Three moles of an ideal diatomic gas is taken through a...

Name: Three moles of an ideal diatomic gas is taken through a cyclic process ABCA as shown in the P-T diagram. During the process AB pressure and temperature of the gas varies such that PT² = const. If TB = 300K, calculate: (a) the work done by the gas in the path AB. (b) the heat absorbed or released by the gas in each of the processes. Give answer in terms of the gas constant R.
Uploaded: 2021-12-14T07:51:52-08:00
Duration: 6 min 59 s
Channel: Pranay Shrivastava
Description: Three moles of an ideal diatomic gas is taken through a cyclic process ABCA as shown in the P-T diagram. During the process AB pressure and temperature of the gas varies such that PT² = const. If TB = 300K, calculate: (a) the work done by the gas in the path AB. (b) the heat absorbed or released by the gas in each of the processes. Give answer in terms of the gas constant R.

Three moles of an ideal diatomic gas is taken through a cyclic process ABCA as shown in the P-T diagram. During the process AB pressure and temperature of the gas varies such that PT² = const. If TB = 300K, calculate: (a) the work done by the gas in the path AB. (b) the heat absorbed or released by the gas in each of the processes. Give answer in terms of the gas constant R.

Transcript

00:01 In this question we have given 3 moles of an ideal atomic gas and it is taken through a cyclic process and we have given the pt diagram for this process we have given that for process a to b the variation is pt square is equal to constant now we have to find the work done by the gas in path a b and heat absorbed or released by the gas in each process so given that number of moles is 3 and this is is diatomic gas and for process ab this is p t square is equal to constant and we have given the temperature at point b that is 300 kelvin so as we have so we can say that for a b process we have this equation so we can say that p a t a square is equal to p b t b square so from here we can find the value of t a so we can say that the pressure at a is this is p1 into t a square the pressure at b is to p1 into this is 300 whole square so from here we get the value of t way that is 4 to 4 .264 kelvin now as we can say that we have p t square is equal to constant and as we know that this can be written as p into in terms of volume if we put this is p square v square is equal to constant as we know that pv is equal to n r t or we can say that t is directly proportional to pv so in the replace t we can put this so from here we can say that p this is p cube into v square is equal to constant or we can say this is pv to the power 2 by 3 is equals to constant so this is the polytropic process with index that is n is equal to 2 by 3 now we know that for work done we have to calculate work done through process path ab so we can say that work done for path ab which is polytropic so that is given why that is wab is equal to this is n r t2 minus n r t 1 upon 1 minus n is the number of mole so this is 3 into r into 300 minus 4 24 point 264 upon 1 minus 2 by 3 so we can say that this will comes out to be minus 1111 .11 .18 .376 r.

04:05 Sorry, this is minus 11118 .376 r.

04:17 So this is the answer for first part of the problem.

04:20 Now in the second part we have to calculate the heat for each process...

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