Question

Calculate the pressure, in atmospheres, required to compress a sample of helium gas from 27.3 L (at 1.00 atm) to 3.36 L at constant temperature. Enter your answer in the box provided. atm

Answer 1

Answer:

8.125 atm

Explanation:

Hello,

Boyle's law states that at constant temperature:

$P_1V_1=P_2V_2$

In this case:

$P_1=1 atm, V_1=27.3L, V_2=3.36L$

Solving for $P_2$:

$P_2=\frac{P_1V_1}{V_2}=\frac{1atm*27.3L}{3.36L}\\P_2=8.125atm$

Best regards!