When the state of a system changes from `A` to `B` adiabatically the work done on the system is `322` joule. If the state of the same system is changed from `A` to `B` by another process, and heat required is `50` calories of heat is required then find work done on the system in this process? `(J = 4.2 J//cal)`

Explanation

To solve the problem step by step, we will use the first law of thermodynamics and the properties of internal energy. Here’s the detailed solution:

### Step 1: Understand the Adiabatic Process
In the adiabatic process from state A to state B, the work done on the system is given as 322 joules. Since it is an adiabatic process, there is no heat transfer (ΔQ = 0).

### Step 2: Apply the First Law of Thermodynamics for the Adiabatic Process
The first law of thermodynamics states:
\[ \Delta Q = \Delta U + \Delta W \]
Since ΔQ = 0 for the adiabatic process, we can write:
\[ 0 = \Delta U + \Delta W \]
This implies:
\[ \Delta U = -\Delta W \]
Given that the work done on the system (ΔW) is 322 joules, we have:
\[ \Delta U = -322 \text{ joules} \]
Thus, the change in internal energy (ΔU) is:
\[ \Delta U = 322 \text{ joules} \]

### Step 3: Understand the Second Process
In the second process from A to B, it is given that 50 calories of heat is required. We need to find the work done on the system in this process.

### Step 4: Convert Calories to Joules
To convert the heat from calories to joules, we use the conversion factor:
\[ 1 \text{ calorie} = 4.2 \text{ joules} \]
Thus, for 50 calories:
\[ \Delta Q = 50 \text{ calories} \times 4.2 \text{ J/cal} = 210 \text{ joules} \]

### Step 5: Apply the First Law of Thermodynamics for the Second Process
Now, we apply the first law of thermodynamics again for the second process:
\[ \Delta Q = \Delta U + \Delta W \]
Substituting the known values:
\[ 210 \text{ joules} = 322 \text{ joules} + \Delta W \]

### Step 6: Solve for Work Done (ΔW)
Rearranging the equation to find ΔW:
\[ \Delta W = 210 \text{ joules} - 322 \text{ joules} \]
\[ \Delta W = -112 \text{ joules} \]

### Conclusion
The work done on the system in the second process is:
\[ \Delta W = -112 \text{ joules} \]
This indicates that work is done by the system in the second process.

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