Molecular Weight
1,000 meters above the surface of the Earth, the local speed of sound in -1°C atmospheric air is 344 m/s. If the ratio of specific heats is 1.4, what is the molecular weight of air in kg/kmol? Note the universal gas constant is 8,314 J/(kmol·K).
Expand Hint
The local speed of sound in an ideal gas is:
$$$c=\sqrt{kRT}$$$
where
$$k$$
is the ratio of specific heats,
$$R$$
is the specific gas constant, and
$$T$$
is the absolute temperature.
Hint 2
The specific gas constant is:
$$$R=\frac{\bar{R}}{(mol.wt_i)}$$$
where
$$\bar R$$
is the universal gas constant.
To find the local speed of sound in an ideal gas:
$$$c=\sqrt{kRT}$$$
where
$$k$$
is the ratio of specific heats,
$$R$$
is the specific gas constant, and
$$T$$
is the absolute temperature.
$$$344\frac{m}{s}=\sqrt{(1.4)R(-1+273K)}$$$
$$$(344m/s)^2=(1.4)R(272K)$$$
$$$R=\frac{118,336m^2}{380.8K\cdot s^2}=310.756\:\frac{m^2}{s^2 \cdot K}$$$
Recall that the specific gas constant is:
$$$R=\frac{\bar{R}}{(mol.wt_i)}$$$
where
$$\bar R$$
is the universal gas constant.
$$$310.756\frac{m^2}{s^2 \cdot K}=\frac{8,314J}{(kmol\cdot K)(mol.wt_i)}$$$
Remember the base SI units for a joule is
$$kg\cdot m^2/s^2$$
. Thus,
$$$310.756=\frac{8,314kg}{(kmol)(mol.wt_i)}$$$
$$$mol.wt_i=\frac{8,314kg}{(kmol)(310.756)}=26.75\:\frac{kg}{kmol}$$$
26.75 kg/kmol
Time Analysis
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