A gas can be taken from A to B via two different paths ACB and ADB.
When path ACB is used, 60J of heat flows into the system and 30J of work is done by the system. If path ADB is used, work done by the system is 10J. The heat flow into the system in path ADB is}
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This type of problem is a direct application of the fact that internal energy (U) is a state function. The key is to calculate \(\Delta\)U using the path for which complete information is given, and then use that value of \(\Delta\)U to find the missing variable for the other path.
Step 1: Understanding the Concept:
The change in internal energy ($\Delta U$) of a system is a state function. This means $\Delta U$ is identical for any path taken between the same initial state (A) and final state (B). Step 2: Key Formula or Approach:
Apply the First Law of Thermodynamics: $\Delta U = q + w$.
(Using IUPAC sign convention: $q$ is positive if heat flows {into} the system; $w$ is negative if work is done {by} the system). Step 3: Detailed Explanation:
1. Analyze Path ACB:
Heat added to system, $q_{ACB} = +60$ J.
Work done by the system, $w_{ACB} = -30$ J.
Calculate internal energy change:
$\Delta U = q_{ACB} + w_{ACB} = 60 + (-30) = 30$ J.
Because $\Delta U$ is a state function, $\Delta U_{A \rightarrow B}$ is $30$ J regardless of the path.
2. Analyze Path ADB:
We know $\Delta U = 30$ J.
Work done by the system, $w_{ADB} = -10$ J.
Use the First Law to find the unknown heat flow ($q_{ADB}$):
$\Delta U = q_{ADB} + w_{ADB}$
$30 = q_{ADB} + (-10)$
$q_{ADB} = 30 + 10 = 40$ J.
Since $q_{ADB}$ is positive, 40 J of heat flows into the system. Step 4: Final Answer:
The heat flow into the system via path ADB is 40J.
