Author: **SpudBlaster15** » Thu Nov 22, 2007 2:14 am

Sigh... I could show you this in a much simpler sequence of calculations (canceling out variables that divide to become the number 1), but with comments like "My logic says lower temp, lower vapor pressure, that means less psi it the meter to get the same amount of gas" I doubt you would understand unless I put it into real world perspective.

The ideal gas law: PV = nRT. P = pressure, V = volume, n = number of moles of the gas, R = the gas constant (8.314 472), T = temperature.

For ease of calculation, we will assume that the meter volume is 0.0001m<sup>3</sup>, and the chamber volume is 0.01m<sup>3</sup>. The chamber pressure will be assumed to be constant at absolute atmospheric level, about 101,352.932 pascals, and the meter pressure will be assumed to be 300000.00000 pascals, about 43.5psia. Temperature will be the only variable.

First, rearrange to n = PV/RT

We will assume a temperature of 300 kelvin, probably common summer shooting conditions. (It doesn't get that warm around here too often, but oh well)

n<sub>chamber</sub> = ((101352.932)(0.010000))/((8.314472)(300.00000))

n<sub>meter</sub> = ((300000.00000)(0.000100))/((8.314472)(300.00000))

This is a ratio of (((101352.932)(0.010000))/((8.314472)(300.00000)))/(((300000.00000)(0.000100))/((8.314472)(300.00000))) = 33.7843:1

Now we will decrease the temperature to 272 kelvin, or 0*C.

n<sub>chamber</sub> = ((101352.932)(0.010000))/((8.314472)(272.00000))

n<sub>meter</sub> = ((300000.00000)(0.000100))/((8.314472)(272.00000))

This is a ratio of (((101352.932)(0.010000))/((8.314472)(272.00000)))/(((300000.00000)(0.000100))/((8.314472)(272.00000))) = 33.7843:1

The ratios are identical, meaning that temperature will effect all gases in an identical manner.

And this is exactly why we don't overcomplicate meter pressure calculations by using PV = nRT.

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