How do you correct samples of gases to standard temperature?

Question

grimsnightmare · Answer

I have;
622 cm^3 at 15.0 C
and another 
55.6 dm^3 at 110 K

teddyiswatshecallsme · Answer

V2 = 589.6cm³  for #1
&
138.0 dm³  for #2

aaronq · Answer

Use the formula:  $\large \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

For example, for the first set of conditions:

$V_1$=622 cm^3   $T_1$= 15.0 C + 273 = 288 K

$V_2$= unknown  $T_2$= 273 K

$\large \dfrac{622~ cm ^3}{288 ~K}=\dfrac{V_2}{273~K}\rightarrow V_2=\dfrac{273~K*622~ cm ^3}{288 ~K}=589.6~cm^3 $

grimsnightmare · Answer

Thank you so much for explaining that! Sorry I didn't respond earlier I was working on something else.