2.09 ×10^24 molecules of an ideal gas with energy E =...

2.09 \times 10^{24} molecules of an ideal gas with energy $E = \frac{3}{2}N k_B T$ is subjected to a cyclic, quasi-static (i.e., reversible) process shown in the figure. What is the change in the entropy (in J/K) of the gas as it moves from state A to state C? 2.91 \times 10^{24} molecules of an ideal gas with energy $E = \frac{3}{2}N k_B T$ is subjected to a cyclic, quasi-static (i.e., reversible) process shown in the figure. What is the change in the entropy (in J/K) of the gas as it moves from state A to state D?

Transcript

00:06 So we know that, okay, delta s equal to we have integrate dq, right? here, dq, v over t, right? is that clear? so now, here we can write equal to integrate from t1 to t2, then we can.

00:49 Can write here actually cv d t over t right plus from we will integrate to v2 we can write pdv over t right so because as we know pv equal to nr t therefore p equal to we can write right is that clear so now we can write here delta s equal to cv ln t2 over t1 right plus integrate from v1 to v2 and here we can write r dv over v is that clear so now first we can write delta s equal to cv ln t2 over t1 plus nr ln v2 over v1 right so now further we can write here 7 .88 ln here will be 150 over 50 plus to multiply with 1 .9872 by ln 115 over 100 right okay so now here it will cut zero zero cancel here so now we will get here equal to 7 .888 and here and here we 1 .0986 plus we can write 3 .974x multiply with 0 .4054.

03:54 Is that clear? so now we will get here delta s equal to 8 .6569 plus 1 .6969 plus 1 .1 1 .6112 right so now we will get here value of delta as equal to if you will plus both then we will get one 0 .12 sorry not 1 2 actually here 1 0 .2 10 .2 6 81 calorie per more kilby and this will be answered for first actually so now we will solve second question right so here clear in first entropy change for two most ideal gas is 1 0 .2681 calorie per mole kelvin so now second we can solve now delta as as we know so we know that as constant pressure and constant volume so here delta has equal to we can write n c p l n t2 over t1 right this is the formula so now we can write here 3 multiply with 5 over 2 according to question here we can write r ln right r ln and here 600 if we change in kelvin and here we can put 300 right so now here we can write 15 over 2 r ln okay and here if we cut then we will get 2 so now here equal to 15 over 2 multiply with 1 .9872, okay, multiply with 0 .6931.

06:31 Is that clear? so now we will get here 20 .64549 over 2...

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