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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Equilibrium
Phase equilibrium occurs when multiple phases of a substance coexist at stable proportions. This concept is critical in understanding the behavior of substances as they change states under different conditions.
At the phase equilibrium point, the rates of phase transition between solid, liquid, and gas are equal, having no net change in the quantity of each phase. For instance, ice will neither gain nor lose mass in a container with already existing water at a stable temperature of 0°C.
The most intriguing aspect is the **triple point**, where solid, liquid, and gas phases all coexist in equilibrium. Each substance has a unique triple point, characterized by a specific temperature and pressure. The triple point is fundamental for defining basic calibration points in thermometry.
Understanding phase equilibrium helps you to predict how a substance behaves as it surpasses its typical state boundaries. It ensures materials are stable and predictable under certain pressures and temperatures.
  • The concept demonstrates how a slight change can shift the equilibrium and alter the phase composition.
  • It assists in indicating the conditions necessary for phase transitions, essential for scientific and industrial processes.
Melting Point
The melting point is the temperature at which a solid becomes a liquid under standard atmospheric pressure. It marks a specific type of phase transition.
When the temperature reaches the melting point, the solid state's molecular structure breaks down due to increased kinetic energy, allowing the molecules to move more freely. Consequently, the substance turns into a liquid while keeping the temperature constant until the phase transition is complete.
Factors like pressure can significantly impact the melting point. Under higher pressures than atmospheric, some substances may require increased temperatures to melt.
The melting point is essential in specifying pure substances and identifying materials. It helps us determine their quality since impurities often lower the melting point, creating a melting range instead of a sharp temperature.
Practical uses of melting points span various industries:
  • In pharmaceuticals, it's used for drug identification and purity testing.
  • In food processing, it ensures safe storage and processing conditions for ingredients and products.
  • In materials science, it helps in selecting materials that withstand environmental and mechanical stresses.
Boiling Point
Boiling point is the temperature at which a liquid turns into vapor at a given pressure, usually standard atmospheric pressure. At this point, the vapor pressure of the liquid equals the external pressure exerting on it.
When a liquid reaches its boiling point, the molecules gain enough energy to overcome the liquid's surface tension and enter the gaseous phase. The temperature remains consistent during this phase transition until all the liquid has converted into vapor.
Unlike the general perception, the boiling point isn't always a fixed number. It changes with varying pressures; for instance, water boils at a lower temperature at high altitudes due to decreased atmospheric pressure.
The boiling point serves numerous practical applications:
  • In culinary practices, understanding boiling point helps in preparing and preserving food.
  • In distillation, it's crucial for separating liquid mixtures based on boiling point differences.
  • In chemical engineering, it assists in designing reactors where controlled evaporation and condensation are needed.
Grasping the boiling point concept provides insight into how different pressures affect phase changes and guides the selection of appropriate temperatures for various scientific experiments.

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Most popular questions from this chapter

We have learned that the enthalpy of vaporization of a liquid is generally a function of temperature. If we wish to take this temperature variation into account, we cannot use the Clausius-Clapeyron equation in the form given in the text (that is, equation 12.2 ). Instead, we must go back to the differential equation upon which the Clausius-Clapeyron equation is based and reintegrate it into a new expression. Our starting point is the following equation describing the rate of change of vapor pressure with temperature in terms of the enthalpy of vaporization, the difference in molar volumes of the vapor \(\left(V_{g}\right),\) and liquid \(\left(V_{1}\right),\) and the temperature. $$\frac{d P}{d T}=\frac{\Delta H_{\mathrm{vap}}}{T\left(V_{\mathrm{g}}-V_{1}\right)}$$ Because in most cases the volume of one mole of vapor greatly exceeds the molar volume of liquid, we can treat the \(V_{1}\) term as if it were zero. Also, unless the vapor pressure is unusually high, we can treat the vapor as if it were an ideal gas; that is, for one mole of vapor, \(P V=R T\). Make appropriate substitutions into the above expression, and separate the \(P\) and \(d P\) terms from the \(T\) and \(d T\) terms. The appropriate substitution for \(\Delta H_{\text {vap }}\) means expressing it as a function of temperature. Finally, integrate the two sides of the equation between the limits \(P_{1}\) and \(P_{2}\) on one side and \(T_{1}\) and \(T_{2}\) on the other. (a) Derive an equation for the vapor pressure of \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{l})\) as a function of temperature, if \(\Delta H_{\mathrm{vap}}=\) \(15,971+14.55 T-0.160 T^{2}\left(\text { in } J m o l^{-1}\right)\) (b) Use the equation derived in (a), together with the fact that the vapor pressure of \(\mathrm{C}_{2} \mathrm{H}_{4}(1)\) at \(120 \mathrm{K}\) is 10.16 Torr, to determine the normal boiling point of ethylene.

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In DNA the nucleic acid bases form hydrogen bonds between them, which are responsible for the formation of the double-stranded helix. Arrange the bases guanine and cytosine to give the maximum number of hydrogen bonds.

A \(2.50 \mathrm{g}\) sample of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is sealed in a \(5.00 \mathrm{L}\) flask at \(120.0^{\circ} \mathrm{C}\) (a) Show that the sample exists completely as vapor. (b) Estimate the temperature to which the flask must be cooled before liquid water condenses.

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