Answer to Question #191203 in General Chemistry for Kayla

Question

A gas with an initial pressure of 15.7 atm was allowed to expand to a volume of 15.2 L and a new pressure of 6.6 atm. What was the original volume (in L) of the gas?

Answer 1

$P_1V_1 =P_2V_2$

$15.7atm×V_2=6.6atm×15.2L$

$V_2=\frac {6.6atm ×15.2L}{15.7atm}$

$V_2=6.4L$