Question

1. a sample of natural gas (methane) is initially at a pressure of 190 kpa and a volume of 40.0 l. the new pressure of the same amount of gas when the volume is changed to 15.0 l and the temperature is held constant is

1.97×10^ - 3kpa
71.3kpa
507kpa
0.0140kpa

Explanation:

Step1: Identify the gas - law formula

Boyle's law is $P_1V_1 = P_2V_2$ for a gas at constant temperature.

Step2: Assign the given values

$P_1=190$ kPa, $V_1 = 40.0$ L, $V_2=15.0$ L. We need to find $P_2$.

Step3: Rearrange the formula to solve for $P_2$

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

Step4: Substitute the values

$P_2=\frac{190\times40.0}{15.0}=\frac{7600}{15.0}\approx507$ kPa.

Answer:

507 kPa