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I can understand upto where the ideal gas equation is used to calculate volume of steam. But on trying to find out $ \Delta V$ , i saw that there are no gases on the reactant side so then $P\Delta V=(\Delta n)\times R\times T$ where $\Delta n$ would be $1.39$. Then $\Delta V$ is simply the volume of the steam i.e., $42.51$. But when calculating the work a value of $42.8$ has been used for $\mathrm dV$. Why is that done? Secondly, there seems to be two different values for $\Delta$H. First they have written it to be $\pu{12470.6 cal}$. and then while calculating $\Delta U$ they have used the value of $\pu{13500 calories}$. (That value seems to have come from the given data: If we multiply $9.72\times 10^3$ (molar $\Delta H_{vap}$) by $1.39$ (number of moles) we get something around $13500$). How did those two different values appear? I also want to tell here that this is not a hw question for me, as you can see this is a solved example, so you can consider it as my attempt to understand thermochemistry better

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  • $\begingroup$ Using units will very often reduce errors in calculations and help those you may ask for help $\endgroup$ – A.K. Oct 15 '18 at 15:16
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The $\Delta V$ is supposed to be the volume of the vapor minus the volume of the liquid. The volume of the liquid is only 0.025 liters, which they neglect.

I don't know where they get the 12470.6 from. I'm guessing this is an error.

(0.082)(373)(25)/18=42.48. You solve it basically the way you said.

Δ(PV)=42.48 liter−atm =1029 cal=42.48 liter-at = 1029 cal

So, ΔU=13500−1029=12471 cal

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  • $\begingroup$ but sir how did it become 42.8 from 41.52. Even if we consider the volume of liquid, as you said we must subtract, but here they have added 0.39 . Does it have any connection with 1.39? $\endgroup$ – Maxwell's Ghost Oct 15 '18 at 14:53
  • $\begingroup$ shouldn't we consider only the substances in gaseous phase(after you said that $\Delta$V is volume of vapour minus volume of liquid) $\endgroup$ – Maxwell's Ghost Oct 15 '18 at 14:57
  • $\begingroup$ I don't know where they get the 42.8 liters from. That seems to me to be an error also. If the initial volume of the liquid were taken into account, it should have been 42.48 - 0.025 =42.45 liters. As far as your second questions is concerned, the initial state is liquid and the final state is vapor. So ΔV should be the volume of the vapor minus the volume of the liquid. In their calculation, they neglect the volume of the liquid compared to the volume of the vapor. $\endgroup$ – Chet Miller Oct 15 '18 at 15:38
  • $\begingroup$ sir , how did we get 42.48(in your comment)? so then how do i solve this problem $\endgroup$ – Maxwell's Ghost Oct 15 '18 at 16:06
  • $\begingroup$ (0.082)(373)(25)/18=42.48. You solve it basically they way you said. $\Delta (PV)=42.48\ liter-atm\ =1029\ cal$. So, $\Delta U=13500-1029=12471\ cal$ $\endgroup$ – Chet Miller Oct 15 '18 at 16:21

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