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 - 79. °c and 13.9 atm. what will be the state of the sample? suppose the temperature is held constant at - 79. °c but the pressure is decreased by 6.9 atm. what will happen to the sample? suppose, on the other hand, the pressure is held constant at 13.9 atm but the temperature is decreased by 171. °c. what will happen to the sample?

Explanation:

Step1: Convert temperature to Kelvin

First, convert - 79°C to Kelvin. The conversion formula is $T(K)=T(^{\circ}C)+273.15$. So, $T=-79 + 273.15=194.15K$.

Step2: Analyze first - condition

At $T = 194.15K$ and $P = 13.9atm$, from the phase - diagram, the sample is in the liquid state.

Step3: Calculate new pressure for second - condition

The new pressure is $P_{new}=13.9 - 6.9=7atm$. At $T = 194.15K$ and $P = 7atm$, from the phase - diagram, the sample will be in the gas state. So it will vaporize (change from liquid to gas).

Step4: Calculate new temperature for third - condition

The new temperature is $T_{new}=-79-171=-250^{\circ}C$. Convert it to Kelvin: $T_{new}=-250 + 273.15 = 23.15K$. At $P = 13.9atm$ and $T = 23.15K$, from the phase - diagram, the sample will be in the solid state. So it will freeze (change from liquid to solid).

Answer:

Vaporize; Freeze