chem isu part 3

Gay-Lussac's law states that when the volume is constant, the pressure of a gas increases with the absolute temperature. This can cause pressurized canisters to explode. In a mathematical example, if the initial pressure is 1 atm and the final pressure is 1.5 atm at a temperature of 250 K, the initial temperature can be calculated using the equation 1 divided by T1 equals 1.5 divided by 250. The initial temperature in this case would be 0.06 K. As for Gay-Lussac's law, it is the law of combining volumes of gas. It states that when the volume is kept at a constant, k, the pressure of a given mass of gas, p, varies directly with the absolute temperature of gas, T. The equation for this would be initial pressure divided by initial temperature equals final pressure divided by final temperature. Now a real life example that relates to this law. When heating something like deodorant or a spray paint can, it increases the pressure exerted by the gases on the container that can result in an explosion. This is the reason why many pressurized canisters have warning labels. Now to use Gay-Lussac's law in a mathematical example. The pressure of gas in a can when heated reaches a temperature of 250 k at 1.5 atm. What was the initial temperature of gas if its initial pressure was at 1 atm? In this problem, you are given your initial pressure which is 1 atm, your final pressure 1.5 atm, and final temperature which is 250 k. Now after putting all your numbers into the equation, it should look like 1 divided by your initial temperature equals 1.5 divided by 250. Then after you would need to cross multiply, so that would get you 250 T1 equals 1.5. Then you would want to get T1 by itself, so you would divide 1.5 with 250 on both sides of the equal sign. And after doing that, you should be able to get an initial temperature of 0.06 k.