Answer to Question #191203 in General Chemistry for Kayla

Question #191203

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?

Expert's answer

$P_1V_1 =P_2V_2$

$15.7atm×V_2=6.6atm×15.2L$

$V_2=\frac {6.6atm ×15.2L}{15.7atm}$

$V_2=6.4L$

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