SI units in chemistry — kg, mol, K, Pa; derived units (J, L): Exercises

Before we start, one picture to fix the "unit-building" idea in your head.

Figure — SI units in chemistry — kg, mol, K, Pa; derived units (J, L)

Read it like this: the base bricks on the left (kg, m, s, mol, K) are decreed. Everything on the right (N, J, Pa, L) is built by snapping bricks together through a defining equation. No memorising — you read the unit off the formula.

Level 1 — Recognition

Recall Solution L1·1

WHAT/WHY: we recall which units are decreed independently vs built from others.

(a) amount of substance → mole (mol), base.
(b) temperature → kelvin (K), base.
(c) mass → kilogram (kg), base.
(d) pressure → pascal (Pa), derived ( $Pa = kg m^{- 1} s^{- 2}$ ). The odd one out.

Recall Solution L1·2

Energy = work = force × distance. Force is $ma$ so $[F] = kg m s^{- 2} = N$ . Then $1 J = N \cdot m = kg m^{2} s^{- 2} .$ WHY: we just multiplied force's units by a length — that's what "force acting over distance" means.

Recall Solution L1·3

$T_{K} = T_{^{\circ} C} + 273.15 = 25 + 273.15 = 298.15 K .$ WHY: kelvin and Celsius have the same step size; we only shift the zero to absolute zero.

Level 2 — Application

Recall Solution L2·1

$n = \frac{m}{M} = \frac{8.0}{16} = 0.50 mol .$ WHY: $M$ is grams per mole, so dividing grams by it cancels grams and leaves mol.

Recall Solution L2·2

$N = n N_{A} = 0.50 \times 6.022 \times 1 0^{23} = 3.011 \times 1 0^{23} molecules .$ WHY: one mole holds $N_{A}$ particles; scale that by the number of moles (pure proportionality).

Recall Solution L2·3

$p = \frac{F}{A} = \frac{600}{0.020} = 30000 Pa = 3.0 \times 1 0^{4} Pa .$ Convert: $1 bar = 1 0^{5} Pa$ , so $p = 30000/1 0^{5} = 0.30 bar$ . WHY: the same push over a bigger area gives less pressure, so we divide by area.

Level 3 — Analysis

Recall Solution L3·1

Step 1 — fix the volume unit. $R$ carries joules, and $1 J = Pa \cdot m^{3}$ , so volume must be in $m^{3}$ : $250 mL = 0.250 L = 0.250 \times 1 0^{- 3} m^{3} = 2.50 \times 1 0^{- 4} m^{3} .$ Step 2 — solve $p V = n R T$ for $n$ . $n = \frac{p V}{R T} = \frac{( 1.00 \times 1 0 ^{5} ) ( 2.50 \times 1 0 ^{- 4} )}{( 8.314 ) ( 273.15 )} = \frac{25.0}{2271.0} \approx 1.10 \times 1 0^{- 2} mol .$ WHY: every quantity is now in coherent SI, so joules cancel cleanly and the answer is pure mol. See Ideal Gas Law.

Recall Solution L3·2

Unit check first (the Dimensional Analysis habit): $kg \cdot (m/s)^{2} = kg m^{2} s^{- 2} = J . ✓$ So the answer can be an energy. Now the number: $K E = \frac{1}{2} m v^{2} = \frac{1}{2} (5.3 \times 1 0^{- 26}) (480)^{2} = \frac{1}{2} (5.3 \times 1 0^{- 26}) (2.304 \times 1 0^{5}) \approx 6.11 \times 1 0^{- 21} J .$ WHY: checking units before trusting the number catches slips early — this links to Kinetic Theory of Gases.

Level 4 — Synthesis

Recall Solution L4·1

Idea: density $ρ = m / V$ and $n = m / M$ . Substitute $n$ into $p V = n R T$ : $p V = \frac{m}{M} R T \Rightarrow M = \frac{m R T}{p V} = \frac{ρR T}{p} .$ Step 1 — SI-ify the density. $ρ = 1.25 g/L = 1.25 kg/m^{3}$ (since $g/L = kg/m^{3}$ — a neat coincidence of the two $\times 1 0^{- 3}$ factors cancelling). Step 2 — plug in (all SI). $M = \frac{( 1.25 ) ( 8.314 ) ( 273.15 )}{1.013 \times 1 0 ^{5}} = \frac{2839.0}{1.013 \times 1 0 ^{5}} \approx 0.0280 kg/mol = 28.0 g/mol .$ Step 3 — identify. $28 g/mol$ → nitrogen $N_{2}$ (or CO). See Molar Mass and Molecular Mass. WHY: we fused three tools — density, $n = m / M$ , and $p V = n R T$ — into one clean formula, then trusted units to hand us kg/mol.

Recall Solution L4·2

Part A: $250 J = 250 kg m^{2} s^{- 2}$ — the joule is that base-unit spelling, nothing to compute. Part B: work at constant pressure is $W = p Δ V$ , so $Δ V = \frac{W}{p} = \frac{250}{1.0 \times 1 0 ^{5}} = 2.5 \times 1 0^{- 3} m^{3} = 2.5 L .$ Unit check: $\frac{J}{Pa} = \frac{kg m ^{2} s ^{- 2}}{kg m ^{- 1} s ^{- 2}} = m^{3} . ✓$ WHY: the same 250 J is describable as base units or as pressure×volume — derived units are just different LEGO builds of the same energy.

Level 5 — Mastery

Recall Solution L5·1

At fixed $V, n$ : $\frac{p _{2}}{p _{1}} = \frac{T _{2}}{T _{1}}$ — only with absolute temperature (so $p \propto$ average kinetic energy from Kinetic Theory of Gases). Convert to kelvin: $T_{1} = 27 + 273.15 = 300.15 K$ , $T_{2} = 327 + 273.15 = 600.15 K$ . $\frac{p _{2}}{p _{1}} = \frac{600.15}{300.15} \approx 2.00.$ The wrong path: using Celsius gives $327/27 \approx 12.1$ — absurd, because Celsius zero is arbitrary (freezing water), so ratios of °C values are physically meaningless. WHY: kelvin's zero is zero motion, so doubling $T$ genuinely doubles thermal push. This is the whole reason Ideal Gas Law insists on K.

Recall Solution L5·2

Rule: in multiplication/division the result keeps the fewest significant figures among the inputs (Significant Figures and Measurement Uncertainty). Here the fewest is $3$ (from $ρ$ ). Full value: $M = 0.02801 kg/mol$ . Rounded to 3 s.f.: $M = 28.0 g/mol .$ WHY: reporting $28.014$ g/mol would fake a precision the $1.25$ density never had — the answer is only as sharp as its bluntest input.

Recall Solution L5·3

$V \to 0$ : $p = n R T / V \to \infty$ . Squeezing the same particles into vanishing space makes pressure blow up — matches intuition (and why real gases eventually stop being ideal).
$n = 0$ : no particles, so $p = 0$ for any $V, T$ . An empty container pushes on nothing.
Unit of $R$ : from $R = p V / (n T)$ , $[R] = \frac{Pa \cdot m ^{3}}{mol \cdot K} = \frac{J}{mol \cdot K} = J mol^{- 1} K^{- 1} . ✓$ WHY: limiting cases and a unit check together prove the formula is well-behaved everywhere, not just at "nice" inputs.

Recall One-line self-test

Why must volume be in $m^{3}$ (not L) when $R$ is in $J mol^{- 1} K^{- 1}$ ? ::: Because $1 J = Pa \cdot m^{3}$ ; only with $V$ in $m^{3}$ do the joules cancel. $8.0 g$ of $CH_{4}$ is how many mol? ::: $0.50 mol$ . Pressure ratio when a fixed-volume gas goes $27 \to 327^{\circ} C$ ? ::: $\approx 2.00$ (using kelvin).

Connections

The Mole Concept — the $n = m / M$ and $N = n N_{A}$ steps used in L2–L4.
Ideal Gas Law — L3, L4, L5 all live inside $p V = n R T$ .
Dimensional Analysis — the unit-check-first habit throughout.
Molar Mass and Molecular Mass — identifying the gas in L4·1.
Significant Figures and Measurement Uncertainty — L5·2.
Kinetic Theory of Gases — why $p \propto T$ needs kelvin (L5·1).