Four paneful gases

One must remember that pressure is simply a measure of the relative amount of gas molecules which hit the sides of a container. At the same amount of moles, if a gas is physically larger (i.e. has a larger atomic number), the intermolecular forces of attraction (specifically, London dispersion forces) between the molecules will be larger. Thus, the measured pressure will decrease as well, since less gases at any given moment will be able to collide with the sides of the container. This gives measurements which deviate from the ideal gas law.

Even if hydrogen bonding could be possible with certain molecules with a higher electronegativity (which would increase IMFs by an even larger amount), truly significant hydrogen bonding is not possible with any of the elements found in these gases. Therefore, one can look at the molar masses of each of the gases to determine roughly which elements are larger, and thus have the most London dispersion forces. Using this method, tetraborane is the largest, and thus, the container with tetraborane will have the lowest pressure. $\boxed { { B }_{ 4 }{ H }_{ 10 } }$ is the answer.

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Furthermore, one may also look at the melting and boiling points to determine the amount of intermolecular forces. Using this method, tetraborane has the highest melting and boiling points. Therefore, it requires the most energy to break the intermolecular forces of attraction, and these forces, thus, cause a more significant deviation from the pressure of an ideal gas under the same conditions.

I would also like to mention that the higher the Van der Waals constant is for a gas, the lesser that gas will behave like an ideal gas. However, the Van der Waals constant for tetraborane was not immediately available to me on the internet, so I doubt this fact might have helped.