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 59. °c and 1.4 atm. what will be the state of the sample? suppose the temperature is held constant at 59. °c but the pressure is decreased by 0.8 atm. what will happen to the sample? suppose, on the other hand, the pressure is held constant at 1.4 atm but the temperature is increased by 164. °c. what will happen to the sample?

Explanation:

Step1: Convert temperature to Kelvin

First, convert $59^{\circ}C$ to Kelvin. The conversion formula is $K = ^{\circ}C+ 273.15$. So, $T = 59 + 273.15=332.15\ K$ and the pressure $P = 1.4\ atm$. Locate this point on the phase - diagram. We see that at $T = 332.15\ K$ and $P = 1.4\ atm$, the substance is in the liquid state.

Step2: Calculate new pressure

The new pressure when decreased by $0.8\ atm$ is $P_{new}=1.4 - 0.8=0.6\ atm$ with $T = 332.15\ K$ still. Locate this new point on the phase - diagram. Since the original state was liquid and the new point is in the gas region, the substance will vaporize (change from liquid to gas).

Step3: Calculate new temperature

The new temperature when increased by $164^{\circ}C$ from the original $59^{\circ}C$ is $T_{new}=59 + 164=223^{\circ}C$. Convert this to Kelvin: $T_{new}=223+273.15 = 496.15\ K$ with $P = 1.4\ atm$. Locate this new point on the phase - diagram. Since the original state was liquid and the new point is in the gas region, the substance will vaporize (change from liquid to gas).

Answer:

Liquid
It will vaporize
It will vaporize