Question

The air in a hot-air balloon at 763 torr is heated from 14.0°C to 31.0°C. Assuming that the moles of air and the pressure remain constant, what is the density of the air at each temperature? (The average molar mass of air is 29.0 g/mol.)

Answer 1

Answer:

1.237g/L = density at 14°C

1.167g/L = density at 31°C

Explanation:

Density is the ratio between mass and volume of a determined substance

From ideal gas law:

PV = nRT:

Where P is pressure, V volume, n moles, R constant gas law, and T absolute temperature

P/RT = n/V

You can obtain moles / L. now:

moles / L × (molar mass of the gas (g/mol)

g/L

You can obtain the density of the gas in g/L

763torr are:

763torr × (1atm / 760torr) = 1.004 atm

14°C and 31.0°C in absolute temperature are:

14°C + 273.15 = 287.15K

31°C + 273.15 = 304.15K

Replacing, density at 14°C:

1.004atm / (0.082atmL/molK×287.15K) × 29g/mol = density

1.237g/L = density at 14°C

And at 31.0°C:

1.004atm / (0.082atmL/molK×304.15K) × 29g/mol = density

1.167g/L = density at 31°C