Ambient pressure, temperature, and relative humidity at a location are 101 kPa, 300 K, and 60%, respectively. The saturation pressure of water at 300 K is 3.6 kPa. The specific humidity of ambient air is _________ g/kg of dry air.

The correct option is 13.6.

Analysis of the Given Data:
Based on the question details and the step-by-step solution shown in the provided image, we have the following parameters for the ambient air:
• Total ambient pressure (or barometric pressure), $P=101\text{ kPa}$
• Dry-bulb temperature of the air, $T=300\text{ K}$
• Relative humidity, $\varphi =60\%=0.6$
• Saturation pressure of water vapor at 300 K, $P_{vs}=3.6\text{ kPa}$

Step 1: Calculate the Partial Pressure of Water Vapor ($p_v$)
The relative humidity ($\varphi$) is defined as the ratio of the partial pressure of water vapor ($p_v$) to the saturation pressure of water vapor ($p_{vs}$) at the same temperature:
$$\varphi =\frac{p_v}{p_{vs}}$$
Rearranging the formula to find the partial pressure of water vapor:
$$p_v=\varphi \times p_{vs}$$
Substituting the given values:
$$p_v=0.6\times 3.6\text{ kPa}=2.16\text{ kPa}$$

Step 2: Calculate the Specific Humidity ($\omega$)
Specific humidity (or humidity ratio, $\omega$) is the ratio of the mass of water vapor to the mass of dry air in a given volume. Using the ideal gas equation and molecular weights of water vapor (approx. 18 g/mol) and dry air (approx. 28.97 g/mol), specific humidity is calculated as:
$$\omega =0.622\times \frac{p_v}{P-p_v}$$
Now substitute the values of total pressure P and partial vapor pressure $p_v$:
$$\omega =0.622\times \frac{2.16}{101-2.16}$$
$$\omega =0.622\times \frac{2.16}{98.84}$$
$$\omega \approx 0.013593\text{ kg of water vapor / kg of dry air}$$

Step 3: Convert the Specific Humidity to g/kg of Dry Air
To convert specific humidity from kg/kg to g/kg, multiply the value by 1000:
$$\omega =0.013593\times 1000\text{ g/kg of dry air}\approx 13.6\text{ g/kg of dry air}$$
Thus, the specific humidity of the ambient air is 13.6 g/kg of dry air.