Question

500 mL of He at 98 kPa expands to 750 mL. Find P2.

Answer 1

Answer: $147 kPa$

Explanation:

According to Boyle's law, which mathematically describes the thermodynamic process in which the pressure of a gas has the tendency to increase as the volume of the container decreases, we have:

$P_{1}V_{1}=P_{2}V_{2}$ (1)

Where:

$P_{1}=98kPa=98000 Pa$ is the initial pressure of Helium (He)

$V_{1}=500 mL=500(10)^{-3}L$ is the initial volume of Helium (He)

$V_{2}=750 mL=750(10)^{-3}L$ is the final volume of Helium (He)

$P_{2}$ is the final pressure of Helium (He)

Now, we need to isolate $P_{2}$ from (1):

$P_{2}=\frac{P_{1}V_{1}}{V_{2}}$ (2)

$P_{2}=\frac{(98000 Pa)(500(10)^{-3}L)}{750(10)^{-3}L}$ (3)

Finally:

$P_{2}=147000 Pa=147 kPa$ This is the final pressure of Helium