Question 903: Thermodynamics & KTG

Question

The equation of state for 2 g of oxygen at a pressure ' $P$ ' and temperature ' $T$, when occupying a volume ' $V$ ' will be

Accepted Answer

The equation of state for a gas can be described using the ideal gas law:
$ pV = \mu RT $
where:
$ p $ is the pressure,
$ V $ is the volume,
$ \mu $ is the number of moles of the gas,
$ R $ is the ideal gas constant, and
$ T $ is the temperature.
The number of moles ($ \mu $) is calculated using the formula:
$ \mu = \frac{	ext{Given mass}}{	ext{Molecular mass}} = \frac{m}{M} $
For this specific problem:
The given mass of oxygen ($ m $) is 2 g.
The molecular mass ($ M $) of oxygen ($\mathrm{O}_2$) is 32 g/mol.
Substituting these values into the formula, we have:
$ \mu = \frac{2 \, 	ext{g}}{32 \, 	ext{g/mol}} = \frac{1}{16} \, 	ext{mol} $
Now, substituting $\mu$ back into the ideal gas equation:
$ pV = \frac{1}{16} RT $
Therefore, the equation of state for 2 g of oxygen under the given conditions is:
$ pV = \frac{1}{16} RT $

Answer

$p V=16 R T$

Answer

$p V=R T$

Answer

$p V=\frac{1}{16} R T$

Answer

$p V=2 R T$

Thermodynamics & KTG

The equation of state for 2 g of oxygen at a pressure ' P ' and temperature ' T, when occupying a volume ' V ' will be

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