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Hello students, in this question we have been given an ideal gas in two states, state one and state two.
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In state one, v1 is given to be 18 .25 liters.
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T1 is given to be 15 .9 degree celsius.
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Now t1 in kelvin can be written as 273 plus 15 .9 which is equal to 28.
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8 .9 kelvin.
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Similarly for state 2 we have not been given v2 here actually but t 2 is given which is 40 .7 degrees celsius now t 2 in kelvin can be written as 273 plus 40 .7 which is equal to 313 .7 kelvin.
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Now, we have also been given when the ideal gas is going from state 1 to state 2, p is constant and number of moles is constant.
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So, using the ideal gas equation where pv is equal to nrt, where p is our pressure, v is our volume, n is our moles of gas, r is our idle gas constant, and and t is our temperature in kelvin.
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Now, we will write the ideal gas equation for state 1 and state 2.
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So, state 1 will have p b1 equal to nr t1.
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And for state 2, we will have p v2 equal to nr t 2.
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Now we'll rearrange this equation and write it as b1 by t1 equal to n r by p here for state 2 will be v2 by t2 equal to nr by p now we know that n r by p is constant at state 1 and state 2 thus we can write b1 by t 1 is equal to v2 by t 2 now we do not know the value of v2 so we'll rearrange the equation to find the value of v2 so v2 is equal to v1 into t2 divide by t1 now i will put the values for each one of them so v1 is given to be 18 .25 liters t2 is given to be 313 .7 calvin divided by t1 which is 288 .9 kelvin now, when we solve this, we will get the value of volume at state 2, which is coming out to be 19 .8166, which is approximately equal to 19 .817 liters.
03:41
So for question number 1, we are getting the value volume equal to 19 .817 liter when temperature is 40 .7 degrees celsius given that p and n are constant.
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Okay, now we'll move on to the next part.
04:09
In this question, we are given pressure p1 at 60 atm and temperature t1 at 25 degrees celsius and volume v1 equal to 10 .8 liters.
04:27
Now, for state 2, p2 is given as 370 mm of hg...