If #2/5 L# of a gas at room temperature exerts a pressure of #15 kPa# on its container, what pressure will the gas exert if the container's volume changes to #7/2 L#?

Question

Lily Adams · Accepted Answer

When the volume increases to #"7/2 L"#, the pressure decreases to #"1.7 kPa"#.

Boyle's law applies to this problem; it states that, provided that temperature and mass stay constant, a gas's volume is inversely proportional to its pressure, meaning that as volume increases, pressure will decrease and vice versa. The Boyle's law equation is: #P_1V_1=P_2V_2#, where #P# is pressure, and #V# is volume. Arrange the information: Recognized #P_1="2/5 L"# #V_1="15 kPa"# #V_2="7/2 L"# Not sure #P_2# Resolution Rearrange the equation to isolate #P_2#. Plug in the known values and solve. #P_2=(P_1V_1)/V_2# #P_2=(15"kPa"xx2/5color(red)cancel(color(black)("L")))/(7/2color(red)cancel(color(black)("L")))="1.7 kPa"#

Isabella Alvarez · Answer

undefined Using Boyle's Law, we can solve for the new pressure:

$P_1 \times V_1 = P_2 \times V_2$

Where:
$P_1 =$ initial pressure (15 kPa)
$V_1 =$ initial volume ($2/5$ L)
$P_2 =$ final pressure (unknown)
$V_2 =$ final volume ($7/2$ L)

Plugging in the values:

$15 \, \text{kPa} \times \frac{2}{5} \, \text{L} = P_2 \times \frac{7}{2} \, \text{L}$

Solving for $P_2$:

$P_2 = \frac{15 \, \text{kPa} \times \frac{2}{5} \, \text{L}}{\frac{7}{2} \, \text{L}}$

$P_2 = \frac{15 \, \text{kPa} \times 2}{5 \times 7}$

$P_2 = \frac{30}{35}\, \text{kPa}$

$P_2 = \frac{6}{7}\, \text{kPa}$

Therefore, the gas will exert a pressure of $\frac{6}{7}$ kPa if the container's volume changes to $\frac{7}{2}$ L.