Real gases — deviations from ideality, compressibility factor Z: Exercises

Symbols used throughout (all defined in the parent, repeated so you never guess):

$P$ = pressure, $V$ = volume, $n$ = number of moles, $T$ = absolute temperature.
$R = 0.0821 atm\cdotL\cdotmol^{- 1} K^{- 1}$ = the gas constant.
$V_{m} = V / n$ = molar volume (the volume one mole occupies).
$Z = \frac{P V _{m}}{R T} = \frac{P V}{n R T}$ = the compressibility factor (real-world $P V$ divided by ideal-world $R T$ ).
$a$ = attraction strength constant; $b$ = excluded volume per mole.

Level 1 — Recognition

L1.1

State the value of $Z$ for a perfectly ideal gas, and say what $Z = 1.4$ and $Z = 0.7$ each tell you physically.

Recall Solution L1.1

WHAT $Z$ is: it is the ratio "what we measured" $\div$ "what the ideal law predicted". If they agree, the ratio is exactly $1$ .

Ideal gas: $Z = 1$ at every $P$ and $T$ .
$Z = 1.4 > 1$ : the real $P V_{m}$ is $40%$ bigger than $R T$ → the gas is harder to compress than ideal → the finite-size / repulsion effect is winning (typical at high $P$ ).
$Z = 0.7 < 1$ : the real $P V_{m}$ is $30%$ smaller than $R T$ → the gas is easier to compress → attraction is winning (typical at low $P$ ).

L1.2

A gas has $Z = 1$ over a whole range of low pressures at one particular temperature. What is that temperature called?

Recall Solution L1.2

That is the Boyle temperature $T_{B}$ . At $T_{B}$ the two corrections (size pushing $Z$ up, attraction pulling $Z$ down) cancel at low pressure, so the gas mimics an ideal gas over a band of low pressures. See Boyle Temperature.

Level 2 — Application

L2.1

$0.5 mol$ of a gas occupies $V = 10 L$ at $P = 1 atm$ and $T = 300 K$ . Compute $Z$ and interpret it.

Recall Solution L2.1

WHAT we do: plug straight into $Z = \frac{P V}{n R T}$ — measured over predicted. $Z = \frac{( 1 ) ( 10 )}{( 0.5 ) ( 0.0821 ) ( 300 )} = \frac{10}{12.315} = 0.812$ Interpret: $Z < 1$ → attractions dominate at this low pressure → the gas is squeezed a little tighter than the ideal law expects.

L2.2

$2 mol$ of $N_{2}$ at $T = 250 K$ is found to have $Z = 0.92$ when $P = 50 atm$ . Find the actual volume $V$ .

Recall Solution L2.2

WHY rearrange: we know $Z$ , $P$ , $n$ , $T$ and want $V$ . Start from $Z = \frac{P V}{n R T}$ and solve for $V$ : $V = \frac{Z n R T}{P} = \frac{( 0.92 ) ( 2 ) ( 0.0821 ) ( 250 )}{50}$ $V = \frac{37.766}{50} = 0.7553 L$ The ideal law alone would have given $V_{ideal} = n R T / P = 41.05/50 = 0.821 L$ ; the real gas sits at $0.92$ of that — physically smaller because attractions pull it inward.

L2.3

Compute the Boyle temperature of $N_{2}$ given $a = 1.35 atm\cdotL^{2} mol^{- 2}$ , $b = 0.0387 L\cdotmol^{- 1}$ .

Recall Solution L2.3

WHY this formula: $T_{B}$ is the temperature where $b - \frac{a}{R T} = 0$ , i.e. the size correction equals the attraction correction. Solving gives $T_{B} = \frac{a}{R b}$ . $T_{B} = \frac{1.35}{( 0.0821 ) ( 0.0387 )} = \frac{1.35}{0.0031773} \approx 424.9 K$ Above this temperature $N_{2}$ shows $Z > 1$ from the start; below it, $Z$ dips under 1 first.

Level 3 — Analysis

L3.1

From van der Waals, $Z = \frac{V _{m}}{V _{m} - b} - \frac{a}{R T V _{m}}$ . For $CO_{2}$ ( $a = 3.59$ , $b = 0.0427$ ) at $T = 300 K$ and molar volume $V_{m} = 0.5 L\cdotmol^{- 1}$ , compute $Z$ and say which term wins.

Recall Solution L3.1

WHAT each term is: first term = size/repulsion (always $> 1$ ), second term = attraction (always subtracts). $size term = \frac{0.5}{0.5 - 0.0427} = \frac{0.5}{0.4573} = 1.09338$ $attraction term = \frac{3.59}{( 0.0821 ) ( 300 ) ( 0.5 )} = \frac{3.59}{12.315} = 0.29151$ $Z = 1.09338 - 0.29151 = 0.8019$ WHY $Z < 1$ : the attraction subtraction ( $0.29$ ) beats the size excess ( $0.09$ ), so at this moderate compression $CO_{2}$ 's strong attraction (large $a$ ) drags $Z$ well below 1. See Intermolecular Forces.

L3.2

Using the low-pressure expansion $Z \approx 1 + \frac{1}{V _{m}} (b - \frac{a}{R T})$ , decide the sign of the initial slope of $Z$ -vs- $P$ for $H_{2}$ ( $a = 0.244$ , $b = 0.0266$ ) at $T = 298 K$ .

Recall Solution L3.2

WHY this expansion: at low $P$ , $V_{m}$ is large so $b / V_{m} ≪ 1$ and $\frac{V _{m}}{V _{m} - b} \approx 1 + \frac{b}{V _{m}}$ . The bracket $(b - \frac{a}{R T})$ decides whether $Z$ rises or falls first. $\frac{a}{R T} = \frac{0.244}{( 0.0821 ) ( 298 )} = \frac{0.244}{24.4658} = 0.009973$ $b - \frac{a}{R T} = 0.0266 - 0.009973 = + 0.016627 > 0$ Conclusion: the bracket is positive, so the initial slope of $Z$ vs $P$ is positive → $H_{2}$ has $Z > 1$ from the very start at room temperature. Its tiny $a$ means size wins immediately. Look at figure s01 below.

Figure — Real gases — deviations from ideality, compressibility factor Z

Level 4 — Synthesis

L4.1

A gas has $T_{B} = 110 K$ and $b = 0.0391 L\cdotmol^{- 1}$ . (a) Find $a$ . (b) At $T = 300 K$ (well above $T_{B}$ ), predict whether $Z$ starts above or below 1, and justify with the bracket $(b - \frac{a}{R T})$ .

Recall Solution L4.1

(a) WHY invert $T_{B}$ : from $T_{B} = \frac{a}{R b}$ we get $a = T_{B} R b$ . $a = (110) (0.0821) (0.0391) = 0.35311 atm\cdotL^{2} mol^{- 2}$ (b) At $T = 300 K$ : $\frac{a}{R T} = \frac{0.35311}{( 0.0821 ) ( 300 )} = \frac{0.35311}{24.63} = 0.014336$ $b - \frac{a}{R T} = 0.0391 - 0.014336 = + 0.024764 > 0$ Positive bracket → $Z$ starts above 1. Consistency check: we are at $T = 300 K > T_{B} = 110 K$ , and the rule says above $T_{B}$ , $Z > 1$ from the start. ✔ The two methods agree.

L4.2

Two gases at the same $T$ and low $P$ : Gas A has $Z = 1.05$ , Gas B has $Z = 0.95$ . Both have the same $b$ . Explain which has the larger $a$ , and connect this to their critical temperatures.

Recall Solution L4.2

Reasoning through the bracket $Z - 1 \approx \frac{1}{V _{m}} (b - \frac{a}{R T})$ , same $V_{m}$ , same $b$ , same $T$ :

Gas A: $Z - 1 = + 0.05 > 0 \Rightarrow b > \frac{a _{A}}{R T} \Rightarrow a_{A}$ is small.
Gas B: $Z - 1 = - 0.05 < 0 \Rightarrow b < \frac{a _{B}}{R T} \Rightarrow a_{B}$ is large. So $a_{B} > a_{A}$ — Gas B attracts more strongly. Link to critical constants: stronger attraction (larger $a$ ) means the gas liquefies more easily and has a higher critical temperature $T_{c}$ . Since $T_{c} \propto \frac{8 a}{27 R b}$ , Gas B (bigger $a$ , same $b$ ) has the higher $T_{c}$ . See Critical Constants and Liquefaction.

Level 5 — Mastery

L5.1

For $1 mol$ of a van der Waals gas at $T = 273 K$ , $P = 100 atm$ , with $a = 1.36$ , $b = 0.0318$ , the molar volume is found (by solving vdW) to be $V_{m} = 0.1975 L\cdotmol^{- 1}$ . Compute $Z$ two independent ways — directly from $Z = P V_{m} / R T$ , and from the two-term vdW expression — and confirm they match. Then state which effect dominates.

Recall Solution L5.1

Method 1 — direct definition: $Z = \frac{P V _{m}}{R T} = \frac{( 100 ) ( 0.1975 )}{( 0.0821 ) ( 273 )} = \frac{19.75}{22.4133} = 0.8812$ Method 2 — two-term vdW: $size term = \frac{V _{m}}{V _{m} - b} = \frac{0.1975}{0.1975 - 0.0318} = \frac{0.1975}{0.1657} = 1.19191$ $attraction term = \frac{a}{R T V _{m}} = \frac{1.36}{( 0.0821 ) ( 273 ) ( 0.1975 )} = \frac{1.36}{4.42663} = 0.30723$ $Z = 1.19191 - 0.30723 = 0.88468$ Match: both give $Z \approx 0.88$ (tiny difference from the rounded $V_{m}$ ). ✔ Which wins: even at $100 atm$ the attraction subtraction ( $0.31$ ) still slightly beats the size excess ( $0.19$ ), so $Z < 1$ . This gas has not yet reached the pressure where size takes over. See figure s02.

L5.2

Sketch qualitatively (and reason out the key features of) $Z$ vs $P$ at $273 K$ for $H_{2}$ , $N_{2}$ , and $CO_{2}$ . State, for each, the starting slope sign and the eventual high- $P$ behaviour.

Recall Solution L5.2

Use $Z - 1 \approx \frac{1}{V _{m}} (b - \frac{a}{R T})$ for the start, and the fact that every gas rises above 1 at very high $P$ .

$H_{2}$ (tiny $a$ ): bracket positive → starts above 1, slope up, stays and rises above 1. Never dips.
$N_{2}$ (moderate $a$ , $T_{B} \approx 425 K > 273 K$ ): we are below its Boyle temperature, so bracket negative → dips below 1 first (shallow minimum), then climbs back above 1 at high $P$ .
$CO_{2}$ (large $a$ ): strongly negative bracket → deep dip below 1 (deepest minimum of the three), then eventually rises above 1 at very high $P$ . Universal high- $P$ rule: as $P \to$ large, $V_{m} \to b$ , the term $\frac{V _{m}}{V _{m} - b} \to \infty$ , so $Z \to \infty$ — all three curves head upward. See figure s03.

Recall One-line summary you should be able to say aloud

$Z$ is measured-over-ideal; the bracket $(b - \frac{a}{R T})$ sets the low- $P$ slope, $T_{B} = a / R b$ is where that bracket is zero, and no matter what, high pressure sends every gas above 1.