Question: Air in human lungs has a temperature of 37.0ºC and a saturation vapor density of 44.0 g/m3. (a) If 2.00 L of air is exhaled and very dry air inhaled, what is the maximum loss of water vapor by the person? (b) Calculate the partial pressure of water vapor having this density, and compare it with the vapor pressure of 6.31×103 N/m2.
The maximum water vapor exhaled by the person is 0.088 g and the partial pressure of water vapor is equal to the vapor pressure.
Density is the ratio of mass to volume. Therefore, mass is the product of density and volume.
The loss of water vapor from the person’s body is equal to the mass of oxygen exhaled by the person. Here, volume and density are given.
Therefore, 0.088 g of water vapor was lost from the person.
According to the gas law,
Here, is the pressure, is the volume, is the number of moles, is the universal gas constant and is the temperature.
The value of density and temperature are given.
The value of the universal gas constant is 8.314 JK-1mol-1 and the molar mass of water is 18g. The partial pressure of water vapor can be calculated by substituting these values in the expression of pressure.
This value is almost equal to the vapor pressure which is equal to 6.31×103 N/m2. So, partial pressure of water vapor and vapor pressure are equal.
A high-pressure gas cylinder contains of toxic gas at a pressure of and a temperature of . Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature () to reduce the leak rate and pressure so that it can be safely repaired. (a) What is the final pressure in the tank, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? (b) What is the final pressure if one-tenth of the gas escapes? (c) To what temperature must the tank be cooled to reduce the pressure to (assuming the gas does not change phase and that there is no leakage during cooling)? (d) Does cooling the tank appear to be a practical solution?
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