Pressure Units and Conversions — atm, bar, torr, Pa, psi
Gas-law numericals (PV = nRT, combined gas law, partial pressures) routinely hand you a pressure in one unit and expect an answer, or a value of R, in another. Get the unit conversion wrong and every downstream number is wrong even though your method was correct. This article is not about the gas laws themselves — it's specifically the conversion factors between the five pressure units you'll actually meet: atmosphere (atm), bar, torr (equivalently mmHg), pascal (Pa) and pound-force per square inch (psi).
1 atm = 1.01325 bar
1 atm = 760 torr = 760 mmHg
1 atm = 14.6959 psi
1 bar = 100000 Pa (exactly, by definition)
What each unit is: the pascal (Pa) is the SI base unit — one newton per square metre — and is what you must convert to before using R = 8.314 J mol⁻¹ K⁻¹ in PV = nRT with volume in m³. The atmosphere (atm) is defined relative to a standard sea-level pressure and is what most Indian textbooks default to. The bar is close to but not identical to 1 atm (1 bar is slightly less), and is common in engineering and meteorology. The torr (named after Torricelli) equals the pressure exerted by 1 mm of mercury under standard conditions, which is why torr and mmHg are used interchangeably. psi is common on pressure gauges and in engineering contexts, rare in Indian school/competitive chemistry.
kPa = 4.2 × 101.325 = 425.565 kPa.
bar = 4.2 × 1.01325 = 4.2557 bar.
Check: since 1 bar ≈ 100 kPa, 4.2557 bar should be close to 425.57 kPa — it is (425.565 kPa vs 425.57 kPa; the tiny gap is rounding).
Step 1 — convert to atm: 570 ÷ 760 = 0.75 atm.
Step 2 — use R = 0.0821 L·atm·mol⁻¹·K⁻¹ (the atm-litre form, matched to these units):
n = PV ÷ RT = (0.75 × 2.0) ÷ (0.0821 × 300) = 1.5 ÷ 24.63 = 0.0609 mol (3 s.f.).
P = 0.75 atm × 101325 = 75993.75 Pa.
V = 2.0 L = 2.0 × 10⁻³ m³.
n = PV ÷ RT = (75993.75 × 2.0×10⁻³) ÷ (8.314 × 300) = 151.9875 ÷ 2494.2 = 0.0609 mol.
Same answer either way — proof that the two R values and their matching unit systems are self-consistent, not two different physical facts.
atm = 32 ÷ 14.6959 = 2.178 atm.
- Using R = 0.0821 L·atm·mol⁻¹·K⁻¹ with a pressure still in torr or Pa — R's numeric value is only valid with the matching unit system (atm + litre, or Pa + m³, or kPa + m³, and so on).
- Confusing 1 atm = 760 mmHg with 1 atm = 760 Pa — a very common slip that's off by roughly a factor of 100.
- Treating 1 bar and 1 atm as identical — they're close (1 atm = 1.01325 bar) but not the same, and the difference matters in a precise numerical answer.
- Forgetting that "gauge pressure" (what a tyre gauge shows) is pressure above atmospheric, not absolute pressure — a genuine physics distinction, separate from the pure unit-conversion factors above.
| From | To | Multiply by |
|---|---|---|
| atm | Pa | 101325 |
| atm | bar | 1.01325 |
| atm | torr (mmHg) | 760 |
| atm | psi | 14.6959 |
| bar | Pa | 100000 (exact) |
| torr | atm | ÷760 |
Need instant conversion between these units? The calculator suite's unit converter has a dedicated Pressure category.
Open the Pressure Converter →For structured practice on gas-law numericals like the ones above, ABC Chemistry runs Class 11–12 chemistry coaching at its Gurugram centre and online across India — see abcchemistry.in for details.