9A. A sample of hydrogen at 1.56 atm had it's pressure decreased to 0.73 atm producing a new volume of 751 mL. What was its original volume? *

351.43mL

Explanation:

To calculate the original volume of hydrogen gas in this question, the Boyle's law equation will be used. Boyle's law equation is:

P1V1 = P2V2

Where; P1 = initial pressure

V1 = initial volume

P2 = final pressure

V2 = final volume

According to this question, the P1= 1.56atm, V1 = ?, P2 = 0.73atm, V2 = 751mL

Hence;

P1V1 = P2V2

1.56 × V1 = 0.73 × 751

1.56 V1 = 548.23

V1 = 548.23/1.56

V1 = 351.43mL

Therefore, the original volume of hydrogen gas is 351.43 mL.

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