Question

The above diagram is known as phase diagram. The phase diagram for a substance is shown in the figure where $T$ is temperature in Kelvin and $P$ is pressure in atmospheres. For example if a substance at a particular temperature and pressure lies in gas area , the substance exists only as a gas.
However, if the temperature-pressure pairing is on the line segment between gas and liquid, the substance exists as both a gas and a liquid.. If $T>0$ and $P>0$, which of the following systems of inequalities would describe the temperature and pressure ranges in which this substance exists only as a liquid?

• A. $P<\frac{17}{2}T-1,860;P>\frac{5}{18}T-\frac{460}{9}$
• B. $P>\frac{17}{2}T-1,860;P<\frac{5}{18}T-\frac{460}{9}$
• C. $P<\frac{17}{2}T-1,860;P>\frac{11}{200}T-\frac{460}{9}$
• D. $P\ge \frac{17}{2}T-1,860;P\le \frac{11}{200}T-\frac{460}{9}$

Answer 1

Answer: (A)

$P<\frac{17}{2}T-1,860;P>\frac{5}{18}T-\frac{460}{9}$

The equation of line joining points $(220,10)$ and $(400,60)$ is $(P-10)=\frac{5}{18}(T-220)$

$\Rightarrow P=\frac{5}{18}T+-\frac{460}{9}$

the equation of line joining points $(220,10)$ and $(230,95)$ is $(P-10)=\frac{17}{2}(T-220)$

$\Rightarrow P=\frac{17}{2}T-1860$

For the substance to exist only in liquid region $P<\frac{17}{2}T-1860$ and $P>\frac{5}{18}T-\frac{460}{9}$

Therefore correct option is $A$