Question

study the following phase diagram of substance x. use this diagram to answer the following questions. suppose a small sample of pure x is held at −234. °c and 1.7 atm. what will be the state of the sample? select suppose the temperature is held constant at −234. °c but the pressure is decreased by 1 atm. what will happen to the sample? select suppose, on the other hand, the pressure is held constant at 1.7 atm but the temperature is increased by 237. °c. what will happen to the sample? select

Answer:

Sub - Question 1: Determine the state at \(-234^\circ C\) and \(1.7\) atm

Step 1: Convert temperature to Kelvin

The formula to convert Celsius to Kelvin is \(T(K)=T(^\circ C)+273.15\). For \(T = - 234^\circ C\), we have \(T(K)=-234 + 273.15=39.15\ K\) (approx \(39\ K\)). But wait, looking at the phase diagram, the temperature axis starts at \(0\ K\) and goes to \(600\ K\). Wait, maybe I made a mistake. Wait, \(-234^\circ C\) to Kelvin: \(-234+273 = 39\ K\)? Wait, no, \(273.15\) is more accurate, but maybe the diagram has a different scale? Wait, the x - axis is temperature in K, with marks at \(0\), \(200\), \(400\), \(600\). Wait, maybe the temperature conversion was wrong. Wait, \(-234^\circ C\) is \(T = 273-234=39\ K\)? But the phase diagram's first grid line after \(0\) is maybe \(200\)? Wait, no, maybe the question has a typo? Wait, no, let's re - check. Wait, if the temperature is \(-234^\circ C\), \(T(K)=-234 + 273.15 = 39.15\ K\). But the phase diagram's x - axis is from \(0\) to \(600\ K\), with \(200\), \(400\), \(600\) marked. Wait, maybe the temperature is \(-234^\circ C\) which is \(39\ K\), but the phase diagram's solid - liquid - gas regions: the solid region is on the left (low temperature), liquid in the middle, gas on the right. The pressure is \(1.7\) atm. Looking at the phase diagram, at low temperature (around \(39\ K\)) and \(1.7\) atm, which is in the solid region? Wait, no, maybe I miscalculated the temperature. Wait, \(-234^\circ C\) is \(39\ K\), but the phase diagram's x - axis: the first curve (solid - gas) starts at \((0,0)\) and goes up, the solid - liquid curve is vertical? Wait, no, the phase diagram has solid, liquid, gas regions. The solid region is where temperature is low and pressure is high? Wait, no, the phase diagram: the left - most region is solid (low temperature, can be low or high pressure? Wait, the y - axis is pressure (atm), x - axis is temperature (K). The solid - liquid boundary: if the solid - liquid line has a negative slope? Wait, no, in the diagram, the solid region is on the left (low temperature), liquid in the middle (moderate temperature and pressure), gas on the right (high temperature, low pressure) or high temperature and high pressure? Wait, the gas region is at the bottom right? Wait, no, the labels: solid is on the left, liquid in the middle, gas on the bottom right? Wait, the curve between solid and gas is increasing, between liquid and gas is increasing, and between solid and liquid is a vertical or steep line. Wait, the point where all three meet is the triple point. Let's assume that \(-234^\circ C=39\ K\) (approx) and pressure \(1.7\) atm. Looking at the phase diagram, at \(T = 39\ K\) (low temperature) and \(P = 1.7\) atm (moderate pressure), which is in the solid region? Wait, no, maybe the temperature is \(-234^\circ C\) which is \(39\ K\), but the phase diagram's x - axis: the first major mark is \(200\ K\). Wait, maybe I made a mistake in temperature conversion. Wait, \(-234^\circ C\) to Kelvin: \(T(K)=-234 + 273.15 = 39.15\ K\). But the phase diagram's x - axis is from \(0\) to \(600\ K\), with \(200\), \(400\), \(600\) as grid marks. So at \(T = 39\ K\) (left of \(200\ K\)) and \(P = 1.7\) atm (above the solid - gas curve), the state is solid? Wait, no, the solid - gas curve starts at \((0,0)\) and goes up. So at \(P = 1.7\) atm and \(T = 39\ K\), which is above the solid - gas curve, so it's solid? Wait, no, the solid region is where temperature is low and pressure is high relative to the solid - gas curve. So the state is solid? Wait, no, maybe the temperature is \(-234^\circ C\) which is \(39\ K\), and pressure \(1.7\) atm: looking at the phase diagram, the solid region is on the left (low temperature), liquid in the middle (between the two curves), gas on the right (high temperature, low pressure) or high temperature, high pressure? Wait, the liquid region is between the solid - liquid curve and the liquid - gas curve. So if at \(T = 39\ K\) (low) and \(P = 1.7\) atm (moderate), it's in the solid region.

Sub - Question 2: Temperature constant at \(-234^\circ C\) ( \(39\ K\) ), pressure decreased by \(1\) atm (new pressure \(=1.7 - 1=0.7\) atm)

Step 1: Analyze the phase change

At constant temperature (\(39\ K\)), when pressure decreases from \(1.7\) atm to \(0.7\) atm, we move vertically down on the phase diagram. The solid - gas curve is at lower pressure for a given temperature. So crossing the solid - gas curve, the solid will sublime (change from solid to gas) because we are moving from the solid region to the gas region at constant temperature by decreasing pressure.

Sub - Question 3: Pressure constant at \(1.7\) atm, temperature increased by \(237^\circ C\)

Step 1: Calculate the new temperature

Initial temperature \(T_1=-234^\circ C\), increase by \(237^\circ C\), so \(T_2=-234 + 237 = 3^\circ C\). Convert to Kelvin: \(T_2=3 + 273.15 = 276.15\ K\) (approx \(276\ K\)).

Step 2: Analyze the phase change

At constant pressure (\(1.7\) atm), when temperature increases from \(39\ K\) to \(276\ K\), we move horizontally to the right on the phase diagram. We will cross the solid - liquid curve (if the solid - liquid curve is at some temperature between \(39\) and \(276\ K\) at \(1.7\) atm) or the solid - gas or liquid - gas curve. Looking at the phase diagram, the solid - liquid curve is steep, and the liquid - gas curve is increasing. At \(P = 1.7\) atm, as temperature increases from low (solid region) to moderate, we cross the solid - liquid curve, so the solid will melt (change from solid to liquid) or if we cross the liquid - gas curve, it will boil, but since the temperature increase is to \(276\ K\) (moderate), it's likely melting (solid to liquid) or maybe vaporization, but more likely melting as we move from solid to liquid region.

Final Answers (assuming standard phase diagram analysis):

1. State at \(-234^\circ C\) and \(1.7\) atm: \(\boldsymbol{\text{solid}}\)
2. When pressure decreased by \(1\) atm at \(-234^\circ C\): \(\boldsymbol{\text{sublime (solid to gas)}}\)
3. When temperature increased by \(237^\circ C\) at \(1.7\) atm: \(\boldsymbol{\text{melt (solid to liquid)}}\) (or boil if the temperature crosses liquid - gas curve, but based on the diagram's scale, more likely melt)