The density of O3 gas which has pressure 730 mm at 50°C is

To find the density of O3 gas under the given conditions, we can use the ideal gas law equation, which is:

PV = nRT

Where: P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the given conditions to the appropriate units.

Pressure (P) = 730 mmHg. We need to convert this to atmospheres because the value of R we will be using is in atmospheres. 1 atm = 760 mmHg, so P = 730/760 = 0.96 atm.

Temperature (T) = 50°C. We need to convert this to Kelvin by adding 273.15. So, T = 50 + 273.15 = 323.15 K.

We are looking for the density, which is mass/volume. We can express mass as the number of moles (n) times the molar mass (M). So, the density (D) = nM/V.

We can express n/V from the ideal gas law as P/RT, so the density equation becomes D = PM/RT.

The molar mass (M) of O3 (ozone) is approximately 48 g/mol.

The ideal gas constant (R) we will use is 0.0821 L·atm/(K·mol).

Substituting these values into the density equation gives:

D = (0.96 atm * 48 g/mol) / (0.0821 L·atm/(K·mol) * 323.15 K) = 1.79 g/L.

So, the density of O3 gas under the given conditions is approximately 1.79 g/L.