### Constant R

Posted:

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

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=127&t=59115

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Posted: **Tue Feb 11, 2020 2:07 pm**

When do you use the constant R=8.314 vs 8.206x10^-2 and what are the units for each

Posted: **Tue Feb 11, 2020 2:10 pm**

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.

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.

Posted: **Tue Feb 11, 2020 2:11 pm**

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.

Posted: **Mon Feb 17, 2020 2:41 pm**

What circumstances should we use the difference values of R?

Posted: **Mon Feb 17, 2020 3:06 pm**

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).

Posted: **Mon Feb 17, 2020 4:04 pm**

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.

Posted: **Mon Feb 17, 2020 4:16 pm**

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

Posted: **Mon Feb 17, 2020 4:38 pm**

The units for each are on the equation sheet. Depending on that you decide which to use.

Posted: **Mon Feb 17, 2020 5:04 pm**

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.

Posted: **Mon Feb 17, 2020 5:35 pm**

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

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

Posted: **Tue Feb 18, 2020 6:23 pm**

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.

Posted: **Tue Feb 18, 2020 7:26 pm**

Which version of the gas constant (R) applies to your equation depends on the units present in the equation.

Posted: **Tue Feb 18, 2020 8:53 pm**

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.

Posted: **Tue Feb 18, 2020 8:55 pm**

Just look at the units needed in the problem and choose the R with those units same units

Posted: **Tue Feb 18, 2020 9:01 pm**

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.

Posted: **Tue Feb 18, 2020 9:37 pm**

You use 8.314 when the units are J/(mol*K). You use 0.08206 when the units are (atm*L)/(mol*K).