Suppose 1.00 mol of methane is combusted irreversibly at constant room temperature (25 degrees C). Calculate the work associated with this process.
In the solution, it indicates the process as:
CH4(g) + 2O2(g) --> 2H2O(l) + CO2(g)
and that delta n(gas) = -2.00
Isn't H2O water vapor (gas) not liquid? If not, can someone explain why it's a liquid?
Final 2011 Question 1B Combustion of Methane
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