On a pressure against temperature diagram for a pure substance, the vaporisation curve ends at
Answer: A
Above the critical point liquid and vapour are no longer distinguishable, so there is nothing left for the curve to separate. The triple point is the other end of the same curve, where the solid phase joins the equilibrium.
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State whether the following is true or false.
At the critical point the densities of the saturated liquid and the saturated vapour become equal.
Answer: B
Correct answer: True
The critical point is where the saturated liquid and saturated vapour lines meet, so the two phases become identical there - same density, same refractive index, and no meniscus. The latent heat of vaporisation falls to zero for the same reason, since there is no longer any distinction between the phases to overcome.
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A fluid held above its critical temperature
Answer: C
Correct answer: cannot be liquefied by compression alone
Above the critical temperature no amount of pressure will produce a distinct liquid phase; the fluid simply becomes denser continuously, without any phase boundary appearing. Cooling below the critical temperature is a necessary condition for liquefaction. It also does not behave as an ideal gas - a supercritical fluid can be as dense as a liquid.
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Match each region of the PVT surface with what is found there.
List I
A. Above the critical temperature
B. Inside the saturation dome
C. Left of the saturated liquid line
D. Right of the saturated vapour line
List II
1. a subcooled liquid
2. a single supercritical fluid
3. a superheated vapour
4. liquid and vapour in equilibrium
Answer: D
Locating a state on the diagram decides which relations may be used, since inside the dome pressure and temperature are no longer independent while outside it they are. Reading the region first prevents most errors in property lookup.
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Assertion (A): Inside the two-phase region, pressure and temperature are not independent.
Reason (R): Along the saturation curve one fixes the other, so a further property such as quality is needed to locate the state.
Answer: A
Correct answer: Both A and R are true, and R is the correct explanation of A
Both are true and the reason gives the consequence. Inside the two-phase envelope the saturation relation ties pressure and temperature together, so specifying both fixes nothing beyond which saturation state it is. A second, independent property is needed - the quality, or equivalently the specific volume - to say where between the saturated liquid and saturated vapour the mixture lies.
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A wet steam mixture has a quality of 0.8. If the saturated liquid volume is 0.001 and the saturated vapour volume is 0.20 cubic metres per kilogram, what is the specific volume of the mixture?
Answer: B
The mixture volume is the liquid volume plus the quality times the difference between them, which is one thousandth plus eight tenths of about one hundred and ninety-nine thousandths. Quality always multiplies the difference, never the vapour value alone.
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The compressibility factor is defined as
Answer: C
Written as pressure times volume over the gas constant and temperature, it measures how far a gas departs from ideality at the conditions given. Values below one mean attraction dominates and above one that repulsion does.
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Consider the following statements about the compressibility factor.
1. It equals one for an ideal gas at all conditions.
2. It falls below one at moderate pressures for most gases.
3. It can never exceed one for any real gas.
Answer: D
The first two are correct. The third is wrong, because at high pressure the molecules are pushed close enough that repulsive forces dominate and the factor climbs above one, sometimes well above. Hydrogen and helium show this at quite ordinary pressures.
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The principle of corresponding states asserts that
Answer: A
Correct answer: all fluids behave alike at the same reduced conditions
The principle states that the compressibility factor is approximately a universal function of reduced temperature and reduced pressure, so a chart built from one fluid can be used for another. It is an approximation rather than a law, and it is best for small non-polar molecules - the limitation the acentric factor was introduced to address.
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Assertion (A): A third parameter, the acentric factor, is added to the corresponding states correlation.
Reason (R): Two-parameter corresponding states works well for simple spherical molecules but poorly for polar or elongated ones.
Answer: B
Correct answer: Both A and R are true, and R is the correct explanation of A
Both are true and the reason gives the motivation. Two-parameter corresponding states assumes spherical, non-polar molecules, and argon, krypton and methane do follow it closely. Elongated or polar molecules deviate systematically, so Pitzer’s acentric factor - defined from the reduced vapour pressure at a reduced temperature of 0.7 - is added as a third parameter to correct for exactly that departure.
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