In gas laws, when you are given pressure in mmhg, do you have to change this to actual atm pressure? V=nRT/P?

Question

If you are given 626mmhg for pressure, to solve for volume, in the above formula,do you have to change this into atm pressure, or can you use either in the P part of the equation? Which would be 626mmhg/760mmhg for the conversion to atmoshere pressure. If I knew this, the gas laws would be a little easier. Thanks for help, I like chemistry, but it can get a little hairy!

Answer 1

You just have to make sure that all your units are the same. If you want to use mmHg, then use a gas constant that is in mmHg. If you want to use atm, then use a gas constant that is in atm. I typically use R=0.082 (L*atm)/(mol * K) If you want to do it in mmHg, use R=62.36 (L*mmHg)/(mol*K).

The gas constant exists in all sorts of forms depending on what units you want to use. Decide on the units you want your answer to be in, then choose the right gas constant, and then convert the things you are given in the problem to the appropriate units if necessary. Check out the wiki link for all the different gas constants.

Answer 2

hi
It all depends in which unit the R constant (ideal gas constant) is given. There are even different values for R, depending of the units to be used, even if all the values represent exactly the same.
R can be, as an example 0.08205784 L·atm·K-1·mol-1, or 62.3637 L·mmHg·K-1·mol-1. As you can see, in the second case, you wouldn't have to convert mmHg to atm, but you had to do it if you were using the 1st value.

About the convertions, it's pretty easy. If you are given a mmHg value, to convert it to atm you just have to divide it by 760 (because 760mmHg=1atm). Same way, to convert atm to mmHg multiply the value by 760. In your case, 626mmHg would be 626/760=0,82368atm

hope that has helpful

Answer 3

Whatever pressure unit is used for P1, then P2 must be in the same unit (and vice-versa). It is not necessary to change to atmospheres as any unit that represents atmospheric pressure or a part thereof, is OK as long as both sides of the calculation use the same unit.
(Choices are...mmHg (or torr), cmHg, inchesHg ("Hg), atm, psi, Pa or kPa, bar, mbar (millibar), inches of water ("H2O)...
(I Can't think of any more).

I enjoy Chemistry too, but as all gases are not Ideal Gases, I prefer to use the 3 Gas laws of Boyle, Charles and Gay Lussac from which the Combined Gas Law is derived and, I've found if used correctly, these laws will give the same result as the Ideal Gas Law.

Answer 4

You have TWO choices. (1) You can convert to atmospheres (2) You can use the equation with mm of Hg with the "R" having pressure units in mm Hg. Choice (1) is easier, since you don't have to know different R values. Personally, I work with atmospheres, knowing how to convert pressures to that unit.