Q:

Temperature of chlorine gas at given pressure and volume

Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of $\ce{Cl2}$ has is $\pu{9.22 L}$ at $\pu{1124 torr}$ and $\pu{24 ^\circ C}$. At what temperature will the volume be $\pu{15.00 L}$ if the pressure is $\pu{8.76e2 torr}$ ? 

I found the moles of the sample to be $\pu{0.559 mol}$. However, now I'm stuck at the last part of the question. Can I use $\pu{0.559 mol}$ to solve for the temperature when the volume is $\pu{15 L}$ and the pressure is $\pu{8.76e2 torr}$?

A:

Assuming at these conditions it's safe to use ideal gas law ($pV = nRT$) for chlorine, and given the same gas sample ($n = \text{const}$, hence $nR = \text{const}$), let's express the unknown temperature $T_2$:
$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \to \quad T_2 = \frac{p_2V_2}{p_1V_1} \cdot T_1$$
Generally, I would recommend not to do intermediate calculations, especially if you are not asked to do so. Construct the mathematical expression first, convert the units if necessary, and plug in the values when you are done. Note that in this case you only need to convert temperature units.

 $$T_2 = \frac{\pu{876 torr} \cdot \pu{15.00 L}}{\pu{1124 torr} \cdot \pu{9.22 L}} \cdot \pu{297 K} = \pu{377 K}~\text{or}~\pu{104 ^\circ C}$$ 

