Three moles of an ideal gas at `127^(@)C` expands isothermally untill the volume is doubled. Calculate the amount of work done and heat absorbed.

Explanation

To solve the problem of calculating the work done and heat absorbed during the isothermal expansion of an ideal gas, we can follow these steps:

### Step 1: Convert the temperature to Kelvin
The given temperature is \(127^{\circ}C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the given value:
\[ T = 127 + 273 = 400 \, K \]

### Step 2: Identify the initial and final volumes
Let the initial volume be \(V_1\). Since the volume is doubled during the expansion, the final volume \(V_2\) is:
\[ V_2 = 2V_1 \]

### Step 3: Use the formula for work done in isothermal expansion
The work done \(W\) during an isothermal expansion of an ideal gas can be calculated using the formula:
\[ W = nRT \ln\left(\frac{V_2}{V_1}\right) \]
Substituting the values:
- \(n = 3 \, \text{moles}\)
- \(R = 8.314 \, \text{J/(mol·K)}\)
- \(T = 400 \, K\)
- \(\frac{V_2}{V_1} = 2\)

Now, substituting these values into the formula:
\[ W = 3 \cdot 8.314 \cdot 400 \cdot \ln(2) \]

### Step 4: Calculate the natural logarithm
The natural logarithm of 2 is approximately \(0.693\). Thus:
\[ W = 3 \cdot 8.314 \cdot 400 \cdot 0.693 \]

### Step 5: Perform the calculation
Calculating the work done:
\[ W = 3 \cdot 8.314 \cdot 400 \cdot 0.693 \approx 6912 \, J \]

### Step 6: Determine the change in internal energy
For an ideal gas undergoing an isothermal process, the change in internal energy (\(ΔU\)) is zero because the temperature remains constant:
\[ ΔU = 0 \]

### Step 7: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ ΔQ = ΔU + ΔW \]
Since \(ΔU = 0\), we have:
\[ ΔQ = ΔW \]
Thus, the heat absorbed \(Q\) is equal to the work done:
\[ Q = W = 6912 \, J \]

### Final Answers
- The amount of work done is \(6912 \, J\).
- The amount of heat absorbed is also \(6912 \, J\).

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