Request a table showing the relationship between the density of hydrogen fluoride gas and temperature

This post was last edited by ylb913 on 2016-10-24 at 20:33. 1. The molecular weight of HF is 20; at standard pressure (one atmosphere, or 1 atm, or 101325 Pa), its density at 0°C is 20/22.4 = 0.893 kg/m3. 2. The density at other temperatures under standard pressure is given by 0.893*273/(273*t); using 273.15 yields more accurate results. However, when the pressure is not at atmospheric pressure, it is necessary to multiply by that pressure (absolute pressure, with the initial unit being atmospheres, that is, how many atm). ——As the temperature rises, the gas expands and its density decreases; as the pressure increases, the density rises. This is a simple way of understanding the phenomenon, and it’s also very easy to apply in practice. 3. This is based on the molar volume of an ideal gas (22.4 liters per mole), the molecular weight of the gas (with the original unit being grams per liter; by multiplying both the numerator and denominator by 1000, it becomes kilograms per cubic meter), as well as the ideal gas law (PV=nRT). Since n remains constant during state conversions and R is a constant, the formula P2V2/T2=P1V1/T1 arises. 4. R is a constant; what should its value be? According to R=PV/nT: the pressure is in atm, and it is 0.082. Where does this value come from? Assuming P=1 atm, V=22.4, n=1 (where 1 mol corresponds to a volume of 22.4 liters), and t=273.15, then R=22.4/273.15=0.082. Pressure is expressed in kPa; 1 atm = 101.325 MPa. If P is replaced by P = 101.325 kPa, then R = 0.082 * 101.325 = 8.31. 5. What is the biggest practical challenge, then? It’s the issue of how to convert gauge pressure into absolute pressure The main question is what the local atmospheric pressure is exactly? The same place looks different throughout the four seasons; moreover, this is also related to the temperature and humidity of the atmosphere, and in different places, elevation has an impact on air pressure. Therefore, converting gauge pressure to absolute pressure is not such a simple process. In practical applications, in areas where the local atmospheric pressure is relatively close to the standard atmospheric pressure, it isn’t necessary to be overly precise in the calculations. For example, when converting between MPa and atm, if we simply assume that 0.1 MPa equals 1 atm, the error will be 1325/101325 = 1.3%. This error is very small compared to the precision of most measuring instruments. ——When precision is required, it is necessary to measure in real time and use the atmospheric pressure at your location and at that particular time for the calculations.