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Chemistry Help?
A quantity of water was vaporized completely in a piston-cylinder apparatus at 100oC against a constant
pressure of 1.00 atm. During the process, 22.6 kJ of heat flowed into the system which did 1.70 kJ of work
on the surroundings. What is the enthalpy change of the system?
The answer is +22.6 kj but idk how to solve it! Pls explain thanks!
1 Answer
- hfshawLv 73 years ago
In general, assuminlg only p-V work is involved in a process:
dH = δq + V dp
where dH is the differential enthalpy change of a system, δq is the thermal energy added to the system V is the volume of the system, and dP is the differential pressure change.
For a constant-pressure process,dp = 0, so the change in enthalpy of a system undergoing an isobaric (constant-pressure) process is simply equal to the thermal energy added to the system, which in this case, you are told is 22.6 kJ.
