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:

P_1V_1=nR\cdot 311

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

P_2V_2=nRT_2

V_2=8V_1

P_2=\frac{P_1}{7}

Substituting the new values for P and V, you get

\frac 8 7 P_1 V_1=nRT_2

P_1 V_1=\frac 7 8 nRT_2

Now both equations (the one from the smaller tank and the one from the larger) have P_1 V_1 on the left side. Therefore we can set the right sides equal to each other:

P_1V_1=nR\cdot 311

P_1 V_1=\frac 7 8 nRT_2

nR\cdot 311=\frac 7 8 nRT_2

\frac 8 7 311=T_2

T_2 \approx 355 \text K

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.