1. **Problem statement:** The air in a vessel has an initial relative humidity (RH) of 50%. It is compressed isothermally to one-third of its original volume. We need to find the new relative humidity after compression. 2. **Key concepts:** - Relative humidity (RH) is the ratio of the partial pressure of water vapor to the saturation vapor pressure at the same temperature. - In isothermal compression, temperature remains constant. - The saturation vapor pressure remains constant because temperature is constant. - The partial pressure of water vapor changes with volume. 3. **Formula:** $$RH = \frac{p_{v}}{p_{sat}} \times 100\%$$ where $p_v$ is the partial pressure of water vapor and $p_{sat}$ is the saturation vapor pressure. 4. **Initial conditions:** - Initial relative humidity $RH_1 = 50\%$ - Initial volume $V_1$ - Final volume $V_2 = \frac{V_1}{3}$ 5. **Isothermal compression effect:** - Since temperature is constant, $p_{sat}$ remains the same. - The partial pressure of water vapor changes inversely with volume (Boyle's law): $$p_{v2} = p_{v1} \times \frac{V_1}{V_2} = p_{v1} \times 3$$ 6. **Calculate initial partial pressure:** $$p_{v1} = RH_1 \times p_{sat} = 0.5 \times p_{sat}$$ 7. **Calculate new partial pressure:** $$p_{v2} = 3 \times p_{v1} = 3 \times 0.5 \times p_{sat} = 1.5 \times p_{sat}$$ 8. **Calculate new relative humidity:** $$RH_2 = \frac{p_{v2}}{p_{sat}} \times 100\% = \frac{1.5 \times p_{sat}}{p_{sat}} \times 100\% = 150\%$$ 9. **Interpretation:** Relative humidity above 100% means the air is supersaturated and condensation would occur. **Final answer:** The relative humidity after compression is **150%**.