If #13/4 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 #3/8 L#?

Question

Emily Abbate · Accepted Answer

The gas will exert a pressure of #138.7 kPa#

We can obtain the final pressure via Boyle's Law:

Let's identify the known and unknown variables:

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

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

All we have to do is rearrange the equation to solve for the final pressure. We do this by dividing both sides by #V_2# in order to get #P_2# by itself like this:
#P_2=(P_1xxV_1)/V_2#

Insert the given values into the equation to solve for #P_2#:

#P_2= (16kPa xx 13/4cancel"L")/(3/8\cancel"L")# = #138.7 kPa#

Xavier Adame · Answer

undefined To find the new pressure of the gas, we can use the formula for the relationship between pressure and volume for a given amount of gas at constant temperature:

$$P_1 \times V_1 = P_2 \times V_2$$

Where:
- $P_1 = 16 \text{ kPa}$ (initial pressure)
- $V_1 = 13/4 \text{ L}$ (initial volume)
- $V_2 = 3/8 \text{ L}$ (final volume)
- $P_2$ is the pressure we want to find

Plugging in the values and solving for $P_2$:

$$16 \text{ kPa} \times 13/4 \text{ L} = P_2 \times 3/8 \text{ L}$$

$$52 \text{ kPa} \cdot \text{L} = P_2 \times 3/8 \text{ L}$$

$$52 \text{ kPa} = P_2 \times 3/8$$

$$P_2 = 52 \text{ kPa} \div (3/8)$$

$$P_2 = 52 \text{ kPa} \times 8/3$$

$$P_2 = 416 \text{ kPa} / 3$$

$$P_2 = 138.67 \text{ kPa}$$

So, the gas will exert a pressure of approximately $138.67 \text{ kPa}$ when the container's volume changes to $3/8 \text{ L}$.