This is a suggested solution to GCE O Level Pure Physics examination in year 2012 November, Paper 2, section A, question 4a.

## Suggested Solution:

Given the density of air in the cylinder is 0.0012 g/cm^{3}.

Hence, the mass of air is=0.0012 g/cm^{3} x 30 cm^{3 }=0.036 g

Therefore, the density of cold air = 0.036 g /(20 cm^{3 })=0.0018 g/cm^{3}

## Explanation:

In this question, it was stated that when the piston is placed in room temperature, the air in the cylinder was as follows:

And in cold freezer, the piston showed that the volume of air has reduced to:

From the above diagram, the volume of air at room temperature is 30 cm^{3} and the volume of air in cold freezer is 20 cm^{3}. Candidates are required to calculate the density of cold air.

Since the question also stated that the density of air in the cylinder at room temperature is 0.0012 g/cm^{3}, we can calculate the mass of the air at room temperature using the volume of air at room temperature and the density of air at room temperature with the following formula:

Note that the mass of air at room temperature is the same as mass of air in cold freezer because mass does not change. Once we can get the mass of air, we can calculate the density of cold air with the volume of cold air as 20 cm^{3}.