1 mole of certain gas expands isothermally and reversibly from 1 dm^3 to 10 dm^3 and performs -4308 J work. Calculate the temperature at which expansion of gas takes place [R = 8.314 JK^ -1 mol^ -1 ]

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 225 K.

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

W = -4308

J,

n = 1

mol,

R = 8.314

J K^{-1}

mol^{-1}

,

V_1 = 1

dm^3

, and

V_2 = 10

dm

^3

:

-4308 = -2.303 \times1 \times8.314 \times

T

\times \log_{10} \left( \dfrac{10}{1} \right)<br><br>

-4308 = -19.147

\times

T

\times1

T = \dfrac{4308}{19.147}<br><br>

T

\approx225

K

Answer:

225

K.

1 mole of certain gas expands isothermally and reversibly from 1...

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