Cooling at constant pressure — Moist air at 101.3 kPa, 40°C, and 50% relative humidity is cooled to 30°C with no condensation. The cooled air will have:

Introduction:Psychrometric changes under sensible cooling at constant pressure illustrate how relative humidity responds to temperature. This scenario begins at 40°C and 50% RH and cools to 30°C without reaching the dew point (so no moisture is removed).

Given Data / Assumptions:

• Total pressure fixed at 101.3 kPa.
• No condensation occurs; moisture content is unchanged.
• Ideal-gas behavior for air–vapor mixture.

Concept / Approach:Specific humidity w and vapor partial pressure p_v remain constant when no water is added or removed. As temperature decreases, the saturation vapor pressure p_sat(T) decreases. Relative humidity is RH = p_v / p_sat(T). Since p_v is constant while p_sat(T) decreases, RH must increase.

Step-by-Step Solution:Note initial: T1 = 40°C, RH1 = 50% → dew point around the high 20s °C.Cool to T2 = 30°C > dew point → no condensation → w, p_v constant.Because p_sat(30°C) < p_sat(40°C), RH2 = p_v / p_sat(30°C) > RH1.

Verification / Alternative check:On a psychrometric chart, a horizontal move left from 40°C to 30°C (constant humidity ratio) crosses to a higher RH curve, confirming the conclusion.

Why Other Options Are Wrong:

• (a) Dew point is set by p_v; since p_v is unchanged, dew point is the same, not higher.
• (b) Absolute humidity is unchanged (no moisture removal or addition).
• (d) Wet-bulb typically falls with dry-bulb during sensible cooling absent evaporation.
• (e) RH does not remain the same; it increases.

Common Pitfalls:Confusing dew point (depends on p_v) with dry-bulb temperature; assuming RH always decreases with cooling.

Final Answer:A higher relative humidity