If #3 L# of a gas at room temperature exerts a pressure of #2 kPa# on its container, what pressure will the gas exert if it the container's volume changes to #1 L#?

Question

Olivia Askew · Accepted Answer

The gas will exert a pressure of #6kpa#

Let's identify the known and unknown variables:

#color(yellow)("Knowns:")#
- Initial Volume
- Final Volume
- Initial Pressure

#color(orange)("Unknowns:")#
- Final Pressure

We can obtain the answer using Boyle's Law:

The letters i and f represent the initial and final conditions, respectively.

All we have to do is rearrange the equation to solve for the final pressure.

We do this by dividing both sides by #V_f# in order to get #P_f# by itself like this:
#P_f=(P_ixxV_i)/V_f#

Now all we do is plug in the values and we're done!

#P_f= (2kPa xx 3cancel"L")/(1\cancel"L")# = #6kPa#

Benjamin Asiedu · Answer

undefined To solve this problem, you can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when the temperature remains constant.

Boyle's Law equation: P1V1 = P2V2

Where:
P1 = initial pressure (2 kPa)
V1 = initial volume (3 L)
P2 = final pressure (what we need to find)
V2 = final volume (1 L)

Substitute the given values into the equation:

2 kPa × 3 L = P2 × 1 L

Solve for P2:

2 kPa × 3 L / 1 L = P2

P2 = 6 kPa

So, the pressure the gas will exert if the container's volume changes to 1 L is 6 kPa.