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 99. °c and 11.6 atm. what will be the state of the sample? suppose the temperature is held constant at 99. °c but the pressure is decreased by 4.3 atm. what will happen to the sample? suppose, on the other hand, the pressure is held constant at 11.6 atm but the temperature is decreased by 135. °c. what will happen to the sample?

Explanation:

Step1: Convert temperature to Kelvin

First, convert $99^{\circ}C$ to Kelvin. The conversion formula is $T(K)=T(^{\circ}C)+ 273.15$. So, $T = 99 + 273.15=372.15K$.

Step2: Determine initial state

Locate the point with $T = 372.15K$ and $P = 11.6atm$ on the phase - diagram. We can see that the sample is in the liquid state.

Step3: Calculate new pressure for second part

The new pressure is $P_{new}=11.6 - 4.3=7.3atm$ with $T = 372.15K$ constant. Looking at the phase - diagram, as we decrease the pressure from $11.6atm$ to $7.3atm$ at $372.15K$, the sample will change from liquid to gas (vaporization).

Step4: Calculate new temperature for third part

The new temperature is $T_{new}=99 - 135=- 36^{\circ}C$. Convert it to Kelvin: $T_{new}=-36 + 273.15 = 237.15K$ with $P = 11.6atm$ constant. Looking at the phase - diagram, as we decrease the temperature from $372.15K$ to $237.15K$ at $11.6atm$, the sample will change from liquid to solid (freezing).

Answer:

Liquid
It will vaporize
It will freeze