An air compressor has a pressure of #"5200 Torr"# and contains #"200 L"# of compressed air. If the container ruptures, what is the volume of air that escapes through the rupture?

Question

Owen Ambrose · Accepted Answer

The volume of air that escapes through the rupture is #~~"1000 L"#.

This is an example of Boyle's law, which states that the volume of a given amount of gas varies inversely with the applied pressure when temperature and mass are kept constant. This means that as the volume increases, the pressure increases, and vice-versa. The equation to use is: #P_1V_1=P_2V_2#, where: #P# is pressure and #V# is volume. We aren't given altitude, so I'm going to use the pressure at sea level for #P_2#. When the hose ruptured, the pressure would have been immediately decreased to that of the air pressure at the altitutde of the air compressor. Organize data: Known #P_1="5200 torr"# #V_1="200 L"# #P_2="760.00 torr"# Unknown #V_2# Solution Rearrange the equation to isolate #V_2#. Plug in the known data and solve. #V_2=(P_1V_2)/(P_2)# #V=(5200color(red)cancel(color(black)("torr"))xx200"L")/(760.00color(red)cancel(color(black)("torr")))="1000 L"# to one significant figure due to #"200 L"#.

Natalie Anton · Answer

undefined To calculate the volume of air that escapes through the rupture, you can use Boyle's Law, which states that the product of pressure and volume is constant when temperature remains constant.

Initial pressure (P₁) = 5200 Torr
Initial volume (V₁) = 200 L

If the container ruptures, the pressure becomes atmospheric pressure (usually around 1 atm). Let's convert 5200 Torr to atm:
5200 Torr ÷ 760 Torr/atm ≈ 6.84 atm

So, the final pressure (P₂) = 1 atm

Using Boyle's Law:

P₁ * V₁ = P₂ * V₂

5200 Torr * 200 L = 1 atm * V₂

V₂ = (5200 Torr * 200 L) / 1 atm

Now, you have the volume of air that escapes through the rupture.