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    jeemainoptimist's Avatar
    jeemainoptimist Posts: 2, Reputation: 1
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    #1

    Jan 15, 2015, 02:52 AM
    Total pressure of vessel
    (Q)22 g of dry ice is placed in an an empty 600ml closed vessel at 298K.Find final pressure inside the vessel , if all CO2 gets evaporated?
    Now by applying pv= nrt ,the value of p comes out to be 20.4atm , but in the solution shown , it adds 1 atm i.e atm pressure , so the answer comes out be 21.4atm . Now why do we need to add atm pressure , the Q tells us to find the total pressure inside the vessel , isn't the atm pressure acting only on the container , and not on its contents. Please correct me if am wrong.
    kreysiz's Avatar
    kreysiz Posts: 37, Reputation: 2
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    #2

    Jul 19, 2015, 11:12 AM
    dry ice is solid CO2.
    Molar mass of CO2 = 44 g/mole
    Find the number of moles of CO2 and apply Ideal Gas Equation to find the total pressure in the vessel.
    tickle's Avatar
    tickle Posts: 23,796, Reputation: 2674
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    #3

    Jul 19, 2015, 11:22 AM
    Quote Originally Posted by kreysiz View Post
    dry ice is solid CO2.
    Molar mass of CO2 = 44 g/mole
    Find the number of moles of CO2 and apply Ideal Gas Equation to find the total pressure in the vessel.
    Kreysiz, you must stop answering homework questions.

    What is your point in doing so if it is against site rules?
    kreysiz's Avatar
    kreysiz Posts: 37, Reputation: 2
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    #4

    Jul 20, 2015, 02:09 AM
    Quote Originally Posted by tickle View Post
    Kreysiz, you must stop answering homework questions.

    What is your point in doing so if it is against site rules?
    See, I'm new to the site and am unable to distingiush between homework questions and answerable questions. Please guide me on this.
    smoothy's Avatar
    smoothy Posts: 25,492, Reputation: 2853
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    #5

    Jul 20, 2015, 07:41 AM
    Its simple if you read the post. It can't be anything BUT a homework question. Also the original poster needs to provide what they believe the answer is... and many times how they arrived at that answer. If they don't... they get no help... if they do... and they pick an incorrect method of arriving at the answer... you can guide them in the correct way to get the answer... but you do not simply hand them the answer. If they don't get the answer through their own work... they learn nothing.

    And MOST people posting homework questions are to lazy to do that. Some of them might actually be trying to cheat on tests.

    Look at it this way... would YOU want to hire someone who cheated their way to graduating in their field or someone that actually knows what they are doing? It's clear you worked hard to gain your knowledge... certainly you know people that haven't. Handing them their answers helps them... and hurts people like you who studied and worked hard honestly.

    Look at it like being a tutor.....you aren't helping them learn if you do their work for them.

    Hope that explains it a bit better.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Jul 20, 2015, 02:01 PM
    Personally I think it would be fine to offer a hint for a problem like this, but not do any of the calculations. An acceptable hint for this might be "Have you heard of the Ideal Gas Law?" and if the OP responds back asking for further help in understanding how to apply the it this case then the conversation can go from there.

    Sometimes it can be indeed be difficult to distinguish homework from "real world" problems - there have been cases here where a person asks a question which is clearly not homework, and yet the first response is the knee-jerk "we don't do homework." In general if the problem is simple - meaning it can be solved by applying high school level physics theory and math - then it's almost certainly a homework. question. Real world questions tend to be much messier, and often the question can't even be answered without further information.

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