# How do significant figures work on this computation?

This isn't a homework question although it may seem like one. We've never really discussed significant figures and it was just introduced to us when we were learning how to solve word problems regarding gas laws.

Find $V_2$ when $V_1=\pu{400 mL}$ and $T_1=\pu{32^\circ C}$ and $T_2=\pu{-12^\circ C}$.

My computation goes like this:

$$V_2=\frac{T_2V_1}{T_1}=\frac{(\pu{261 K})(\pu{400 mL})}{\pu{305 K}}=\pu{342.295082 mL}$$

As per the given, the least number of significant figures is one (from $\pu{400 mL}$). How do I approximate $\pu{342.295082 mL}$ to one significant figure?

• 400 is three significant figures. 0 is as good a figure as any other. – Ivan Neretin Sep 16 '18 at 6:00
• 342 the number with least figures determines result figures in this operation. Usually 400 should be written with a bar above the zeros. If we took 400 with just one figure, the result would be 300. – user43021 Sep 16 '18 at 6:00
• @IvanNeretin No it isn't. The zeros might be significant. And to solve this question (i.e. report a sensible answer), you'll have to decide for one or the other. – Karl Sep 16 '18 at 12:59
• Technically $400$ has 1 significant figure. $400.$ has 3. Though for this problem I think assuming 3 significant figures in $\pu{400mL}$ is reasonable – A.K. Sep 16 '18 at 18:27

Assuming the precision of the given volume, $V_1 = \pu{400 mL}$, is only to 1 significant figure (i.e. only the hundreds place digit of "4" is precise), then your final answer would be rounded to $V_2 = \pu{300 mL}$.*
If we were to say that $V_1$ had 3 significant figures, then the final answer would round to the ones place, resulting in $V_2 = \pu{342 mL}$. However, the value would technically need to be written as "$\pu{400\!. mL}$," including the decimal place to denote the precision going out to the ones place, in order to imply 3 significant figures instead of 1.