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Question: How do you calculate molar volume of oxygen gas?...

How do you calculate molar volume of oxygen gas?

Explanation

Solution

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=nRTpV = 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 CT = 0^\circ {\text{ C}} and pressure, p=1 barp = 1{\text{ bar}}.
The ideal gas equation is, pV=nRTpV = nRT
To calculate volume VV in litres, following units must be followed:
unit of pressure pp- atmosphere,
unit of number of moles nn- mol,
unit of universal gas constant RR- L atm mol1 K1{\text{L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}},
unit of temperature TT- Kelvin
In the ideal gas equation,
Pressure p=1 barp = 1{\text{ bar}}and we know that 1 bar=105 Pa1{\text{ bar}} = {10^5}{\text{ Pa}}
Also, 1.01325×105 Pa=1 atm1.01325 \times {10^5}{\text{ Pa}} = 1{\text{ atm}}
105 Pa=11.01325 atm=0.987 atm\Rightarrow {10^5}{\text{ Pa}} = \dfrac{1}{{1.01325}}{\text{ atm}} = 0.987{\text{ atm}}
1 bar=0.987 atm\Rightarrow 1{\text{ bar}} = 0.987{\text{ atm}}
p=0.987 atm\Rightarrow p = 0.987{\text{ atm}}
Number of moles, n=1 moln = 1{\text{ mol}}
Universal gas constant, R=0.0821 L atm mol1 K1R = 0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}
Temperature, T=0 C=273.15 KT = 0^\circ {\text{ C}} = 273.15{\text{ K}}
Volume, V=?V = ?
From the ideal gas equation, V=nRTpV = \dfrac{{nRT}}{p}
Substituting the values of known variables in this equation:
V=1×0.0821×273.150.987 LV = \dfrac{{1 \times 0.0821 \times 273.15}}{{0.987}}{\text{ L}}
Simplifying the numerator, we get,
V=22.4260.987 LV = \dfrac{{22.426}}{{0.987}}{\text{ L}}
Solving the above expression, we get,
V=22.721 LV = 22.721{\text{ L}}
Now, approximating the above value up to two values after decimal, we get,
V=22.72 LV = 22.72{\text{ L}}

Hence, the molar volume of oxygen gas at S.T.P. is 22.72 L22.72{\text{ L}}

Note: The definition of S.T.P. was changed in 19821982. Before that, S.T.P. was defined as T=0 CT = 0^\circ {\text{ C}} and pressure, p=1 atmp = 1{\text{ atm}}.
The value of the universal gas constant , RR should be taken carefully. If all the other units are in S.I., then R=8.314 J K1 mol1R = 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}. In this case, we will get the required volume in m3{{\text{m}}^3}.