- Van 't Hoff factor i
- 1
- Molar concentration M (mol/L)
- 0.1mol/L
- Temperature T (K)
- 310K
2.5438atm
Open with these values2.5438atm
Result: 2.5438 atmΠ = i × M × R × T, with R = 0.082057 L·atm/(mol·K) and the temperature in kelvin, gives the pressure in atmospheres. 0.1 mol/L glucose at 310 K is 2.5438 atm, or 257.75 kPa. Multiply by 101.325 for kilopascals; add 273.15 to a Celsius reading first.
Held fixed: Van 't Hoff factor i 1.00, Molar concentration M (mol/L) 0.1000 mol/L.
| Temperature T (K) (K) | Result (atm) |
|---|---|
| 100.00 | 0.8206 |
| 200.00 | 1.6411 |
| 300.00 | 2.4617 |
| 310.00Your value | 2.5438 |
| 400.00 | 3.2823 |
| 500.00 | 4.1029 |
2.5438atm
Open with these values24.4653atm
Open with these values15.2626atm
Open with these valuesΠ = i × M × R × T, R = 0.082057 L·atm/(mol·K)
| i, M, T | Same pressure in kPa | Π in atm |
|---|---|---|
| 1, 0.05, 273.15 | 113.55 | 1.1207 |
| 1, 0.1, 310 | 257.75 | 2.5438 |
| 3, 0.2, 310 | 1546.48 | 15.2626 |
| 2, 0.5, 298.15 | 2478.95 | 24.4653 |
| 2, 1, 300 | 4988.66 | 49.2342 |
Multiply the van 't Hoff factor by the molarity, the gas constant R = 0.082057 L·atm/(mol·K), and the absolute temperature. For 0.1 mol/L glucose at 310 K that is 1 × 0.1 × 0.082057 × 310 ≈ 2.5438 atm.
The equation mirrors the ideal gas law, which uses absolute temperature measured from absolute zero. Convert Celsius by adding 273.15, so body temperature at 37 °C is 310.15 K.
It is how many particles one formula unit produces on dissolving: 1 for non-electrolytes such as glucose, 2 for NaCl, 3 for CaCl₂. More particles mean more pressure at the same concentration, which is why the calculator asks rather than assuming a substance.
Multiply the atmospheres by 101.325 for kilopascals, or by 1.01325 for bar. So 2.5438 atm is about 257.75 kPa.
It is at its best for dilute, ideal solutions. In concentrated electrolytes ion pairing lowers the effective van 't Hoff factor, so the calculated pressure is an upper estimate.
Information, not professional advice.
Diese Seite gibt es auch auf Deutsch.
Zu Deutsch wechseln