In a conventional stack sampling train for particulate matter, water vapor is removed from the sampled gases by condensation and desiccation to protect the gas volume meter. If 90 g of water was removed in a particular test, estimate the partial volume of this water vapor [m3] at the stack conditions of 180°C and 95 kPa.
Moles of water = n = mass/molar mass = 90/18 = 5 moles
Let volume is V liters
Temperature (T) = 180°C = 273 + 180 = 453 K
Pressure (P) = 95 kPa = 95 * 1000 * 9.869 * 10^-6 = 0.938 atm
Gas constant (R) = 0.0821 L atm / mol K
Now, using ideal gas equation,
PV = n R T
V = nRT / P = 5 * 0.0821 * 453 / 0.938 = 198.34 liters
We know that,
1 liter = 0.001 m^3
So, 198.34 L = 198.34 * 0.001 = 0.19834 meter^3 .... Answer
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In a conventional stack sampling train for particulate matter, water vapor is removed from the sampled...