Determine the potential deaths resulting from the following exposure to chlorine:
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3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane () with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Stoichiometric calculations:
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From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.
Mole of 815.74 grams of propane =
Mole of 815.74 grams of propane = 18.49750567 moles
Mole of 1,006.29 grams of oxygen =
Mole of 1,006.29 grams of oxygen = 31.4465625 moles
Going by the mole ratio, it appears propane is limiting while oxygen is in excess.
From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:
18.49750567 moles x 4 = 73.99 moles.
Using the ideal gas equation:
PV = nRT
v = (73.99 x 0.08206 x 623) ÷ 0.96
v = 3940.2
Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane () with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.
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Answer:
The answer to your question is: T2 = 235.44 °K
Explanation:
Data
V1 = 3.15 L V2 = 2.78 L
P1 = 2.40 atm P2 = 1.97 atm
T1 = 325°K T2 = ?
Formula
Process
T2 = (P2V2T1) / (P1V1)
T2 = (1.97x 2.78x 325) / (2.40 x 3.15)
T2 = 1779.895 / 7.56
T2 = 235.44 °K