Boyle’s, Charles’s and Gay-Lussac’s laws are each a special case of one underlying relationship. This solves that combined equation for any of its six values.
How it works
Combined gas law
P₁V₁ / T₁ = P₂V₂ / T₂
Holding temperature constant reduces this to Boyle’s law (P₁V₁ = P₂V₂); holding pressure constant gives Charles’s law; holding volume constant gives Gay-Lussac’s law.
Why temperature must be in kelvin here
The relationship is a direct proportion between the ratios PV/T, and a direct proportion is only meaningful measured from an actual zero point. Celsius’s zero is arbitrary (water’s freezing point); kelvin’s zero is absolute zero, a genuine physical floor. Using Celsius numbers directly — treating 0°C as if it meant “no temperature” the way 0 K actually does — produces a plausible-looking but wrong answer, which is why this calculator’s temperature fields are in kelvin throughout.
How to use this calculator
- Choose which of the six values you want solved.
- Enter the other five, with both temperatures in kelvin.
Frequently asked questions
What’s the difference between this and the ideal gas law calculator?
The ideal gas law (PV = nRT) needs to know the actual amount of gas (moles) and gives an absolute state. This combined law compares two states of the same fixed amount of gas to each other, and moles cancels out of the equation entirely — useful when you don’t know or don’t need the mole count.
Why do Boyle’s, Charles’s and Gay-Lussac’s laws all reduce to this one equation?
Because each of them is this same relationship with one variable held deliberately fixed — Boyle’s law is this equation at constant temperature, Charles’s at constant pressure, Gay-Lussac’s at constant volume. Learning the combined form covers all three special cases at once.
Can pressure and volume be in any units?
Yes, as long as you’re consistent — atmospheres and litres, or pascals and cubic metres, or any other matching pair — since the pressure and volume units cancel out when comparing state 1 to state 2, unlike temperature, which must be an absolute scale.
Why does the calculator decline a temperature of exactly 0 K?
Dividing by zero temperature is mathematically undefined in this equation, and absolute zero is also physically unreachable, so it isn’t a meaningful input in either sense.
Does this account for real gas behaviour, like the ideal gas law calculator’s own caveat?
No — the same “ideal” assumptions apply here too. This is most accurate for ordinary pressures and temperatures, and least accurate near a gas’s liquefaction point.
