Water can be in three main aggregate states: liquid, firm and gaseous. Steam, in turn, is nonsaturated and saturated - having the identical temperature and pressure with the boiling water. If temperature of water vapor at increase in pressure exceeded 100 degrees Celsius, then such steam is called superheated. Often when studying a school course of physics or when carrying out technological process there is a task: to determine pressure of water vapor under some specific conditions.
Instruction
1. Let's say you were given the following task: in a certain metal vessel filled in water in quantity which is equal to a quarter of its volume. After that the vessel was closed hermetically and heated to temperature equal to 500 OS. If to present that all water which was in a vessel evaporated what pressure of this steam will be? At first the vessel contained one water (its quantity which turned into gaseous state is insignificant a little therefore it can be neglected). Designate its weight as m, and volume as V1. Therefore, density of water will be calculated on a formula: ρ1 = m/V1.
2. After heating in a vessel there was one water vapor of the same mass of m, but the bigger volume of V2 borrowing four times. Therefore, density of water vapor is equal: ρ2 = ρ1/4.
3. Now transfer temperature from degrees Celsius to degrees Kelvin. 500 degrees Celsius are approximately equal to 773 degrees Kelvin (273 + to Shopping center).
4. Write the universal equation of Mendeleyev-Klapeyrona. Of course, strongly heated water vapor cannot be considered at all as ideal gas which condition it describes, but the error in calculations will be rather small. P2V2 = mRT/µ or, transforming it taking into account that V2 is four times more, than V1: 4P2V1 = mRT/µ. Where P2 – that pressure of water vapor which you need to find; R – universal gas constant approximately equal 8.31; T – temperature in degrees Kelvin (773); and µ – the molar mass of water (or water vapor), equal 18 grams/mol (0.018 kg/mol).
5. Thus, at you the formula turns out: P2 = mRT/4V1 µ. However as initial volume of V1 = m/ρ1, total type of the equation is as follows: P2 = ρ1RT/4µ. Having substituted the known sizes in a formula, and knowing what water density is equal to, calculate the required size of pressure of water vapor.