Given information:
- Volume of air = 300 m³
- Initial dry bulb temperature = 30°C
- Initial wet bulb temperature = 25°C
- Final dry bulb temperature = 40°C
- Air pressure = 1.01325 bar

To solve this problem, we need to use the following formula:

Q = m × cp × Δt

Where Q is the amount of heat added, m is the mass of the air, cp is the specific heat of air at constant pressure, and Δt is the change in temperature.

Calculation of amount of heat added:
- First, we need to calculate the mass of air using the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
- Converting the given temperatures to Kelvin:
Initial dry bulb temperature = 30 + 273.15 = 303.15 K
Initial wet bulb temperature = 25 + 273.15 = 298.15 K
Final dry bulb temperature = 40 + 273.15 = 313.15 K
- Assuming the air to be dry, we can calculate the number of moles using the ideal gas law:
n = PV/RT
n = (1.01325 × 10⁵ Pa) × (300 m³) / [(8.314 J/mol.K) × (303.15 K)]
n = 36.26 moles
- Using the molar mass of air (28.97 g/mol), we can calculate the mass of air:
m = n × M
m = 36.26 moles × 28.97 g/mol
m = 1.05 kg
- Finally, we can calculate the amount of heat added:
Q = m × cp × Δt
cp of air at constant pressure is approximately 1005 J/kg.K
Q = (1.05 kg) × (1005 J/kg.K) × (40 - 30) K
Q = 10,525.5 J or 10.5 kJ (rounded to one decimal place)

Calculation of relative humidity:
- Relative humidity is the ratio of the actual amount of water vapor present in the air to the maximum amount of water vapor that can be present in the air at a given temperature and pressure.
- We can use the psychrometric chart to find the relative humidity.
- First, we need to find the properties of the air at the initial condition (point A) and the final condition (point B) on the chart.
- At point A, the dry bulb temperature is 30°C and the wet bulb temperature is 25°C. Following the lines on the chart, we can find that the humidity ratio (mass of water vapor per unit mass of dry air) is approximately 0.0172 kg/kg.
- At point B, the dry bulb temperature is 40°C. Following the lines on the chart, we can find that the humidity ratio is approximately 0.0117 kg/kg.
- The maximum amount of water vapor that can be present in the air at point B can be found by extrapolating the line from point A to the saturation curve (the curve where the air is fully saturated with water vapor). This gives a humidity ratio of approximately 0.032 kg/kg.