course PHY 232
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Good job.
In this experiment we set up a 3-liter bottle about 1/5th full of soda and with a three-holed stopper in the top. A tube runs from the soda, through the stopper, and about 40cm beyond. A second tube runs from the stopper out, it has a sealed end. Also the second tube has alcohol in it sealing of an air chamber. The third hole is just sealed off. This creates a system with no escape for air and sealed at 1 atm.With no force acting upon the system, the length of the air chamber i
with enough pressure on the bottle to raise the column of soda to 50cm in heing, the length of the air chamber becomes 23.95cm.
Our goal is to find the measure for atm in pascals (N/m²), so we need to find the difference in pressure of the system in pascals and as a ratio. To find the ratio we can go back to the Ideal Gas Law
PV=nRT; n,R,T are constand so that implies that PV = k
From here we can set up the ratio as P1V1=P2V2=k. Since we have our V1 and V2 we can set up a ratio for P2:P1
P2/P1 = V1/V2 = 25/23.95 = 1.044; which means P2 is equal to 1.044 atm and the change in P is .044 atm.
Now to find the difference in ,pascals, we use the definiton of pressure which is P=pyh; or density*force_gravity*height.
y_0=0 so the change in pressure will just be the pressure at the second height.
P=pgy=1000kg/m³(9.8m/s²)(.5m)= 4900 Pa
now that we know .044 atm = 4900 Pa, we know that 1 atm = 111363.63 Pa"