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Constant R

Posted: Tue Feb 11, 2020 2:07 pm
by Haley Dveirin 1E
When do you use the constant R=8.314 vs 8.206x10^-2 and what are the units for each

Re: Constant R

Posted: Tue Feb 11, 2020 2:10 pm
by Jessica Booth 2F
When do you use the constant R=8.314 vs 8.206x10^-2 and what are the units for each
The units are R = 8.314 J/(K mol) and 8.206x10^-2 L atm/(K mol). You use 8.314 whenever you are working with J and you use 8.206x10^-2 when using atm and L.

Re: Constant R

Posted: Tue Feb 11, 2020 2:11 pm
by JesseAuLec1Dis1G
You just have to pay attention to the units of each. Just make sure your calculations cancel out units so you end up with the units you want when you've finished all of your calculations.

Re: Constant R

Posted: Mon Feb 17, 2020 2:41 pm
by Natalie Benitez 1E
What circumstances should we use the difference values of R?

Re: Constant R

Posted: Mon Feb 17, 2020 3:06 pm
by Benjamin Feng 1B
Usually when you are solving for pressure or volume using PV=nrt you use the one with liters and atm (.08205). If you are solving for work, it is more common to use the one with Joules in the units (8.3145).

Re: Constant R

Posted: Mon Feb 17, 2020 4:04 pm
by MAC 4G
Natalie Benitez 1E wrote:What circumstances should we use the difference values of R?


The best way to figure out which value of R to use, I would look at what units you're given from the question to see what you can cancel out. Similarly, you can also look at what units you need in your final answer as to what you are looking for to determine how the units will cancel out to give you your desired units.

For example, if you are looking for an answer in joules, and you are given temperature (in K) and moles (or a value you can convert to these units) you would use R = 8.314 J/(K mol). So, the K and moles will cancel out leaving you with Joules.

Re: Constant R

Posted: Mon Feb 17, 2020 4:16 pm
by Brian J Cheng 1I
As many have said, look at the units you need for a certain calculation and choose the R value accordingly. For example: you're given values in Liters, pressure, temperature, and mols, then you would use R = 0.08206 L*atm/K*mol

Re: Constant R

Posted: Mon Feb 17, 2020 4:38 pm
by pmokh14B
The units for each are on the equation sheet. Depending on that you decide which to use.

Re: Constant R

Posted: Mon Feb 17, 2020 5:04 pm
by Kishan Shah 2G
The best way to determine when to use which instead of memorizing them is to look at the units to see which units you are working with and what you want to end up with.

Re: Constant R

Posted: Mon Feb 17, 2020 5:35 pm
by Emil Velasco 1H
You can find the units used on the equation sheet

Just look for what units the problem gives you and what you are trying to get. Using R with the right units allows you to cancel out the things you don't need in order to get what you want

Re: Constant R

Posted: Tue Feb 18, 2020 6:23 pm
by 305421980
To determine which version of R is used I always look at what other units are being used in the problem and which correspond to the R constant units on the equation sheet.

Re: Constant R

Posted: Tue Feb 18, 2020 7:26 pm
by EllieSchmidtke_4I
Which version of the gas constant (R) applies to your equation depends on the units present in the equation.

Re: Constant R

Posted: Tue Feb 18, 2020 8:53 pm
by Juana Abana 1G
To determine when to use which it is best to look at the units to see which units will cancel out and also see what units you will be left with.

Re: Constant R

Posted: Tue Feb 18, 2020 8:55 pm
by Jack Riley 4f
Just look at the units needed in the problem and choose the R with those units same units

Re: Constant R

Posted: Tue Feb 18, 2020 9:01 pm
by Esha Chawla 2E
Haley Dveirin 1E wrote:When do you use the constant R=8.314 vs 8.206x10^-2 and what are the units for each


To determine which R to use, look at the units provided in the problem. The units for R=8.314 are J/K*mol and the units for R = 0.08206 are L*atm/K*mol.

Re: Constant R

Posted: Tue Feb 18, 2020 9:37 pm
by Aarushi Solanki 4F
You use 8.314 when the units are J/(mol*K). You use 0.08206 when the units are (atm*L)/(mol*K).