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 \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \))

Answer 1

Correct Answer: 28721

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].

Answer 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, \( x = 28721 \).