If the triple point pressure of a substance is greater than 1 atm, which two of the following conclusions are valid? (a) The solid and liquid states of the substance cannot coexist at equilibrium. (b) The melting point and boiling point of the substance are identical. (c) The liquid state of the substance cannot exist. (d) The liquid state cannot be maintained in a beaker open to air at 1 atm pressure. (e) The melting point of the solid must be greater than \(0^{\circ} \mathrm{C}\) (f) The gaseous state at 1 atm pressure cannot be condensed to the solid at the triple point temperature.

Short Answer

Expert verified
Hence, options (d) and (f) are the valid conclusions.

Step by step solution

01

Rule out Invalid Conclusions

Based on the definition and understanding of a triple point, a higher triple-point pressure does not directly affect the ability for the solid and liquid states of the substance to coexist at equilibrium (option a), nor does it mean that the melting point and boiling point are identical (option b). Also, it does not make the melting point of the solid necessarily below or above \(0^{\circ} \mathrm{C}\) (option e). These options can be ruled out.
02

Evaluate Remaining Conclusions

Remaining to consider are options (c), (d), and (f). Option (c) is invalidated by the very definition of the triple point -- for the substance to reach its triple point, the liquid state must exist. Option (f) is plausible, because if the gaseous state is at 1 atm and cannot be condensed to the solid at the triple point temperature, it implies that to get the solid state, we need to increase the pressure which aligns with our condition that the triple point pressure is above 1 atm.
03

Evaluate Last Statement

Option (d), saying that the liquid state cannot be maintained in a beaker open to air at 1 atm pressure, also indeed make sense in light of our condition: in an open beaker at 1 atm, we cannot reach the over-1-atm triple point pressure necessary to maintain a liquid state.

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