# R in PV=nRT (1 Viewer)

#### ColdMint123

##### New Member
It appears that when solving questions involving this equation, the value of R depends on units of pressure and volume and sometimes temperature (which is almost always kelvin).

How do we know which value of R to use in a given question and do we have to memories each of these values?

#### mikikieko12

##### New Member
No you don't need to memorise the values, its usually very straightforward (as they usually give you the gas constant if its not the one in the formula sheet). However, there is a common pattern for the variables (check image/attachment).

Usually temperature is always in Kelvins when subbing into the equation, volume may need to be altered depending on what unit it is (volume needs to be m^3 or in L). For pressure there are many conversions (might need to memorise some if you are not familiar).

Pressure conversions:
Chemistry standard is the kilopascal (kPa) / Physics standard is the pascal (Pa)
Pa = 0.001 kPa
MPa = 1000 kPa
1 bar = 100 kPa
1 ATM = 101.3 kPa
1mm Hg = 1 Torr = 0.1333 kPa
1mm H2O = 0.00980 kPa

Also this website is really helpful: https:// www .chemguide.co.uk/physical/kt/idealgases.html

Hope this helps!

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#### blyatman

##### Well-Known Member
That's because R is not dimensionless: E.g. for air at STP, R = 287 J/(kg K) in SI units. Thus, you need to match your units with the units of the value of R you use. So if you wanted to use units of kJ, kg, K, then R = 0.287 kJ/(kg K).

See the link below for different values of R depending on your choice of units.

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#### CM_Tutor

##### Well-Known Member
R = 8.314 J K-1 mol-1 is the value in SI units. It works for:
• P in Pa, V in m3, n in mol, T, in K
• P in kPa, V in L = dm3, n in mol, T in K
This is the standard value which relates to $\bg_white R = k_B N_A$

R = 0.08206 L atm K-1 mol-1 works for:
• P in atm, V in L, n in mol, T in K
If given a volume in some other unit (like mL or cm3), convert to L or m3 as appropriate. If given a pressure in some other unit (ie. 747 mmHg), convert to a suitable pressure unit and apply above...
• $\bg_white 747 \text{ mmHg } = \cfrac{747}{760} \text{ atm}$
• $\bg_white 747 \text{ mmHg } = \cfrac{747}{760} \times 101.3 \text{ kPa}$
If given a temperature not in K, convert to K. The ideal gas law and underlying theory requires a thermodynamic temperature with a zero that is physically meaningful and corresponding to absolute zero. You cannot have a temperature in Celsius and use PV = nRT because Charles' Law tells us that V is proportional to T only if T = 0 corresponds to the theoretical V = 0 limit for a gas... in other words, if T is in Celsius then $\bg_white PV \neq nRT$.

The ideal gas law can be modified to uses masses in kg rather than chemical amount in moles, but this also means the equation becomes
$\bg_white PV = \cfrac{m}{M} RT$ with the above R values, or a change to PV = mRT with R a gas constant with units of g or kg. (Note: SI would say mas must be in kg, but since we seek moles and M is usually in g mol-1, using m in g is usually appropriate. As with all cases, keeping track of units means making sure that they suit the situation rather than applying blanket rules blindly.)

You can have an R in bar or other units, but it is generally easier to change to P to atm or kPa or Pa and use the above Rs.

Addendum / Note: A link above gives R = 8.314 L kPa K-1 mol-1. This is the same as R = 8.314 J K-1 mol-1 as, first:
• 1 J = 1 N m
• 1 m3 Pa = 1 3 N m-2 = 1 N m
• So, 1 J = 1 m3 Pa
Now, looking at the units of R:
• 1 L = 1 dm3 = (0.1 m)3 = 10-3 m3
• 1 kPa = 103 Pa
• So, 1 L kPa = 10-3 . 103 m3 Pa = 1 m3 Pa = 1 J
• And thus, 1 L kPa K-1 mol-1 = 1 J K-1 mol-1
This is why I don't bother to list R with two sets of units (and nor do most data sheets) and just recognise that R will work with either the combination (P in kPa, V in L / dm3) or the SI option (P in Pa, V in m3).

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