A piston filled with `0.04` mole of an ideal gas expands reversibly from `50.0mL` at a constant temperature of `37.0^(@)C` . As it does so, it absorbs `208J` of heat. The value of `q` and `W` for the process will be `(R=8.314J//molK` , `1n7.5=2.01)`

Options

A. q = - 208 J, w = - 208 J
B. q = - 208 J, w = + 208 J
C. q = + 208 J, w = + 208 J
D. q = + 208 J, w = - 208 J

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the type of process
The problem states that the gas expands reversibly at a constant temperature. This indicates that the process is an **isothermal reversible expansion**.

### Step 2: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = Q + W \]
For an isothermal process involving an ideal gas, the change in internal energy (\(\Delta U\)) is zero:
\[ \Delta U = 0 \]
Thus, we can rewrite the equation as:
\[ 0 = Q + W \implies Q = -W \]

### Step 3: Determine the value of \(Q\)
The problem states that the gas absorbs \(208 \, J\) of heat. Therefore:
\[ Q = +208 \, J \]

### Step 4: Calculate the work done \(W\)
From the relationship \(Q = -W\), we can find \(W\):
\[ W = -Q = -208 \, J \]

### Conclusion
The values for the process are:
- \(Q = +208 \, J\)
- \(W = -208 \, J\)

### Final Answer
Thus, the values of \(Q\) and \(W\) for the process are:
- \(Q = 208 \, J\)
- \(W = -208 \, J\)

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