Charles's Law Calculator

Charles's Law is evaluated from Initial Volume, Initial Temperature and Final Volume. The calculation reports Initial Volume, Initial Temperature and Initial Temperature.

Results

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About the Charles's Law Calculator

Charles's Law is treated here as a quantitative relation between Initial Volume, Initial Temperature, Final Volume and Final Temperature and Initial Volume, Initial Temperature, Initial Temperature and Final Volume.

The calculator uses a multi formula configuration. Each reported value is read as a direct evaluation of the stored rules with the declared field formats and units.

Formula basis:
V₁/T₁ = V₂/T₂ (T must be in Kelvin)
V₂ = V₁ x T₂ / T₁
T₂ = T₁ x V₂ / V₁
V₁ = V₂ x T₁ / T₂
T₁ = T₂ x V₁ / V₂
Convert: T(K) = T( degC) + 273.15 = (T( degF) - 32) x 5/9 + 273.15

Interpret the outputs in the order shown by the result fields. Optional inputs affect only the outputs that depend on those variables.

Formula & How It Works

The calculation applies the following relations exactly as recorded in the metadata:

V₁/T₁ = V₂/T₂ (T must be in Kelvin)
V₂ = V₁ x T₂ / T₁
T₂ = T₁ x V₂ / V₁
V₁ = V₂ x T₁ / T₂
T₁ = T₂ x V₁ / V₂
Convert: T(K) = T( degC) + 273.15 = (T( degF) - 32) x 5/9 + 273.15

Each output field is produced by substituting the supplied inputs into the relevant relation and then applying the declared rounding or text format.

Worked Examples

Example 1: Hot Air Balloon — Heating Air for Lift

Inputs

v1: 2800 t1_c: 15 t2_c: 120

Initial Volume: 2,800 L. Initial Temperature: 15 degC. Initial Temperature: 59 degF. Final Volume: 3,820.3019 L. Final Temperature: 120 degC. Final Temperature: 248 degF. Volume Change: 36.44%

With Initial Volume = 2,800, Initial Temperature = 15 and Final Temperature = 120 as the stated inputs, the result is Initial Volume = 2,800 L, Initial Temperature = 15 degC and Initial Temperature = 59 degF. Each value corresponds to the declared output fields.

Example 2: Basketball Inflation — Cold Weather Drop

Inputs

v1: 7.48 t1_c: 22 t2_c: -10

Initial Volume: 7.48 L. Initial Temperature: 22 degC. Initial Temperature: 71.6 degF. Final Volume: 6.669 L. Final Temperature: -10 degC. Final Temperature: 14 degF. Volume Change: -10.84%

With Initial Volume = 7.48, Initial Temperature = 22 and Final Temperature = -10 as the stated inputs, the result is Initial Volume = 7.48 L, Initial Temperature = 22 degC and Initial Temperature = 71.6 degF. Each value corresponds to the declared output fields.

Example 3: Snack Bag on Airplane

Inputs

v1: 0.5 t1_c: 22 t2_c: 22

Initial Volume: 0.5 L. Initial Temperature: 22 degC. Initial Temperature: 71.6 degF. Final Volume: 0.5 L. Final Temperature: 22 degC. Final Temperature: 71.6 degF. Volume Change: 0%

With Initial Volume = 0.5, Initial Temperature = 22 and Final Temperature = 22 as the stated inputs, the result is Initial Volume = 0.5 L, Initial Temperature = 22 degC and Initial Temperature = 71.6 degF. Each value corresponds to the declared output fields.

Example 4: Bread Dough Rising in Oven

Inputs

v1: 0.5 t1_c: 25 t2_c: 190

Initial Volume: 0.5 L. Initial Temperature: 25 degC. Initial Temperature: 77 degF. Final Volume: 0.7767 L. Final Temperature: 190 degC. Final Temperature: 374 degF. Volume Change: 55.34%

With Initial Volume = 0.5, Initial Temperature = 25 and Final Temperature = 190 as the stated inputs, the result is Initial Volume = 0.5 L, Initial Temperature = 25 degC and Initial Temperature = 77 degF. Each value corresponds to the declared output fields.

Common Use Cases

Find new volume of a balloon when temperature changes
Calculate temperature needed to reach a target volume
Understand why hot air balloons rise