A piston of mass M is hung from a massless spring whose restoring force law goes as F = -kx, where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height $ L_0 $ to $ L_1 $, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
A piston of mass M is hung from a massless spring whose restoring force law goes as F = -kx, where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height $ L_0 $ to $ L_1 $, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
Show Hint
- \( 3nRT \ln \left( \frac{L_1}{L_0} \right) + 2Mg(L_1 - L_0) + \frac{k}{3} (L_1^3 - L_0^3) \)
- \( nRT \ln \left( \frac{L_1}{L_0} \right) + \frac{Mg}{2} (L_1 - L_0) + \frac{k}{4} (L_1^4 - L_0^4) \)
- \( nRT \ln \left( \frac{L_1}{L_0} \right) + Mg(L_1 - L_0) + \frac{k}{4} (L_1^4 - L_0^4) \)
- \( nRT \ln \left( \frac{L_1}{L_0} \right) + Mg(L_1 - L_0) + \frac{3k}{4} (L_1^4 - L_0^4) \)
The Correct Option is C
Solution and Explanation
Using WET: Total energy supplied = gravitational potential energy + spring potential energy + work done by gas
\( Mg(L_1 - L_0) + \int_{0}^{L_1-L_0} kx dx + nRT \ln \left( \frac{L_1}{L_0} \right) + W_{ext} = 0 \) \( \frac{k}{4} [x^4]_{0}^{L_1 - L_0} + Mg(L_1 - L_0) + \int_{0}^{L_1-L_0} kx dx + nRT \ln \left( \frac{L_1}{L_0} \right) + W_{ext} = 0 \)
\( \frac{k}{4} (L_1^4 - L_0^4) + Mg(L_1 - L_0) + nRT \ln \left( \frac{L_1}{L_0} \right) + W_{ext} = 0 \) \( W_{ext} = \frac{k}{4} (L_1^4 - L_0^4) + Mg(L_1 - L_0) + nRT \ln \left( \frac{L_1}{L_0} \right) \)
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