Determine the volume of the vessel which contains 0.75 mol of water vapor. The vapor pressure in the vessel is 200 kPa, and the mean square velocity of the particles is 500 m / s
Data: ν – amount of water vapor substance (ν = 0.75 mol); P is the pressure of the taken water vapor in the vessel (P = 200 kPa = 2 * 10 ^ 5 Pa); Vav – the known root-mean-square velocity of molecules (Vav = 500 m / s).
Const: Мвп – molar mass of water vapor (Мвп = 18.02 g / mol ≈ 18 * 10 ^ -3 kg / mol).
To find out the volume of the vessel, we will use the formula: Vkv ^ 2 = 3P / ρvp = 3P * V / mvp = 3P * V / (ν * Mvp), whence we express: V = Vkv2 * ν * Mvp / 3R.
Let’s perform the calculation: V = Vkv ^ 2 * ν * Mvp / 3P = 500 ^ 2 * 0.75 * 18 * 10 ^ -3 / (3 * 2 * 10 ^ 5) = 0.005625 m3 = 5.625 dm3 (5.625 l ).
Answer: The vessel with steam should have a volume of 5.625 dm3 (5.625 l).
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