If #32 L# of a gas at room temperature exerts a pressure of #64 kPa# on its container, what pressure will the gas exert if the container's volume changes to #16 L#?

1 Answer
Jul 9, 2016

Answer:

The Pressure will change to #128kPa#

Explanation:

To solve this problem we can use Boyle's Law comparing the Pressure and Volume relationships

#P_1V_1 = P_2V_2#

#P_1 = 64kPa#
#V_1 = 32L#
#P_2 = ?#
#V_2 = 16L#

#(64kPa)((32L) = P_2(16L)#

Use the multiplicative inverse to isolate the variable for #P_2#

#(64kPa)((32cancelL)/(16cancelL)) = P_2(cancel(16L))/(cancel(16L))#

#(64kPa)(2) = P_2

#128kPa = P_2#

More molecules in a small volume means more collisions and therefore more pressure.