To what pressure would you have to compress 48.0 L of oxygen gas at 99.3 kPa in order to reduce its volume to 16.0 L?

Question

Henry Antrim · Accepted Answer

#298 kPa#

The answer can be determined by using Boyle's Law:

Let's identify our known and unknown variables.

#color(red)("Knowns:"#
#P_1#= 99.3 kpa
#V_1#= 48.0 L
#V_2#= 16.0 L

#color(maroon)("Unknowns:"#
#P_2#

Rearrange the equation to solve for the final pressure by dividing both sides by #V_2# to get #P_2# by itself like this:

#P_2=(P_1xxV_1)/V_2#

Plug in your given values to obtain the final pressure:

#P_2=(99.3\kPa xx 48.0\ cancel"L")/(16.0\cancel"L")# = #298kPa#

Cameron Andrews · Answer

undefined To calculate the pressure to which the oxygen gas must be compressed to reduce its volume to 16.0 L, you can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when the temperature is constant.

P1 * V1 = P2 * V2

Where:
P1 = initial pressure (99.3 kPa)
V1 = initial volume (48.0 L)
P2 = final pressure (unknown)
V2 = final volume (16.0 L)

Rearranging the formula to solve for P2:

P2 = (P1 * V1) / V2
P2 = (99.3 kPa * 48.0 L) / 16.0 L
P2 = (4778.4 kPa*L) / 16.0 L
P2 = 298.65 kPa

Therefore, the pressure to which the oxygen gas must be compressed is approximately 298.65 kPa.