put the value 6 .5 5 8 multiplied by yeah again that kpa multiplied by 211 kelvin divided by 5 3 9 here the calvin calvin is going to be cancelled so what is going to be remaining the thing is is 1388 .38 divided by 539...

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