The main idea is that the vacuum pressure does not directly determine the relative humidity. Relative humidity is based on the partial pressure of water vapor compared with the saturation vapor pressure of water at the same temperature.
At 5°C, the saturation vapor pressure of water is about:
Psat≈0.872 kPaP_{sat}\approx0.872\text{ kPa}
A saturated NaCl solution maintains roughly 75% relative humidity, so the water-vapor partial pressure would be approximately:
PH2O=0.75(0.872)=0.654 kPaP_{H_2O}=0.75(0.872)=0.654\text{ kPa}
Therefore,
RH=PH2OPsat×100%RH=\frac{P_{H_2O}}{P_{sat}}\times100\% RH=0.6540.872×100%≈75%RH=\frac{0.654}{0.872}\times100\%\approx\boxed{75\%}
The chamber volume available to the gas is approximately:
3.785−0.200=3.585 L3.785-0.200=3.585\text{ L}
The ideal gas law can then be used to find how much water vapor is actually present:
PV=nRTPV=nRT n=(0.654)(3.585)(8.314)(278.15)≈0.00101 moln=\frac{(0.654)(3.585)} {(8.314)(278.15)} \approx0.00101\text{ mol}
That is only about:
0.00101(18.0)≈0.018 g0.00101(18.0)\approx0.018\text{ g}
of water vapor.
So, assuming the chamber reaches equilibrium and there is plenty of saturated NaCl solution present, the relative humidity should remain approximately 75%.
The stated vacuum of -0.84 bar would affect the total air pressure in the chamber, but it does not by itself make the RH 75% of that total pressure. The NaCl solution controls the equilibrium water-vapor partial pressure.
So the best answer is approximately:
RH≈75%\boxed{RH\approx75\%}
with the exact value depending slightly on the temperature dependence of saturated NaCl solution.
