The ' x ' moles gas is allowed to expand isothermally and reversibly from 10\ dm^3 to 20\ dm^3 at 300\ K and -17290\ J work is obtained. Find the value of ' x '. [2.303 × R = 19.15\ J K^ -1 mol^ -1 ]

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Quantrex Academy · Accepted Answer

Correct answer: D. 10

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

W = -2.303 n R T \log_{10} \left(\dfrac{V_2}{V_1}\right)<br><br>Substituting the given values:<br><br>

-17290 = -x

\times (2.303 R) \times

T

\times \log_{10} \left(\dfrac{20}{10}\right)<br><br>

-17290 = -x

\times19.15 \times300 \times \log_{10}(2)

-17290 = -x \times19.15 \times300 \times0.301

-17290 = -x \times1729.245

x = \dfrac{17290}{1729.245} \approx10

Answer:

10

The ' x ' moles gas is allowed to expand isothermally and...

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