# Ideal Gas law lab

1. Nov 23, 2006

### mirandasatterley

For the first part of my lab, I have a graph with pressure(atm) as a function of temperature(K).

For the second part of my lab, I have a graph with pressure(atm) as a function of inverse volume(mL^-1)

For both of these, i am supposed to find the value for R, the gas constant.

So far far part a i have: PV = nRT
P = (nR/V)T (in the form y = mx +b),
So the slope of my graph is nR/V
I also know that n is kept constant, because it was a closed system and 1mol= 22.4L. And that the initial volume is 20mL.
And i think I should I pick two points on the line to find the slope and set that equal to nR/V.

And this is where i'm having trouble, i get some value for the slope, m, so
m = nR/V
R = mV/n, here i'm confused at how to find V and n in order to solve for R.

For part b; PV = nRT
P = (nRT) 1/V
So the slope of my graph is nRT
I did the same thing as part one;
m = nRT
R = m/nT, since T was not chnged, I can do that part, but once again, I am confused by finding n.

2. Nov 23, 2006

### Andrew Mason

As I understand your lab experiment you are doing the following:

1) You keep Volume constant, and take measurements of Pressure at various Temperatures. A plot of P as a function of T should be a straight line P = nRT/V with slope m = nR/V and y intercept b=0.

2) Keeping T constant, take measurements of Volume at various Pressures. A plot of P as a function of 1/V should be a straight line: P = nRT(1/V) with slope m = nRT and y intercept b=0.

In 1) you determine V by measurement. You just have to keep n and V constant while you change the temperature.

In 2) you just have to keep n and T constant as you change the volume.

To find n, you have to weigh the gas. If you use the fact that 1 mole occupies 22.414 L at STP you are indirectly using the known value for R:

R = PV/nT = 101325*.022414/1*273.15 = 8.3145 J/mol K

AM

Last edited: Nov 23, 2006