Please help me with Thermodynadmics.
http://i.imgur.com/wBuhxje.png

Question

Please help me with Thermodynadmics.
http://i.imgur.com/wBuhxje.png

photon336 · Answer

$$\Delta H_{v}*mass = (\frac{ kj }{ mol }*moles ) $$

We're given the delta H of vaporization (l->g) but we need the delta H of condensation. 

all we need to do is reverse the values, as in change the sign of the value we're given. 

because think about it: the two processes are the opposite. 

$$100~grams~H_{2}O*(\frac{ mol }{ 18grams }) = 5.6~moles~H_{2}O$$

the entropy is decreasing when going from gas to liquid so the sign should be negative.

$$10~moles_{Water}*-40.67(\frac{ kj }{ mol }) = -407kj$$



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littleredriding · Answer

Thank you, but may I ask where you got the 100 grams from? @Photon336

photon336 · Answer

in your question it said 100 grams for water vapor :)

littleredriding · Answer

Oh, did you switch the 100 C and grams?

photon336 · Answer

well, @LittleRedRiding this what we call a phase change. so the temperature is going to be constant that's why we don't need to use it. also, we know the heat of vaporization already. 
so I used the fact that the total energy we need would be equal to the number of moles* the heat of vaporization. but also notice that I changed the sign too.

littleredriding · Answer

Okay, where did you get the 18 grams from?

photon336 · Answer

alright so 18 grams is the molar mass of water

photon336 · Answer

if you notice the heat of vaporization is in kJ/mol so we need to find the moles of water

littleredriding · Answer

I see! Lol I looked up the molar mass of oxygen.  Also wasn't it 180 g of water vapor?

littleredriding · Answer

@Photon336

photon336 · Answer

yeah, I put that into the calculator and got 10 moles which is right. I looked at it before and I saw it was wrong but mistakenly didn't change it. But i'm confident the final answer is correct.

photon336 · Answer

$$180~grams~H_{2}O*(\frac{ 1~mole }{ 18grams }) = 10~mol~H_{2}O$$

littleredriding · Answer

Okay, thank you. I was just wondering I did something wrong! If you have the time could you help me with another question? I'll tag you into it.

littleredriding · Answer

@Photon336

photon336 · Answer

yeah no problem, strange I didn't see your tag okay go ahead

photon336 · Answer

your answer was right. it's just that you needed to change the sign because when you have condensation you're going from g --> l which means energy is being released.

photon336 · Answer

Make sure you look at the units too. that's a good indication as to whether or not you're doing things right.