3. An ideal gas in a tank at 100 F is allowed to expand quickly into a larger tank. If the volume increased 8 times and the pressure decreased 7 times, what is the new temperature of the gas?

= 53 degrees

Unfortunately no. :(

As I said in the previous answer, you can start with PV=nRT.

The first thing you have to do is convert the temperature to Kelvin. 100F = 311K.

From that, you can apply the ideal gas law to the gas when it was in the smaller tank:



Then you can apply the same formula to the gas after it has expanded into the larger tank:







Substituting the new values for P and V, you get





Now both equations (the one from the smaller tank and the one from the larger) have on the left side. Therefore we can set the right sides equal to each other:











You just have to convert back to Fahrenheit.

By the way, it's always good to do a sanity check on your answer. The gas was allowed to expand to a volume 8 times bigger than before. If the temperature remained the same before and after, the pressure would have dropped by a factor of 8 to compensate. However, in this case the pressure only dropped by a factor of 7. The only way that can happen is if the gas gets warmer, not cooler.