Minimum amount of work done required to compress 5.00 moles of an ideal gas isothermally from 200 L to 40 L is
(a) + 20.1 kJ
(b) – 20.1 kJ
(c) – 20.1 J
(d) + 20.1 J
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Minimum amount of work done required to compress 5.00 moles of an ideal gas isothermally from 200 L to 40 L is
(a) + 20.1 kJ
(b) – 20.1 kJ
(c) – 20.1 J
(d) + 20.1 J
Minimum amount of work done required to compress 5.00 moles of an ideal gas isothermally from 200 L to 40 L is
(a) + 20.1 kJ
(b) – 20.1 kJ
(c) – 20.1 J
(d) + 20.1 J
Answer is : (a) + 20.1 kJ
Work done in isothermal compression can be calculated as follows:
\(W=-nRT\,In\frac{V_2}{V_1}\)
= – 5 x 8.314 x 300 x 2.303 \(log\frac{40}{200}\)
= – 1500 x 8.314 x 2.303 x (- log 5)
= + 20.1 kJ