A graph displays straight lines passing through the origin,...

According to the ideal gas equation, the volume $V$ of a gas of mass $m$ and molar mass $M$ at STP is given by:

$V = \dfrac{m}{M} \times 22.4\text{ L}$

The slope of the $V$ versus $m$ graph is $\dfrac{V}{m} = \dfrac{22.4}{M}$.

Thus, the slope is inversely proportional to the molar mass $M$. The gas with the smallest molar mass will have the steepest slope (line 1).

The molar masses of the given gases are:
$\text{H}_2 = 2\text{ g mol}^{-1}$
$\text{He} = 4\text{ g mol}^{-1}$
$\text{N}_2 = 28\text{ g mol}^{-1}$
$\text{O}_2 = 32\text{ g mol}^{-1}$

Since $M_{\text{H}_2} < M_{\text{He}} < M_{\text{N}_2} < M_{\text{O}_2}$, the order of decreasing slopes is $\text{H}_2 > \text{He} > \text{N}_2 > \text{O}_2$. Therefore, lines 1, 2, 3, and 4 correspond to $\text{H}_2$, $\text{He}$, $\text{N}_2$, and $\text{O}_2$ respectively.

Answer: $\text{H}_2$, $\text{He}$, $\text{N}_2$, $\text{O}_2$