Final answer:
Using Boyle's law (P1 x V1 = P2 x V2), the final pressure of a gas when the volume decreases from 70 mL to 15 mL with an initial pressure of 130 kPa is found to be 606 kPa.
Step-by-step explanation:
To find the final pressure of a gas when its volume decreases, we use Boyle's law, which states that the pressure of a given mass of an ideal gas is inversely proportional to its volume when the temperature and amount of gas are constant. According to Boyle's law, P1 × V1 = P2 × V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively. Given an initial volume (V1) of 70 mL and an initial pressure (P1) of 130 kPa, if the volume is decreased to 15 mL (V2), the final pressure (P2) can be calculated as follows:
P2 = (P1 × V1) / V2
P2 = (130 kPa × 70 mL) / 15 mL
P2 = 606 kPa
This answer aligns with the provided answer; thus, when the volume of a gas is decreased from 70 mL to 15 mL, and the initial pressure is 130 kPa, the final pressure of the gas is 606 kPa.