Question

Question #29d0f

Answer 1

`1)`

No; as a general, rule, don't just take two numbers you see in a question and multiply them together. Look at their units, and see if that makes sense.

In this case, `"500 L"` does not make physical sense. How do you have more than `"100 L"` of air in a `"100-L"` air receiver (unless it exploded)? Units of `"L"cdot"atm"` convert to `"J"`. (So, your calculation gives units of energy, and not volume.)

What you should do instead is recognize that many gases expand to fill their container.

So, no matter what pressure the container is held at, in a `"100-L"` air receiver, at room temperature, there are (likely) `bb("100 L")` of air in there. Only the packing tightness (the density) of the air particles will change, once the particles have distributed themselves evenly.

`2)`

As for pressure changing over the course of 30 seconds, that's too long to assume that it is linear.

In general we can expect that it is too complicated to fit to a linear equation, but we can assume that it does if we remember that it's only an assumption.

When we assume that, if the pressure drops from `"5 atm"` to `"4 atm"`, it was multiplied by `0.8`. Since `P prop 1/V`, the volume was divided by `0.8` (multiplied by `1.25`), so at constant temperature and number of `"mol"`s of gas:

`P_1V_1 = P_2V_2`

`=> V_2 = P_1/P_2V_1`

`= 5/4("100 L")`

`=` `"125 L"`

So, the approximate change in volume over time might be:

`color(blue)((DeltaV)/(Deltat)) ~~ "125 - 100 L"/"30 s" xx "60 s"/"1 min"`

`= color(blue)("50 L"/"min")`

but it could be quite off depending on what was happening in that timeframe. What if you turned off the valve for 5 seconds and then turned it back on? That's not continuous air flow; that's not `"50 L"/"min"` anymore.