If #3/7 L# of a gas at room temperature exerts a pressure of #9 kPa# on its container, what pressure will the gas exert if the container's volume changes to #5/9 L#?

2 Answers
Oct 16, 2016

Answer:

#P_2 = 243/35 kPa#

Explanation:

Let #P_1 = 9color(white).kPa#
Let #V_1 = 3/7color(white).L#
Let #V_2 = 9/5color(white).L#

#P_1V_1 = P_2V_2#

#P_2 = P_1V_1/V_2#

#P_2 = 243/35 kPa#

Oct 16, 2016

Answer:

#243/35#kPa

Explanation:

Boyle's Law for gases states that pressure and volume vary inversely when temperature and mass are held constant. This can be expressed with the formula:

#P_1V_1=P_2V_2#

#P_1=9#kPa

#V_1=3/7 L#

#P_2=#?

#V_2=5/9 L#

#9*3/7=P_2*5/9#

#9/5*9*3/7=P_2*cancel(5)/cancel(9)*cancel(9)/cancel(5)color(white)(aaa)#Multiply both sides by 9/5

#243/35=P_2#

#P_2=243/35#kPa