Question:

If 5 moles of an ideal gas expands from \( 10 \, L \) to a volume of \( 100 \, L \) at \( 300 \, K \) under isothermal and reversible conditions, then work, \( w \), is \( -x \, J \). The value of \( x \) is ______.
(Given \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \))

Updated On: Nov 19, 2024
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Correct Answer: 28721

Solution and Explanation

For an isothermal reversible expansion, the work done \( W \) is given by:

\[ W = -2.303nRT \log \left(\frac{V_f}{V_i}\right) \]

Given:

  • \( n = 5 \, \text{moles} \)
  • \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)
  • \( T = 300 \, \text{K} \)
  • \( V_i = 10 \, \text{L} \)
  • \( V_f = 100 \, \text{L} \)

Substitute into the formula:

\[ W = -2.303 \times 5 \times 8.314 \times 300 \times \log \left(\frac{100}{10}\right) \] \[ W = -2.303 \times 5 \times 8.314 \times 300 \times \log(10) \]

Since \( \log(10) = 1 \):

\[ W = -2.303 \times 5 \times 8.314 \times 300 \] \[ W = -28720.713 \, \text{J} \]

Rounding to the nearest integer:

\[ W = -28721 \, \text{J} \]

Thus, \( x = 28721 \).

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