1.

What is the maximum work that can be obtained by the isothermal expansion of one mole of an ideal gas at 273 K from 2.24 dm3 to 22.4 dm3?

Answer»

w = -2.303 n RT log \(\frac{V_2}{V_1}\)

= -2.303 x 1 x 8.314 x 273 x log \(\frac{22.4}{2.24}\)

= -2.303 x 1 x 8.314 x 273 x 1

= -5227.17 J



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