ThermodynamicsHard

Question

The work involved (w) in an isothermal expansion of n moles of an ideal gas from an initial pressure of ‘P’ atm to final pressure of 1 atm in number of steps such that in every step, the constant external pressure exactly 1 atm less than the initial pressure of gas is maintained, is given as

Options

A.$- nRT\sum_{i = 1}^{i = p - 1}\left( \frac{1}{P + 1 - i} \right)$
B.$- nRT\sum_{i = 1}^{i = P}\left( \frac{1}{P + 1 - i} \right)$
C.$- nRT\sum_{i = 1}^{i = P}\left( \frac{i}{P + 1 - i} \right)$
D.$- nRT\sum_{i = 1}^{i = P - 1}\left( \frac{i}{P + 1 - i} \right)$

Solution

$w = - P_{ext}\left( V_{2} - V_{1} \right)$

$= - P_{2}\left( \frac{nRT}{P_{2}} - \frac{nRT}{P_{1}} \right) $$$= - nRT\left( 1 - \frac{P_{2}}{P_{1}} \right) = - nRT\left( 1 - \frac{P_{1} - 1}{P_{1}} \right) = - nRT \times \frac{1}{P_{1}}$$

$\text{Now, }w_{total} = w_{1} + w_{2} + .... + w_{f} $$${= \left( - nRT \times \frac{1}{P} \right) + \left( - nRT \times \frac{1}{P - 1} \right) + ...... + \left( - nRT \times \frac{1}{2} \right) }{= - nRT\sum_{i = 1}^{i = P - 1}\left( \frac{1}{P + 1 - i} \right)}$$

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