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 \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \))
(Given \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \))
Correct Answer: 28721
Approach Solution - 1
To calculate the work done on an ideal gas during isothermal, reversible expansion, we use the formula:
\( w = -nRT \ln\left(\frac{V_f}{V_i}\right) \), where:
\( n = 5 \) 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 the values into the equation:
\( w = -5 \times 8.314 \times 300 \ln\left(\frac{100}{10}\right) \)
Calculate the natural logarithm:
\( \ln(10) \approx 2.302 \)
Calculate the work:
\( w = -5 \times 8.314 \times 300 \times 2.302 \)
\( w \approx -28721 \, \text{J} \)
Thus, the value of \( x \) is 28721. This value fits perfectly within the provided range [28721, 28721].
Approach Solution -2
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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