If #9/4 L# of a gas at room temperature exerts a pressure of #16 kPa# on its container, what pressure will the gas exert if the container's volume changes to #9/7 L#?

Question

Sebastian Acosta · Accepted Answer

The pressure is #=28kPa#

Using Boyle's Law

#P_1V_1=P_2V_2#

The initial pressure is #P_1=16kPa#

The initial volume is #V_1=9/4L#

The final volume is #V_2=9/7#

The last bit of pressure is

#P_2=V_1/V_2*P_1=(9/4)/(9/7)*16#

#=7/4*16=28kPa#

Leo Abeyta · Answer

undefined To find the new pressure, use Boyle's law:

$$P_1V_1 = P_2V_2$$

Given:
$P_1 = 16 \, kPa$
$V_1 = \frac{9}{4} \, L$
$V_2 = \frac{9}{7} \, L$

Substitute the given values into the equation and solve for $P_2$:

$$P_2 = \frac{P_1V_1}{V_2}$$

$$P_2 = \frac{(16 \, kPa) \times \left(\frac{9}{4} \, L\right)}{\frac{9}{7} \, L}$$

$$P_2 = \frac{144 \, kPa}{\frac{9}{7}}$$

$$P_2 = \frac{144 \, kPa \times 7}{9}$$

$$P_2 = \frac{1008 \, kPa}{9}$$

$$P_2 = 112 \, kPa$$

So, the gas will exert a pressure of $112 \, kPa$ if the container's volume changes to $\frac{9}{7} \, L$.