
How do you calculate molar volume of oxygen gas?
Answer
526.5k+ views
Hint: At a given temperature and pressure, the volume occupied by one mole of oxygen gas can be calculated by using the ideal gas equation. This equation has five variables, therefore, if we know the values of any four variables then the value of the fifth variable can be easily calculated.
Formula used:
\[pV = nRT\]
Complete step by step solution:
We will calculate the volume (in litres) occupied by one mole of oxygen gas at standard temperature and pressure (S.T.P.). S.T.P. is defined as temperature, \[T = 0^\circ {\text{ C}}\] and pressure, \[p = 1{\text{ bar}}\].
The ideal gas equation is, \[pV = nRT\]
To calculate volume \[V\] in litres, following units must be followed:
unit of pressure \[p\]- atmosphere,
unit of number of moles \[n\]- mol,
unit of universal gas constant \[R\]- \[{\text{L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}\],
unit of temperature \[T\]- Kelvin
In the ideal gas equation,
Pressure \[p = 1{\text{ bar}}\]and we know that \[1{\text{ bar}} = {10^5}{\text{ Pa}}\]
Also, \[1.01325 \times {10^5}{\text{ Pa}} = 1{\text{ atm}}\]
\[ \Rightarrow {10^5}{\text{ Pa}} = \dfrac{1}{{1.01325}}{\text{ atm}} = 0.987{\text{ atm}}\]
\[ \Rightarrow 1{\text{ bar}} = 0.987{\text{ atm}}\]
\[ \Rightarrow p = 0.987{\text{ atm}}\]
Number of moles, \[n = 1{\text{ mol}}\]
Universal gas constant, \[R = 0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}\]
Temperature, \[T = 0^\circ {\text{ C}} = 273.15{\text{ K}}\]
Volume, \[V = ?\]
From the ideal gas equation, \[V = \dfrac{{nRT}}{p}\]
Substituting the values of known variables in this equation:
\[V = \dfrac{{1 \times 0.0821 \times 273.15}}{{0.987}}{\text{ L}}\]
Simplifying the numerator, we get,
\[V = \dfrac{{22.426}}{{0.987}}{\text{ L}}\]
Solving the above expression, we get,
\[V = 22.721{\text{ L}}\]
Now, approximating the above value up to two values after decimal, we get,
\[V = 22.72{\text{ L}}\]
Hence, the molar volume of oxygen gas at S.T.P. is \[22.72{\text{ L}}\]
Note: The definition of S.T.P. was changed in \[1982\]. Before that, S.T.P. was defined as \[T = 0^\circ {\text{ C}}\] and pressure, \[p = 1{\text{ atm}}\].
The value of the universal gas constant , \[R\] should be taken carefully. If all the other units are in S.I., then \[R = 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}\]. In this case, we will get the required volume in \[{{\text{m}}^3}\].
Formula used:
\[pV = nRT\]
Complete step by step solution:
We will calculate the volume (in litres) occupied by one mole of oxygen gas at standard temperature and pressure (S.T.P.). S.T.P. is defined as temperature, \[T = 0^\circ {\text{ C}}\] and pressure, \[p = 1{\text{ bar}}\].
The ideal gas equation is, \[pV = nRT\]
To calculate volume \[V\] in litres, following units must be followed:
unit of pressure \[p\]- atmosphere,
unit of number of moles \[n\]- mol,
unit of universal gas constant \[R\]- \[{\text{L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}\],
unit of temperature \[T\]- Kelvin
In the ideal gas equation,
Pressure \[p = 1{\text{ bar}}\]and we know that \[1{\text{ bar}} = {10^5}{\text{ Pa}}\]
Also, \[1.01325 \times {10^5}{\text{ Pa}} = 1{\text{ atm}}\]
\[ \Rightarrow {10^5}{\text{ Pa}} = \dfrac{1}{{1.01325}}{\text{ atm}} = 0.987{\text{ atm}}\]
\[ \Rightarrow 1{\text{ bar}} = 0.987{\text{ atm}}\]
\[ \Rightarrow p = 0.987{\text{ atm}}\]
Number of moles, \[n = 1{\text{ mol}}\]
Universal gas constant, \[R = 0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}\]
Temperature, \[T = 0^\circ {\text{ C}} = 273.15{\text{ K}}\]
Volume, \[V = ?\]
From the ideal gas equation, \[V = \dfrac{{nRT}}{p}\]
Substituting the values of known variables in this equation:
\[V = \dfrac{{1 \times 0.0821 \times 273.15}}{{0.987}}{\text{ L}}\]
Simplifying the numerator, we get,
\[V = \dfrac{{22.426}}{{0.987}}{\text{ L}}\]
Solving the above expression, we get,
\[V = 22.721{\text{ L}}\]
Now, approximating the above value up to two values after decimal, we get,
\[V = 22.72{\text{ L}}\]
Hence, the molar volume of oxygen gas at S.T.P. is \[22.72{\text{ L}}\]
Note: The definition of S.T.P. was changed in \[1982\]. Before that, S.T.P. was defined as \[T = 0^\circ {\text{ C}}\] and pressure, \[p = 1{\text{ atm}}\].
The value of the universal gas constant , \[R\] should be taken carefully. If all the other units are in S.I., then \[R = 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}\]. In this case, we will get the required volume in \[{{\text{m}}^3}\].
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