What kind of pressure do i use for gas laws?

Question

For gas laws like PV=nRT, or PV=PV, or P/T=P/T can I use atms, mm Hg, or Kpa for the pressure, or should i always convert into kPa?

Answer 1

For PV = nRT you want your pressure to be in atm - because the gas law constant (R) has units of Latm/molK

You want your units to work out and cancel such that you can get the missing variable in the problem

When you are dealing with the ratio problems (Charles' law or Boyle's law) it DOES NOT matter what the units for pressure are as long as the units for pressure are the same.

For P1V1 = P2V2 if you are solving for one of the volumes, as long as P1 and P2 are in the same unit, they cancel - so who cares if they are in mm Hg or atm or kPa! If you are given the volumes - the same applies, they can be in mL, L, oz, whatever, as long as they are the same -they will cancel! The ONLY stipulation to that is if you are given P1 in mm Hg, V1 in mL and V2 in mL (therefore you are solving for P2) and you are asked to put your final answer in kPa or atm, at some point (either before you do the math or after you get the answer) you will need to convert to the appropriate unit.

Let's examine a problem, just so you can see that it doesn't matter in the ratio problems

T1 = 273 K T2 = ????
P1 = 945 mm Hg P2 = 762 mm Hg

What is T2?

T2 = P2 x T1/P1

T2 = 945 mm Hg x 273 K/(762 mmHg)

Your mm Hg cancel and you are left with T2 = 338.6 K = 339 K (with correct sig figs)

Let's make the pressures atm instead of mm Hg (but let's make sure that the atm are correct

762 mm Hg x (1 atm/760 mm Hg) = 1.00 atm
945 mm Hg x (1atm/760 mmHg) = 1.24 atm

Let's do the exact same problem after taking the time to do the conversion

P1 = 1.00 atm
P2 = 1.24 atm
T1 = 273 K
T2 = ????

T2 = P2 x T1/P1
T2 = (1.24 atm x 273 K)/1.00 atm = 338.5 K = 339 K

SAME ANSWER!!

Don't waste time converting for ratio problems UNLESS you are asked to report the answer with particular units.

This does NOT apply to the ideal gas law equation. Because the units for R have atm in them, you MUST have pressure units in atm or units won't cancel and you will get the wrong answer!

Answer 2

Prefer to use those units which are associated with gas constant R, when you are using the ideal gas equation:
PV = nRT

Values of R |||||||| Units
8.314472 |||||||| J · K^-1 · mol-1 or L · kPa · K^-1 · mol^-1
0.0820574587||||||| L · atm · K^-1 · mol-1

As a student I would advise to remember these two values of R, because these are used profusely in academics!

For the first value of R use pressures in kPa
For the second value of R use pressures in atm


While using Boyles law and Amagats law which are:
P1V1 = P2V2 &
P1/T1 = P2/T2
use any unit that is required for the sum.
That could be atms, mm Hg, or Kpa , even psi
In such types of sums, it would be absurd to convert everytime to kPa, then you would have to reconvert again, which is just waste of time!

Answer 3

Any of the metric pressure units are OK as they are...
mmHg, Atm, kPa, ..etc.

With Imperial units, gauge pressure (Psig) must be converted to Absolute units (Psia) by adding the gauge pressure to the atmospheric pressure.
e.g. 50psig = 50 + 14.7 = 64.7Psia.

(Absolute Temperature (Kelvin) is a MUST also.
K = °C + 273... °C = K - 273)

Answer 4

It doesn't matter as long as the gas constant (R) you use has the same units. For test purposes, its better to memorize only one gas constant and convert all of your units to fit the one you choose rather than having to memorize all of the gas constants.

Here are a list of all the values for R you can use:
http://en.wikipedia.org/wiki/Gas_constant

Answer 5

You can use any pressure units you're comfortable with. Or use any pressure units that the problem asks for. Just be sure that you have a value for R that goes with the units you use.

Answer 6

always use atmospheres. You may have to convert mmHg to get it.