Question

For a certain gas in a closed container, the pressure has been raised by `0.4%`, and the temperature was raised by `"1 K"`. What temperature did the gas start at?

`a)` `"250 K"`
`b)` `"200 K"`
`c)` `"298 K"`
`d)` `"300 K"`

Answer 1

This is an impossible question. However, if we assume that the question should have written that the vessel is rigid, then I get an initial temperature of `"250 K"`.

---

Well, as usual, when you see pressure, temperature, and "closed vessel" in the same sentence, we assume ideality...

`PV = nRT`

• `P` is pressure in `"atm"`.
• `V` is volume in `"L"`.
• `n` is mols of ideal gas.
• `R = "0.082057 L"cdot"atm/mol"cdot"K"` is the universal gas constant.
• `T` is temperature in `"K"` (as it must be! Why?).

The closed vessel means very little to us; all it says is that the mols of gas are constant. It says nothing about the volume of the vessel being constant, as it could very well be a big fat balloon.

We apparently are given:

`P -> 1.004P`

`T -> T + 1`

`n -> n`

`V -> ??? xx V`

Substitute to get:

`1.004P cdot V = nR(T+1) = nRT + nR`,

where the written variables are all for the initial state and we interpret the question to mean a closed AND rigid vessel. So, we assume that `??? = 1`.

Note that since `PV = nRT`, we can now write:

`1.004nRT = nR(T + 1)`

Then, divide by `nR` to get:

`1.004T = T + 1`

`=> 0.004T = 1`

`=> color(blue)(T = 1/0.004 = "250 K")`

which is one of the given answer choices. It doesn't mean the question can't be revised, but that is probably what the question actually meant.