Combined Gas Law Calculator

Solve Combined Gas Law Problems

Combined Gas Law Formula

Formula

\[ \frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f} \]

Where:

\(P_i\) and \(P_f\) — initial and final absolute pressure
\(V_i\) and \(V_f\) — initial and final volume
\(T_i\) and \(T_f\) — initial and final temperature in kelvins (K)

The combined gas law relates two states of the same amount of gas. Rearrange the equation to find any one of the six values when the other five are known.

Combined Gas Law – Calculation Example

Example: With the calculator’s default values, find the initial volume when a gas starts at 100 kPa and 273 K, then reaches a volume of 1 L at 100 kPa and 293 K.

\[ V_i = \frac{P_f V_f T_i}{P_i T_f} = \frac{(100\,\mathrm{kPa})(1\,\mathrm{L})(273\,\mathrm{K})}{(100\,\mathrm{kPa})(293\,\mathrm{K})} \approx 0.9317\,\mathrm{L} \]

Result: The initial volume is 0.9317 L. At unchanged pressure, the gas occupies more space at the higher final temperature.

How to use the calculator

In I want to calculate, select the unknown: Initial Volume (Vᵢ), Initial Pressure (Pᵢ), Initial Temperature (Tᵢ), Final Volume (Vf), Final Pressure (Pf), or Final Temperature (Tf). The calculator hides the field you selected.
Enter the five known values in the remaining initial and final fields. Use temperatures in K; the temperature fields do not accept Celsius or Fahrenheit as units.
Choose L or mL for each volume field and kPa, atm, or mmHg for each pressure field. If you enter both volumes, give them the same volume unit: this calculator does not convert between L and mL in its calculation. It does convert pressure units internally.
Read the calculated value in the result. A calculated pressure is displayed in kPa; a calculated volume uses the unit selected for the known volume.

What the combined gas law tells you

The combined gas law describes how pressure, volume, and absolute temperature change together for a fixed amount of gas. It states that \(PV/T\) remains constant between an initial and a final state, provided the gas can be treated as ideal. Knowing five of the six initial and final values lets you determine the sixth.

The relationship brings together two simpler patterns. At constant temperature, pressure and volume vary inversely, as described by Boyle’s law: compressing a gas tends to raise its pressure. At constant pressure, volume is proportional to absolute temperature, as described by Charles’ law. The combined gas law handles changes in both pressure and temperature at once.

Interpret the result alongside the conditions you entered. In the example, pressure stays at 100 kPa while temperature rises from 273 K to 293 K. The calculated initial volume is smaller than the known final volume, which fits the expectation that a warmer gas occupies more space if its pressure can remain constant. The law compares two states; it does not describe how quickly the gas changes between them.

If you also need to calculate the amount of gas or relate a single state to its pressure, volume, and temperature, use the Ideal Gas Law Calculator. The combined gas law does not require the amount of gas as an input because it assumes that amount is unchanged.

Assumptions and practical tips

Keep the amount of gas constant. Do not use the combined gas law unchanged if gas enters, escapes, or reacts between the two states.
Use absolute temperature. Enter kelvins, not degrees Celsius. To convert a Celsius temperature to kelvins, add 273.15. Temperatures must be greater than 0 K for a physically meaningful calculation.
Use absolute pressure. A gauge pressure measures pressure relative to its surroundings and cannot be entered directly as though it were absolute pressure. Convert it to absolute pressure first.
Match volume units. Select L for both entered volumes or mL for both. For example, convert 500 mL to 0.5 L before comparing it with a volume entered in L.
Allow for real-gas behavior. The equation is an ideal-gas approximation. Results may be less reliable at high pressures, very low temperatures, or near condensation, where real gases can depart substantially from ideal behavior.

Frequently Asked Questions

How can I solve a combined gas law problem by hand?

Start by identifying the unknown, then cross-multiply and isolate it before substituting numbers. For example, to find final pressure, use \(P_f=P_iV_iT_f/(T_iV_f)\). If a gas starts at 100 kPa and 2 L, then ends at 1 L with the same temperature, the final pressure is 200 kPa. Keeping the units alongside each number helps reveal input errors.

What is a normal pressure, volume, or temperature for a combined gas law problem?

There is no single normal value: it depends on the gas and its container. Atmospheric pressure near sea level is approximately 101.3 kPa (1 atm), but a sealed container may have a very different pressure. A useful check is whether your result fits the situation—for example, halving a gas’s volume at unchanged temperature should double its absolute pressure.

How do I use a temperature given in Fahrenheit?

Convert Fahrenheit to kelvins before using the law: \(T(\mathrm K)=(T(°\mathrm F)-32)\times5/9+273.15\). For example, 68°F is 293.15 K. Do not use a Fahrenheit or Celsius number directly in a temperature ratio: both scales can read zero or below even though the gas still has a positive absolute temperature.

How do I convert gauge pressure to absolute pressure?

Add the surrounding atmospheric pressure to the gauge reading, using the same units for both. For example, a gauge reading of 200 kPa where atmospheric pressure is 101 kPa corresponds to about 301 kPa absolute. A negative gauge reading can still represent positive absolute pressure. If atmospheric pressure is unknown, assuming 101.3 kPa may be unsuitable for precise work.

What happens if the gas is in a rigid container?

A rigid container has the same initial and final volume, so the volume terms cancel. Pressure is then proportional to absolute temperature: \(P_f/P_i=T_f/T_i\). For instance, raising a gas from 300 K to 330 K in a sealed, rigid container predicts a 10% increase in absolute pressure, provided the amount of gas stays fixed and ideal-gas behavior is a reasonable approximation.

Can I enter one volume in liters and the other in milliliters?

Convert the volumes to the same unit first when both are known. The calculation uses their numerical ratio, so entering 1 L and 500 mL as 1 and 500 would give a ratio 1,000 times too large or small. Express them as 1 L and 0.5 L, or as 1,000 mL and 500 mL. Pressure inputs, unlike volume inputs, may use different supported units.

Why did I get a zero, negative, or unexpectedly large result?

Check for a zero denominator, a nonpositive temperature, mismatched volume units, or gauge pressure entered as absolute pressure. A physical gas state in this model requires positive absolute temperature, pressure, and volume. An unexpectedly large answer can also be mathematically consistent—for example, a very small final volume predicts high pressure if temperature does not fall enough to offset the compression.