The kirematic viscosity of oxygen at \(20^{\circ} \mathrm{C}\) ard a pressure of \(150 \mathrm{kPa}(\mathrm{abs})\) is 0.104 stokes. Determine the dynamic viscosity of oxygen at this temperature and pressure.

The standard conditions for gas density refer to the air density at sea level in an average temperature of \(20^{\circ} \mathrm{C}\) (293.15 K). The value is \(1.2754 \frac{kg}{m^3}\). However, these conditions consider the pressure of \(1 \mathrm{atm}\) or \(101.325 \mathrm{kPa}\), while in our case the pressure is \(150 \mathrm{kPa}\). The density of a gas is directly proportional to its pressure, so in order to obtain the density of oxygen under these conditions, we can use the following proportion: \(\frac{P1}{P2} = \frac{\rho1}{\rho2}\). Given that \(P1 = 150 \mathrm{kPa}\), \(P2 = 101.325 \mathrm{kPa}\), and \(\rho1\) is what we want to find, while \(\rho2 = 1.2754 \frac{kg}{m^3}\), we then substitute these values into the equation.

Substituting the values into the equation, we have: \(\frac{150}{101.325} = \frac{\rho1}{1.2754}\). Solving for \(\rho1\), we get \(\rho1 = \frac{150}{101.325} * 1.2754 = \approx1.889 \frac{kg}{m^3}\). So, the density of oxygen under these pressure conditions is approximately \(1.889 \frac{kg}{m^3}\).

The given kinematic viscosity is 0.104 stokes, which needs to be converted to \(\frac{m^{2}}{s}\). As \(1 \, stokes = 10^{-4} \, \frac{m^{2}}{s}\), the velocity in SI units is \(0.104 * 10^{-4} \, \frac{m^{2}}{s}\). Kinematic viscosity \(\nu\) is the ratio of dynamic viscosity \(\mu\) to the fluid's density \(\rho\). Therefore, solving for dynamic viscosity in the equation \(\nu = \frac{\mu}{\rho}\), we get \(\mu = \nu * \rho\). Substituting the values, we find: \(\mu = 0.104 * 10^{-4} \, \frac{m^{2}}{s} * 1.889 \, \frac{kg}{m^{3}} = \approx1.965 * 10^{-5} \, \frac{kg}{m*s}\)