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Molecular Theory of Gases Numericals
Molecular Theory of Gases problems
12 class physics numericals
12 class physics problems

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00:00Numerical number 3 of unit number 15 is, calculate the volume occupied by a gram mole of a gas
00:19at 0 degree centigrade and a pressure of 1 atmosphere. So, here we will calculate the volume
00:46so, V is equal to what? Temperature obtained is 0 degree centigrade and we will convert
01:03to Kelvin. So, Kelvin will convert to 273. So, temperature will be 270
01:15Kelvin. Other words, we can say that 0 degree centigrade is 273 Kelvin. And the second pressure obtained is,
01:30pressure is 1 atm means 1 atmosphere. And universal gas constant R, gas constant capital R ka value,
01:45we use 0.0821 letters into atmosphere means atm divided by mol into Kelvin. And the number of moles,
02:09we have obtained a 1. So, N will be equal to 1. So, we will use general gas equation for its calculation,
02:23which is PV is equal to N or T. So, here we have volume calculate. So, V will be equal to N or T.
02:30So, here we have volume calculate. So, V will be equal to N or T. So, V will be equal to N or T.
02:45Yahaan pa pressure multiply multiply, so, V will be equal to N or T. So, V will be equal to N or T. So,
02:52now we will put values. So, now we will put values. N ka value 1 is 1. Enter R. R ka value 0.0821 into temperature which is 273 Kelvin divided by pressure.
03:12Pressure. Pressure bhi hume 180 m means atmosphere obtained hai. So,
03:19jab hum in ko simplify karenge, to hume volume obtained ho ga 22.41 letters. So, this is our final answer.
03:39Answer.
03:40.
03:41.
03:42.
03:43.
03:44.
03:45.
03:46.
03:47.
03:48.
03:49.
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