First Law of Thermodynamics

In this experiment a can with little liquid was heated, then put in cold ice water. 

We found that the little liquid the can had inside turned to vapor, but once the heated can was put in the cold water that vapor quickly went through a phase change and turned back into a liquid. The can imploded quickly.

This process occurs really fast for the water to be drawn up into the car by the low pressure relative to the outside atmosphere. The only way for the pressure in the can to be equalized so rapidly is for the can to collapse. In this picture, this can had no liquid so when it was heated there was no condensation thus no phase change; the can did not implode. However, water did go into the can. I believe it was because the can expanded when heated (creating greater volume) then decreased in volume when cooled, so when it cooled it sucked in water to make up for the lost volume.

here we discussed charles law which had to do with the relationship between volume and temperature.

this is our second experiment, we used a syringe to measure pressure. In this picture professor mason did the experiment which he then graphed for us.

here we see the linear graph and what we discussed about volume vs. temperature.

next, we performed the experiment. here we have our graph, given values and what we discussed. we found that pressure and volume are inversely proportional. (Boyles's Law)

this is the graph for the syrgine experiment we performed and we found that it was inversely proportional.

the equation we derived was P=A/V

Here we calculated Avogrado's number which turned out to be the factor in the relationship between Boltzman constant and universal gas constant.

The Ideal Gas Law describes all three relationships mathematically in an idealized fashion.  It is given by: 

PV = nRT

where n = the number of moles of gas

R = the Universal Gas Constant given by 8.31 J/mol•K

An alternative statement of the Ideal Gas Law is:

PV = Nk_BT

where N = the number of gas molecules

k_B = Boltzmann’s Constant given by 1.38 × 10^-23 J/K

in our next discussion we talked about a diving bell 

here we have our given values

these are the calculations we did for the diving bell problem. We were asked: What is the pressure of the air in the diving bell when it is submerged ? and How high will the seawater rise in the diving bell when the diving bell is submerged to a depth of 225 m, if the temperature of the air drops to 15.0^oC ?

We found The pressure of the air inside the diving bell must be holding the water from completely flooding the bell, thus the pressure of the air must equal the hydrostatic pressure of the water at depth of 225 m plus the pressure of the atmosphere pushing down on the ocean's surface. for part b its 4.07m.

A marshmallow is placed into a vacuum chamber, what will happen when the pressure is reduced and why?  What will happen when the pressure is allowed back into the chamber and why?

here we see the result of the marshmallow experiment, when pressure decreased the marshmallows expanded. When the pressure was allowed to go back into the chamber the marshmallow returned to a way smaller state than they were before, this is because the air pockets they had were erupted when the pressure was decreased.

A high altitude balloon contains helium whose molecular mass is 4 grams/mole. At its maximum altitude the balloon's volume is 830 m³. The outside temperature and pressure are -51.0 ^oC and 5.40 kPa at the balloon's maximum altitude. Assuming that the helium in the balloon is in equilibrium with the outside air temperature and pressure, i.e., they are the same: 

What is the mass of the helium in the balloon at its maximum altitude?   

here we have our answer to the high altitude balloon problem. relative formula is  then we converted moles to grams