second equation


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connie 2C
Posts: 106
Joined: Thu Jul 11, 2019 12:17 am

second equation

Postby connie 2C » Mon Feb 03, 2020 10:47 pm

in the second equation, would we ever have to use the integral or would we just be using the equation -nRT(V2/V1)?

connie 2C
Posts: 106
Joined: Thu Jul 11, 2019 12:17 am

Re: second equation

Postby connie 2C » Mon Feb 03, 2020 10:48 pm

sorry the equation is -nRTln(V2/V1)*^

RichBollini4G
Posts: 100
Joined: Wed Sep 18, 2019 12:18 am

Re: second equation

Postby RichBollini4G » Tue Feb 04, 2020 2:45 pm

connie 2C wrote:in the second equation, would we ever have to use the integral or would we just be using the equation -nRT(V2/V1)?

I believe it depends on the question.

Vicki Liu 2L
Posts: 101
Joined: Sat Aug 24, 2019 12:15 am

Re: second equation

Postby Vicki Liu 2L » Tue Feb 04, 2020 4:05 pm

I believe the integral was just involved in the process of deriving the equation.

Wendy 1E
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Joined: Sat Aug 17, 2019 12:17 am
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Re: second equation

Postby Wendy 1E » Tue Feb 04, 2020 4:22 pm

Unless asked to derive the equation, you would just use the -nRTln(v2/v1).

Madelyn Romberg 1H
Posts: 102
Joined: Tue Oct 02, 2018 12:16 am

Re: second equation

Postby Madelyn Romberg 1H » Tue Feb 04, 2020 5:44 pm

It is really used for when the pressure isn't constant do to find the work done, it is the integral of the subsequent volume changes. It is also used to derive the first formula.

nicole-2B
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Joined: Fri Aug 30, 2019 12:18 am

Re: second equation

Postby nicole-2B » Tue Feb 04, 2020 6:12 pm

and this equation is used for reversible reactions only?

Sally Qiu 2E
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Joined: Fri Aug 30, 2019 12:18 am

Re: second equation

Postby Sally Qiu 2E » Tue Feb 04, 2020 6:14 pm

if you solve the integral you would get ln( V2/V1).

Matthew Tsai 2H
Posts: 101
Joined: Wed Sep 18, 2019 12:20 am

Re: second equation

Postby Matthew Tsai 2H » Tue Feb 04, 2020 7:57 pm

In this case, the only time you would need to integrate would be if asked to derive the equation, in which case you would need to understand each step of the integral on a conceptual level. For actual calculations you can just use the formula.


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