A solution has [H3O+] = 4.7 x 10-3 M

Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptur

Scrat [10]

Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptured and all the compressed air was released to a volume of 1370 L at atmospheric pressure.

Hank's Garage has an air compressor with a holding tank that contains a volume of 200L (V₁) of compressed air at a pressure of 5200 torr (P₁).

One day a hose ruptured and all the compressed air was released. The final pressure was the atmospheric pressure (1 atm = 760 torr) (P₂).

We can calculate the new volume (V₂) in these conditions using Boyle's law, which states there is an inverse relationship between the volume and the pressure of an ideal gas.

$P_1 \times V_1 = P_2 \times V_2\\\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{5200 torr \times 200L}{760torr} = 1370 L$

Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptured and all the compressed air was released to a volume of 1370 L at atmospheric pressure.

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