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1. For dry air at 1.0000 atm pressure, the densities at –50°C, 0°C, and 69°C are 1.5826 g dm–3 , 1.2929 g dm–3, and 1.0322 g dm–3, respectively. a) Assume a sample of mass 1000 g, and calculate the volume at each temperature. b) From these data, and assuming that air obeys Charles’s law, determine a value for th

Answer :

Answer :

(a) The value of volume of air at temperature [tex]-50^oC,0^oC\text{ and }69^oC[/tex] are [tex]631.87dm^3,773.46dm^3\text{ and }968.80dm^3[/tex] respectively.

Explanation :

(a) The formula used for density is:

[tex]Density=\frac{Mass}{Volume}[/tex]

or,

[tex]Volume=\frac{Mass}{Density}[/tex]

Now we have to calculate the volume at each temperature.

The mass of sample = 1000 g

At temperature [tex]-50^oC[/tex] :

Density = [tex]1.5826g/dm^3[/tex]

[tex]Volume=\frac{1000g}{1.5826g/dm^3}=631.87dm^3[/tex]

At temperature [tex]0^oC[/tex] :

Density = [tex]1.2929g/dm^3[/tex]

[tex]Volume=\frac{1000g}{1.2929g/dm^3}=773.46dm^3[/tex]

At temperature [tex]69^oC[/tex] :

Density = [tex]1.0322g/dm^3[/tex]

[tex]Volume=\frac{1000g}{1.0322g/dm^3}=968.80dm^3[/tex]

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