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If we have one mole monoatomic ideal gas was taken through process AB as shown in the figure then ∆ S is positive and work is also positive but how can we find the value of work and heat absorbed as pressure is changing .kenter image description here

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    $\begingroup$ From what you learned in middle school algebra, what is the equation for the straight line between points A and B in terms of V and T? If you solve this equation for T and substitute it into the ideal gas law, what do you get for P in terms of V? What is the equation for the work in terms of P and V? If you know the work and the change in internal energy, do you know the heat? $\endgroup$ – Chet Miller Dec 18 '15 at 12:53
  • $\begingroup$ Yes but I forget that $\endgroup$ – user101522 Dec 18 '15 at 12:56
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    $\begingroup$ Do you remember it now, or do you need to go back to your old algebra text book? This is not really a chemistry question, it's a math question. $\endgroup$ – Chet Miller Dec 18 '15 at 13:00
  • $\begingroup$ But can you help me $\endgroup$ – user101522 Dec 18 '15 at 13:03
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    $\begingroup$ Can you help yourself? He literally spelt out everything you need to do. Start by finding $p$ in terms of $V$. Google "equation of a straight line through two points" if you need maths help. $\endgroup$ – orthocresol Dec 18 '15 at 13:04
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Since you know the number of moles of ideal gas, you can find the inital and final pressures with the ideal gas law: $PV = nRT$. For entropy you may first take that $U = \frac{3}{2} nRT$, and that $dU = \delta Q + \delta W$. $ $ $\delta Q $ and $ \delta W$ may be substituted by $\delta Q = T \mathrm{d}S$, and $\delta W = -p\mathrm{d}V$. This should be enough for you to solvew the problem.

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  • $\begingroup$ Why you have equate Q = Tds and in this how will we find ds $\endgroup$ – user101522 Dec 18 '15 at 15:41
  • $\begingroup$ You find d$S$ because you know everything else. $\endgroup$ – A.K. Dec 18 '15 at 15:44
  • $\begingroup$ W = -pdV but pressure is changing $\endgroup$ – user101522 Dec 18 '15 at 16:09
  • $\begingroup$ Do it in two steps where you change pressure, then volume and add the two results. $\endgroup$ – A.K. Dec 18 '15 at 16:19

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