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rukai2 · Apr 28, 2011, 11:47 AM

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

jcaron2 · Apr 28, 2011, 12:09 PM

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.

jcaron2 · Apr 28, 2011, 12:12 PM

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.