Calculations related to Clapeyron's equation known also as ideal gas law. Enter known values (e.g. pressure and temperature) and select which value you want to find out (e.g. volume) and we'll show you step-by-step how to transform basic formula and reach your result in desired units.

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Symbolic algebra

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What do you want to calculate today?#

Choose a scenario that best fits your needs

I know number of moles (n), temperature (T) and volume (V) and want to calculate pressure (p)I know number of moles (n), temperature (T) and pressure (p) and want to calculate volume (V)I know pressure (p), volume (V) and temperature (T) and want to calculate number of moles (n)I know pressure (p), volume (V) and number of moles (n) and want to calculate temperature (T)

Calculations data - enter values, that you know here#

Pressure (p)		<=
Volume (V)		=>
Number of moles (n)		<=
Temperature (T)		<=

Units normalization#

Number of moles (n)	Show source $1\ \left[mol\right]$
Temperature (T)	Show source $0\ \left[^\circ C\right]\ =\ \frac{5463}{20}\ \left[K\right]$
Volume (V)
Pressure (p)	Show source $1013.25\ \left[hPa\right]\ =\ 101325\ \left[Pa\right]$

Result: Volume (V)#

Summary

Used formula

Show source

\mathrm{V}=\frac{n \cdot R \cdot \mathrm{T}}{p}

Result

Show source

\frac{1821}{675500}~R

Numerical result

Show source

22.41396207863582531458179126572908956328645447816432272390821614\ \left[dm^3\right]

Result step by step

1	Show source $\frac{1~R \cdot \frac{5463}{20}}{101325}$	Multiply by one	Any number multiplied by one (1) gives the same number: $a \cdot 1 = 1 \cdot a = a$
2	Show source $\frac{R \cdot \frac{5463}{20}}{101325}$	Removed double fractional mark	Divide by fraction is the same as multiply by fraction inverse: $\frac{a}{\frac{c}{b}} = a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$
3	Show source $\frac{R \cdot 5463}{101325 \cdot 20}$	Simplify arithmetic	-
4	Show source $\frac{R \cdot \cancel{5463}}{\cancel{2026500}}$	Cancel terms or fractions	Dividing a number by itself gives one, colloquially we say that such numbers "cancel-out": $\frac{\cancel{a}}{\cancel{a}} = 1$ to find-out the simplest form of fraction we can divide the numerator and denominator by the greatest common divisor (GCD) of both numbers.
5	Show source $\frac{1}{675500} \cdot R \cdot 1821$	Multipled fractions	To multiply two fractions we need to multiply numberators and denominators from first and second fractions: $\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$
6	Show source $\frac{1 \cdot 1821}{675500} \cdot R$	Multipled fractions	To multiply two fractions we need to multiply numberators and denominators from first and second fractions: $\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$
7	Show source $\frac{1 \cdot 1821~R}{675500}$	Multiply by one	Any number multiplied by one (1) gives the same number: $a \cdot 1 = 1 \cdot a = a$
8	Show source $\frac{1821~R}{675500}$	Rearrange coefficients	Multiplication and addition are commutative. Due to this property, we can change arguments order to get better readability of expression.
9	Show source $\frac{1821}{675500}~R$	Result	Your expression reduced to the simplest form known to us.

Numerical result step by step

1	Show source $\frac{1821}{675500} \cdot 8.3144598485$	Multipled fractions	To multiply two fractions we need to multiply numberators and denominators from first and second fractions: $\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$
2	Show source $\frac{1821 \cdot 8.3144598485}{675500}$	Simplify arithmetic	-
3	Show source $\frac{15140.6313841185}{675500}$	Divided fraction	-
4	Show source $0.02241396207863582531458179126572908956328645447816432272390821614$	Result	Your expression reduced to the simplest form known to us.

Units normalization

Show source

0.02241396207863582531458179126572908956328645447816432272390821614\ \left[m^3\right]\ =\ 22.41396207863582531458179126572908956328645447816432272390821614\ \left[dm^3\right]

Some facts#

The perfect gas (also known as ideal gas) is a hypothetical, simplified model approximating the behavior of real gases. A perfect gas is different from the real one, in that its molecules do not interact with each other.
More formally, we say that the perfect gas does not take intermolecular interactions into account.
The ideal gas law was first formulated in 1834 by Benoîta Clapeyron. For this reason, it is also known as the Clapeyron equation.
The ideal gas state equation is usually written in the following form:

$pV = nRT$
where:
- p - pressure in the system,
- V - volume occupied by gas,
- n - number of gas moles in the system,
- R - gas constant,
- T - absolute temperature in the system.
The Clapeyron equation was originally a generalization (synthesis) of the then known empirical laws describing in a rough way the behavior of gases:
- the Boyls law - the gas pressure is inversely proportional to the volume:
  
  $p \propto \dfrac{1}{V}$
- the Charles law - the volume of gas is directly proportional to the temperature:
  
  $V \propto T$
- the Avogadro law - the volume of gas is directly proportional to the number of moles of gas in the vessel:
  
  $V \propto n$
- the Gay-Lussac law - the gas pressure is directly proportional to the temperature:
  
  $p \propto T$

Tags and links to this website#

Tags:

clapeyrons_equation · perfect_gas_equation · ideal_gas_law · general_gas_equation

Tags to Polish version:

rownanie_clapeyrona · rownanie_stanu_doskonalego · prawo_gazu_doskonalego · prawo_gazu_idealnego

What tags this calculator has#

CALCULATOR · CHEMISTRY · PHYSICS · A -> Z (all)

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https://calculla.com/clapeyrons_equation?ioNumberOfMoles=$symbolic(1)&ioTemperature=$symbolic(0)&ioPressure=$symbolic(1013.25)&ioPressureUnitId=hPa&ioVolumeUnitId=dm3&ioTemperatureUnitId=C&ioNumberOfMolesUnitId=mol&ioIntentSchemeId=want_ioVolume

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