# 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) $\text{250 K}$ b) $\text{200 K}$ c) $\text{298 K}$ d) $\text{300 K}$

Jul 29, 2017

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 $\text{250 K}$.

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

$P V = n R T$

• $P$ is pressure in $\text{atm}$.
• $V$ is volume in $\text{L}$.
• $n$ is mols of ideal gas.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant.
• $T$ is temperature in $\text{K}$ (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:

$P \to 1.004 P$

$T \to T + 1$

$n \to n$

V -> ??? xx V

Substitute to get:

$1.004 P \cdot V = n R \left(T + 1\right) = n R T + n R$,

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 $P V = n R T$, we can now write:

$1.004 n R T = n R \left(T + 1\right)$

Then, divide by $n R$ to get:

$1.004 T = T + 1$

$\implies 0.004 T = 1$

$\implies \textcolor{b l u e}{T = \frac{1}{0.004} = \text{250 K}}$

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.