When to use R = 0.082 vs R = 8.314

[arc5005]

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When to use R = 0.082 vs R = 8.314

Anyone have a quick kinda way to memorize/know when to use either of these?

So far in my review I've mostly only been using the R = 0.082, but I did have 1 problem come up where R = 8.314 instead.

 

[Lawpy]

Look at the units for the ideal gas constant. R = 0.082 has units of (L atm)/(mol K), whereas R = 8.314 has units of J / (mol K). If the question has pressure and volume expressed in units of atm and liters respectively, you use the R = 0.082 value. If the pressure is expressed in pascals and volume in cubic meters, you use R = 8.314 value (same if pressure is expressed in kilopascals and volume in liters, because 1 kPa = 1000 Pa and 1 L = 1/1000 m^3).

The two R values are the same thing but the reason for the differences is due to having different units. R = 8.314 is actually the more important value because it is expressed in SI units (note that 1 J = 1 Pa * m^3= 1 N/m^2 * m^3 = 1 N * m).

 

[arc5005]

Look at the units for the ideal gas constant. R = 0.082 has units of (L atm)/(mol K), whereas R = 8.314 has units of J / (mol K). If the question has pressure and volume expressed in units of atm and liters respectively, you use the R = 0.082 value. If the pressure is expressed in pascals and volume in cubic meters, you use R = 8.314 value (same if pressure is expressed in kilopascals and volume in liters, because 1 kPa = 1000 Pa and 1 L = 1/1000 m^3).

The two R values are the same thing but the reason for the differences is due to having different units. R = 8.314 is actually the more important value because it is expressed in SI units (note that 1 J = 1 Pa * m^3= 1 N/m^2 * m^3 = 1 N * m).

Thanks again!

 

[Thor123422]

Just remember that you can easily convert between the two if you just remember a few constants.

Just remember when you do the problem your units will matter, so if you do a quick unit-cancelling you will know if you used the right one or the wrong one.

For instance if they want you to solve for pressure, then P = nRT/V won't cancel units using 8.314 J/(mok*K). You'll end up with P = X Joules/Liter

 

[Lawpy]

Just remember that you can easily convert between the two if you just remember a few constants.

Just remember when you do the problem your units will matter, so if you do a quick unit-cancelling you will know if you used the right one or the wrong one.

For instance if they want you to solve for pressure, then P = nRT/V won't cancel units using 8.314 J/(mok*K). You'll end up with P = X Joules/Liter

The units do cancel with joules/liter. A joule is a unit of work, which for gases is just pressure * volume. 1 J = 1 Pa * m^3 = 1 kPa * 1 L. So 1 joule / liter = 1 kPa * L / L = 1 kPa.

The only time R = 0.082 is used is when the pressure is expressed in atm and volume in liters. R = 8.314 is a much more common and important value when pressure and volume are expressed in SI units, whether Pa and m^3 or kPa and L. The 8.314 J/(mol K) especially links between free energy and temperature, which is why in equations like G = -RTln(K), you use R = 8.314 J/(mol K).

Also the two R values are equivalent as seen by the following:

0.082 L atm/(mol K) * 101325 Pa/1 atm * 1 m^3/1000 L * 1 J/(Pa * m^3) = 0.082 * 101.325 J/(mol K) = 8.31 J / (mol K). It just helps to know both because of time constraints in the MCAT.

 

[Thor123422]

The units do cancel with joules/liter. A joule is a unit of work, which for gases is just pressure * volume. 1 J = 1 Pa * m^3 = 1 kPa * 1 L. So 1 joule / liter = 1 kPa * L / L = 1 kPa.

The only time R = 0.082 is used is when the pressure is expressed in atm and volume in liters. R = 8.314 is a much more common and important value when pressure and volume are expressed in SI units, whether Pa and m^3 or kPa and L. The 8.314 J/(mol K) especially links between free energy and temperature, which is why in equations like G = -RTln(K), you use R = 8.314 J/(mol K).

Also the two R values are equivalent as seen by the following:

0.082 L atm/(mol K) * 101325 Pa/1 atm * 1 m^3/1000 L * 1 J/(Pa * m^3) = 0.082 * 101.325 J/(mol K) = 8.31 J / (mol K). It just helps to know both because of time constraints in the MCAT.

You're going out of your way to explain a lot of things I already agreed with, and that was explained above.

My point was that a quick way to see if you're using the right value is just to do a simple unit cancellation since it only takes a few seconds. You can go through the conversion, but really it's less efficient than just doing a simple dimensional analysis.

 

[Lawpy]

You're going out of your way to explain a lot of things I already agreed with, and that was explained above.

My point was that a quick way to see if you're using the right value is just to do a simple unit cancellation since it only takes a few seconds. You can go through the conversion, but really it's less efficient than just doing a simple dimensional analysis.

Right, I was simply clarifying things (like joules/liter can be reduced to kilopascals). I just think the R = 0.082 value is only used when pressure is in atm and volume in liters. As long as that fact is remembered, it's good because the R = 8.314 value can be used for all of the remaining things.

 

[arc5005]

Just remember that you can easily convert between the two if you just remember a few constants.

Just remember when you do the problem your units will matter, so if you do a quick unit-cancelling you will know if you used the right one or the wrong one.

For instance if they want you to solve for pressure, then P = nRT/V won't cancel units using 8.314 J/(mok*K). You'll end up with P = X Joules/Liter

Right, I was simply clarifying things (like joules/liter can be reduced to kilopascals). I just think the R = 0.082 value is only used when pressure is in atm and volume in liters. As long as that fact is remembered, it's good because the R = 8.314 value can be used for all of the remaining things.

So for a problem like this, how would I know which one to use?

What is the root mean square speed of neon atoms at 27° C?

A. 19.3 m/s
B. 211 m/s
C. 612 m/s
D. 1018 m/s

C) 612 m/s

Because scientists employ the MKS system, the mass must be in terms of kilograms, and temperature must be in units of kelvins. The mass of one mole of neon is 0.020 kg and the temperature is 300 K.

 

[Lawpy]

So for a problem like this, how would I know which one to use?

What is the root mean square speed of neon atoms at 27° C?

A. 19.3 m/s
B. 211 m/s
C. 612 m/s
D. 1018 m/s

C) 612 m/s

Because scientists employ the MKS system, the mass must be in terms of kilograms, and temperature must be in units of kelvins. The mass of one mole of neon is 0.020 kg and the temperature is 300 K.

You would use R = 8.314 J/mol K. You very rarely use R = 0.082 L atm/mol K value unless you are specifically dealing with pressure expressed in atmospheres (atm) and volume in liters (L).

Connecting Gas Properties to Kinetic Theory of Gases