Determine the temperature at which 2 moles of oxygen gas expands isothermally and reversibly from 2\ dm^3 to 20\ dm^3 performs -11490\ J of work [2.303 × R = 19.15\ JK^ -1 mol^ -1 ]

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Quantrex Academy · Accepted Answer

Correct answer: D. 300\ K

The work done during a reversible isothermal expansion is given by:

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

W = -11490\

\text{J},

n = 2\

\text{mol},

V_1 = 2\

\text{dm}^3

,

V_2 = 20\ \text{dm}^3

, and

2.303 R = 19.15\ \text{J K}^{-1} \text{mol}^{-1}:<br><br>

-11490 = -2

\times19.15 \times

T

\times \log_{10} \left(\dfrac{20}{2}\right)<br><br>

-11490 = -38.3

\times

T

\times \log_{10}(10)

-11490 = -38.3 \times

T

\times1

T = \dfrac{11490}{38.3} =300\ \text{K}<br><br>Answer: 300\ \text{K}

Determine the temperature at which 2 moles of oxygen gas expands...

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