Question
The given graph represents the variation of compressibility factor (Z)= p V n R T , for three real gases A, B and C. Identify the only incorrect statement.
BITSAT
Chemistry
States of Matter
2022
BITSAT 2022
BITSAT Chemistry Question (2022) — Solution
Options
- A. For the gas A, a=0 and its dependence on p is linear at all pressure.
- B. For the gas B, b=0 and its dependence on p is linear at all pressure.
- C. For the gas C, which is typical real gas for which neither. a nor b=0. By knowing the minima and point of the intersection, with Z=1, a and b can be calculated.
- D. At high pressure the slope is positive for all real gases.
Answer
B. For the gas B, b=0 and its dependence on p is linear at all pressure.
Step-by-step solution
From the graph it is clear that, the value of ' Z ' decreases with increase of pressure. We can explain as follows on the basis of van der Waals' equation. At high pressure, when ' p ' is large, V will be small and one cannot ignore ' b ' in comparison to V. However, the team a / V^2 may be considered negligible in comparison to ' p ' in van der Waals' equation. (p+ a V^2 )(V-b)=n R T p(V-b)=n R T p V-p b=n R T or p V n R T =1+ p b n R T or Z=1+ p b n R T Thus, Z is greater than 1. As pressure is increased (at constant T ), the factor p b n R T increases. This explains why after minima in the curves, Z increase continuously with pressure. Hence, the only incorrect statement is (b).
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