Question

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 1

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

Explanation :

(a) The formula used for density is:

`Density=(Mass)/(Volume)`

or,

`Volume=(Mass)/(Density)`

Now we have to calculate the volume at each temperature.

The mass of sample = 1000 g

At temperature
`-50^oC` :

Density =
`1.5826g/dm^3`

`Volume=(1000g)/(1.5826g/dm^3)=631.87dm^3`

At temperature
`0^oC` :

Density =
`1.2929g/dm^3`

`Volume=(1000g)/(1.2929g/dm^3)=773.46dm^3`

At temperature
`69^oC` :

Density =
`1.0322g/dm^3`

`Volume=(1000g)/(1.0322g/dm^3)=968.80dm^3`