If #24 L# of a gas at room temperature exerts a pressure of #8 kPa# on its container, what pressure will the gas exert if the container's volume changes to #15 L#?

Question

Nathan Ashcroft · Accepted Answer

#12.8"kPa"# #P_1V_1=P_2V_2# #1L=0.001m^3# #8000*0.024=0.015P_2# #P_2=(8000*0.024)/0.015=12800"Pa"=12.8"kPa"#

Nicholas Agrusa · Answer

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

Boyle's Law equation:
$$ P_1 \cdot V_1 = P_2 \cdot V_2 $$

Where:
- $ P_1 $ is the initial pressure (8 kPa)
- $ V_1 $ is the initial volume (24 L)
- $ P_2 $ is the final pressure (unknown)
- $ V_2 $ is the final volume (15 L)

Substitute the given values into the equation:
$$ 8 \times 24 = P_2 \times 15 $$
$$ 192 = 15P_2 $$
$$ P_2 = \frac{192}{15} $$
$$ P_2 \approx 12.8 \text{ kPa} $$

Therefore, the gas will exert a pressure of approximately 12.8 kPa when the container's volume changes to 15 L.