Minimum amount of work done required to compress 5.00 moles of an idea

1.	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» 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

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»

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

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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

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