Question

One mole of an ideal gas expands isothermally and reversibly to double its volume. If the expansion work done by the system is 1728.85 J, the temperature of the system is _______ K (rounded off to 2 decimal places).

Hint:

For isothermal processes, always use the relation ( W = nRT ln left( frac{V_f}{V_i} right) ). Remember that ( ln(2) approx 0.693 ), which is commonly used in problems involving doubling the volume.

Answer 1

Step 1: Work done during isothermal expansion.

For an isothermal and reversible expansion, the work done is given by:

\[ W = nRT \ln \left( \frac{V_f}{V_i} \right) \]

where:

• \( n = 1 \) (number of moles)
• \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \) (gas constant)
• \( \ln \left( \frac{V_f}{V_i} \right) = \ln (2) \approx 0.693 \)
• \( W = 1728.85 \, \text{J} \) (work done)

Step 2: Rearrange the formula to find \( T \).

Rewriting the equation:

\[ T = \frac{W}{nR \ln \left( \frac{V_f}{V_i} \right)} \] Step 3: Substituting the values. \[ T = \frac{1728.85}{1 \cdot 8.314 \cdot 0.693} \] \[ T = \frac{1728.85}{5.755522} \] \[ T \approx 300.48 \, \text{K} \] Step 4: Conclusion.

The temperature of the system is approximately \( 300.48 \, \text{K} \), which lies between \( 299.90 \, \text{K} \) and \( 301.90 \, \text{K} \).